RD SHARMA Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

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Chapter 4 - Quadratic Equations Exercise 4.82

Question 1

If the equation x2 + 4x + k = 0 has real and distinct root, then

(a) k < 4

(b) k > 4

(c) k ≥ 4

(d) k ≤ 4

Solution 1

We know for the quadratic equation

ax2 + bx + c = 0

condition for roots to be real and distinct is

D = b2 - 4ac > 0       ..........(1)

for the given question

x+ 4x + k = 0

a = 1, b = 4, c = k

from (1)

16 - 4k > 0

k < 4

So, the correct option is (a).

Chapter 4 - Quadratic Equations Exercise 4.83

Question 1

The number of quadratic equations having real roots and which do not change by squaring their roots is

(a) 4

(b) 3

(c) 2

(d) 1

Solution 1

Any quadratic equation having roots 0 or 1 are only possible quadratic equation because on squaring 0 or 1, it remains same.

Hence, 2 solutions are possible, one having roots 1 and 1, while the other having roots 0 and 1.

So, the correct option is (c).

Question 2

If the equation x2 - ax + 1 = 0 has two distinct roots, then

(a) |a| = 2

(b) |a| < 2

(c) |a| > 2

(d) None of these

Solution 2

For the equation x2 - ax + 1 = 0 has two distinct roots, condition is

(-a)- 4 (1) (1) > 0

a- 4 > 0

a> 4

|a| > 2

So, the correct option is (c).

Question 3

begin mathsize 12px style If space the space equation space 9 x squared space plus space 6 kx space plus space 4 space equals space 0 space has space equal space roots comma space then space the space roots space are space both space equal space to
left parenthesis straight a right parenthesis space plus-or-minus space 2 over 3
left parenthesis straight b right parenthesis space plus-or-minus space 3 over 2
open parentheses straight c close parentheses space 0
open parentheses straight d close parentheses space plus-or-minus space 3 end style

Solution 3

begin mathsize 12px style For space quadratic space equation
ax squared space plus space bx space plus space straight c space equals space 0
condition space for space equal space roots space is
straight D space equals space straight b squared space minus space 4 ac space equals space 0 space space space space space space space space space space space.......... open parentheses 1 close parentheses
For space the space given space question
9 straight x squared space plus space 6 kx space plus space 4 space equals space 0
straight a space equals space 9 comma space space space straight b space equals space 6 straight k comma space space space space straight c space equals space 4
from space open parentheses 1 close parentheses
open parentheses 6 straight k close parentheses squared space minus space 4 space open parentheses 9 close parentheses space open parentheses 4 close parentheses space equals space 0
36 straight k squared space minus space 36 space cross times space 4 space equals space 0
36 space open parentheses straight k to the power of 2 space end exponent minus space 4 close parentheses space equals space 0
straight k squared space equals space 4
straight k space equals space plus-or-minus 2
Now comma space equation space is space 9 straight x squared space plus space 6 kx space plus space 4 space equals space 0
rightwards double arrow space 9 straight x squared space plus-or-minus space 12 straight x space plus space 4 space equals 0
rightwards double arrow open parentheses 3 straight x space plus-or-minus space 2 close parentheses squared space equals space 0
rightwards double arrow straight x space equals space plus-or-minus 2 over 3
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 4

begin mathsize 12px style If space ax squared space plus space bx space plus space straight c space equals 0 space has space equal space roots comma space then space straight c space equals
left parenthesis straight a right parenthesis space fraction numerator negative straight b over denominator 2 straight a end fraction
left parenthesis straight b right parenthesis space fraction numerator straight b over denominator 2 straight a end fraction
open parentheses straight c close parentheses space fraction numerator negative straight b squared over denominator 4 straight a end fraction
open parentheses straight d close parentheses space fraction numerator straight b squared over denominator 4 straight a end fraction end style

Solution 4

begin mathsize 12px style If space ax squared space plus space bx space plus space straight c space equals space 0 space has space equal space roots
then
straight b squared space minus space 4 ac space equals space 0
straight b squared space equals space 4 ac
straight c space equals space fraction numerator straight b squared over denominator 4 straight a end fraction
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 5

If the equation ax2 + 2x + a = 0 has two distinct roots, if

(a) a = ± 1

(b) a = 0

(c) a = 0, 1

(d) a = -1, 0

Solution 5

For any quadratic equation

ax + bx + c = 0

having two distinct roots, condition is

b2 - 4ac > 0

For the equation ax2 + 2x + a = 0 to have two distinct roots,

(2)2 - 4 (a) (a) > 0

4 - 4a2 > 0

4(1 - a2) > 0

1 - a2 > 0 since 4 > 0

that is, a2 - 1 < 0

Hence -1 < a < 1, only integral solution possible is a = 0

So, the correct option is (b).

Question 6

The positive value of k for which the equation x2 + kx + 64 = 0 and x2 - 8x + k = 0 will both have real roots, is

(a) 4

(b) 8

(c) 12

(d) 16

Solution 6

For any quadratic equation

ax2 + bx + c = 0

having real roots, condition is

b2 - 4ac ≥ 0     .......(1)

According to question 

x+ kx + 64 = 0 have real root if

k- 4 × 64 ≥ 0

k2 ≥ 256

|k|≥ 16             ........(2)

Also, x2 - 8x + k = 0 has real roots if

64 - 4k ≥ 0

k ≤ 16    .........(3)

from (2), (3) the only positive solution for k is

k = 16

So, the correct option is (d).

Question 7

begin mathsize 12px style The space value space of space square root of 6 space plus square root of 6 space plus square root of 6 space plus space...... end root end root end root is
left parenthesis straight a right parenthesis space 4
left parenthesis straight b right parenthesis space 3
left parenthesis straight c right parenthesis space minus 2
left parenthesis straight d right parenthesis space 3.5 end style

Solution 7

begin mathsize 12px style Let space straight x space equals space square root of 6 space plus square root of 6 space plus square root of 6 space plus space...... end root end root end root space space space space space space space space space space space space space space space space....... left parenthesis 1 right parenthesis
On space squaring space both space sides
straight x squared space equals space 6 space plus space square root of 6 space plus square root of 6 space plus space..... end root end root space space space space space space space space space space space space space space space space space space space space space space space space space space....... left parenthesis 2 right parenthesis
From space left parenthesis 1 right parenthesis space & space left parenthesis 2 right parenthesis
straight x squared space equals space 6 space plus space straight x
straight x squared space minus space straight x space minus space 6 space equals space 0
straight x squared space minus space 3 straight x space plus space 2 straight x space minus space 6 space equals space 0
straight x space open parentheses straight x space minus space 3 close parentheses space plus space 2 left parenthesis straight x space minus space 3 right parenthesis space equals space 0
left parenthesis straight x space plus space 2 right parenthesis space left parenthesis straight x space minus space 3 right parenthesis space equals space 0
straight x space equals space minus 2 space or space straight x space equals space 3
but space straight x space can apostrophe straight t space be space negative comma space hence
box enclose straight x space equals space 3 end enclose
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 8

If 2 is a root of the equation x+ bx + 12 = 0 and the equation x2 + bx + q = 0 has equal roots, then q =

(a) 8

(b) -8

(c) 16

(d) -16

Solution 8

It is given that 2 is a root of equation x + bx + 12 = 0

Hence

(2)+ b(2) + 12 = 0

4 + 2b + 12 = 0

2b + 16 = 0

b = -8        .........(1)

It is also given that x2 + bx + q = 0 has equal root 

so, b2 - 4(q) = 0         ........(2)

from (1) & (2)

(-8)- 4q =0

q = 16

So, the correct option is (c).

Question 9

begin mathsize 12px style If space the space equation space open parentheses straight a squared space plus space straight b squared close parentheses space straight x squared space minus space 2 open parentheses ac space plus space bd close parentheses space straight x space plus space straight c squared space plus space straight d squared space equals space 0 space has space equal space roots comma space then space
open parentheses straight a close parentheses space ab space equals space cd
open parentheses straight b close parentheses space ad space equals space bc
open parentheses straight c close parentheses space ad space equals space square root of bc
open parentheses straight d close parentheses space ab space equals space square root of cd end style

Solution 9

For any quadratic equation

ax2 + bx + c = 0

Having equal roots, the condition is

b - 4ac = 0

For the equation

(a2 + b2) x2 - 2 (ac + bd)x + c2 + d= 0

to have equal roots, we have

(-2(ac + bd))- 4 (c+ d2) (a2 + b2) = 0

4 (ac + bd)- 4 (a2c+ b2c2 + d2a2 + b2d2) = 0

(a2c+ b2d2 + 2abcd) - (a2c+ b2c2 + a2d2+ b2d2) = 0

2abcd - b2c2 - a2d2 = 0

b2c2 - a2d- 2abcd = 0

(bc - ad)= 0

bc = ad

So, the correct option is (b).

Question 10

begin mathsize 12px style If space the space roots space of space the space equation space left parenthesis straight a squared plus space straight b squared right parenthesis straight x squared space minus space 2 straight b left parenthesis straight a space plus space straight c right parenthesis straight x space plus space left parenthesis straight b squared plus space straight c squared right parenthesis space equals space 0 space are space equal comma space then
left parenthesis straight a right parenthesis space 2 straight b space equals space straight a space plus space straight c
open parentheses straight b close parentheses space straight b squared space equals space ac
open parentheses straight c close parentheses space straight b space equals fraction numerator 2 ac over denominator straight a space plus space straight c end fraction
open parentheses straight d close parentheses space straight b space equals space ac end style

Solution 10

For any quadratic equation

ax+ bx + c = 0

having equal roots, condition is

b2 - 4ac = 0

According to question, quadratic equation is

(a2 + b2)x2 - 2b(a + c)x + b2 + c= 0

having equal roots, so

(2b(a + c))2 - 4(a2 + b2) (b2 + c2) = 0

4b2 (a + c)2 - 4(a2b2 + a2c2 + b4 + b2c2) = 0

b2(a2 + c2 + 2ac) - (a2b2 + a2c+ b2c2 + b4) = 0

a2b2 + b2c2 + 2acb2 - a2b2 - a2c- b2c2 - b4 = 0

2acb2 - a2c2 - b4 = 0

a2c2 + b4 - 2acb2 = 0

(ac - b2)2 = 0

ac = b2

So, the correct option is (b).

Question 11

If the equation x2 - bx + 1 = 0 does not possess real roots, then

(a) -3 < b < 3

(b) -2 < b < 2

(c) b < 2

(d) b < -2

Solution 11

For any quadratic equation ax2 + bx + c = 0 having no real roots, condition is

b2 - 4ac < 0

For the equation, x2 - bx + 1 = 0 having no real roots

b2 - 4 < 0

b2 < 4

-2 < b < 2

So, the correct option is (b).

Question 12

If x = 1 is a common root of the equations ax2 + ax + 3 = 0 and x2 + x + b = 0, then ab =

(a) 3

(b) 3.5

(c) 6

(d) -3

Solution 12

begin mathsize 12px style If space straight x space equals space 1 space is space root space of space ax squared space plus space ax space plus space 3 space equals space 0
then
straight a open parentheses 1 close parentheses squared space plus space straight a open parentheses 1 close parentheses space plus space 3 space equals space 0
2 straight a space plus space 3 space equals space 0
straight a space equals space fraction numerator negative 3 over denominator 2 end fraction space space space space space space space space space space space space...... left parenthesis 1 right parenthesis
and space straight x space equals space 1 space is space also space root space of space straight x squared space plus space straight x space plus space straight b space equals 0
then
open parentheses 1 close parentheses squared space plus space open parentheses 1 close parentheses space plus space straight b space equals space 0
straight b space equals space minus 2 space space space space space space space space space.... left parenthesis 2 right parenthesis
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
ab space equals space open parentheses fraction numerator negative 3 over denominator 2 end fraction close parentheses open parentheses negative 2 close parentheses
ab space equals 3
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 13

If p and q are the roots of the equation x-px + q = 0, then

(a) p = 1, q = -2

(b) b = 0, q = 1

(c) p = -2, q = 0

(d) p = -2, q = 1

Solution 13

If p, q are the roots of equation x- px + q = 0, then p and q satisfies the equation

Hence

(p)- p(p) + q = 0

p- p+ q = 0

q = 0

and (q)- p(q) + q = 0

q- p(q) + q = 0

0 = 0

p can take any value

p = -2 and q = 0

So, the correct option is (c).

Question 14

If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax2 + bx + 1 = 0 having real roots is

(a) 10

(b) 7

(c) 6

(d) 12

Solution 14

For the ax+ bx + 1 = 0 having real roots condition is

b- 4(a) (1) ≥ 0

b2 ≥ 4a

For a = 1

b2 ≥ 4

b ≥ 2

b can take value 2, 3, 4

Here, 3 possible solutions are possible            .......(1)

For a = 2

b2 ≥ 8

Here, b can take value 3, 4            ......(2)

Here, 2 solutions are possible

For a = 3

b≥ 12

possible value of b is 4

Hence, only 1 possible solution      ......(3)

For a = 4

b2 ≥ 16

possible value of b is 4

Hence, only 1 possible solution    .......(4)

from (1), (2), (3), (4)

Total possible solutions are 7

So, the correct option is (b).

Chapter 4 - Quadratic Equations Exercise 4.84

Question 1

If one root of the equation x2 + ax + 3 = 0 is 1, then its other root is

(a) 3

(b) -3

(c) 2

(d) -2

Solution 1

x+ ax + 3 = 0

product of roots = 3

One root is 1. Hence other root is 3.

So, the correct option is (a).

Question 2

If (a+ b2) x+ 2(ab + bd) x + c2 + d= 0 has no real roots, then

(a) ad = bc

(b) ab = cd

(c) ac = bd

(d) ad ≠ bc

Solution 2

If any quadratic equation ax+ bx + c has no real roots then b- 4ac < 0    ......(1)

According to the question, the equation is

(a+ b2) x+ 2(ac + bd) x + c2 + d= 0

from (1)

4(ac + bd)- 4(a2 + b2) (c2 + d2) < 0

a2c2 + b2d2 + 2abcd - (a2c2 + a2d2 + b2c2 + b2d2) < 0

a2c2 + b2d2 + 2abcd - a2c2 - a2d2 - b2c2 - b2d2 < 0

2abcd - a2d2 - b2c2 < 0

-(ad - bc)< 0

(ad - bc)> 0

For this condition to be true ad ≠ bc

So, the correct option is (d).

Question 3

begin mathsize 12px style If space the space sum space of space the space roots space of space the space equation space straight x squared space minus space straight x space equals space straight lambda space open parentheses 2 straight x space minus space 1 close parentheses space is space zero comma space then space straight lambda space equals
open parentheses straight a close parentheses space minus 2
open parentheses straight b close parentheses space 2
open parentheses straight c close parentheses space fraction numerator negative 1 over denominator 2 end fraction
open parentheses straight d close parentheses space 1 half end style

Solution 3

begin mathsize 12px style We space know space for space any space quadratic space equation
ax squared space plus space bx space plus space straight c space equals space 0
sum space of space roots space equals fraction numerator negative straight b over denominator straight a end fraction
Now comma space according space to space the space question
straight x squared space minus space straight x space equals space straight lambda space open parentheses 2 straight x space minus space 1 close parentheses
straight x squared space minus space straight x space minus space 2 λx space plus space straight lambda space equals space 0
straight x squared space minus space straight x space open parentheses 1 space plus space 2 straight lambda close parentheses space plus space straight lambda space equals space 0
given space sum space of space roots space of space this space equation space is space zero
Hence comma space fraction numerator negative open parentheses negative open parentheses 1 space plus space 2 straight lambda close parentheses close parentheses over denominator 1 end fraction space equals space 0
space space space space space space space space space space space space space space space space space space 1 space plus space straight lambda 2 space equals space 0
space space space space space space space space space space space space space space space space space space space straight lambda space equals space fraction numerator negative 1 over denominator 2 end fraction
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 4

If x = 1 is a common root of ax+ ax + 2 = 0 and x+ x + b = 0 then, ab =

(a) 1

(b) 2

(c) 4

(d) 3

Solution 4

It is given that x = 1 is root of equation ax2 + ax + 2 = 0

Hence, a(1)2 + a(1) + 2 = 0

           2a + 2 = 0

           a = -1         ......(1)

It is given that x = 1 is also root of x2 + x + b = 0

Hence, (1)+ (1) + b = 0

           b = -2     ......(2)

from (1) & (2)

ab = (-1) (-2)

ab = 2

So, the correct option is (b).

Question 5

begin mathsize 12px style The space value space of space straight c space for space which space the space equation space ax squared plus space 2 bx space plus space straight c space equals space 0 space has space equal space roots space is
left parenthesis straight a right parenthesis space straight b squared over straight a
left parenthesis straight b right parenthesis space fraction numerator straight b squared over denominator 4 straight a end fraction
open parentheses straight c close parentheses space straight a squared over straight b
open parentheses straight d close parentheses space fraction numerator straight a squared over denominator 4 straight b end fraction end style

Solution 5

begin mathsize 12px style For space any space quadratic space equation space ax squared space plus space bx space plus space straight c space equals space 0 space having space equal space roots space condition space is
straight b squared space minus space 4 ac space equals space 0 space space space space space space space space space..... open parentheses 1 close parentheses
According space to space the space question comma space the space quadratic space equation space is
ax squared space plus space 2 bx space plus space straight c space equals space 0
from space left parenthesis 1 right parenthesis
open parentheses 2 straight b close parentheses squared space minus space 4 ac space equals space 0
4 straight b squared space minus space 4 ac space equals space 0
straight b squared space equals space ac
straight c space equals space straight b squared over straight a
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 6

begin mathsize 12px style If space straight x squared space plus space straight k left parenthesis 4 straight x space plus space straight k space minus space 1 right parenthesis space plus space 2 space equals space 0 space has space equal space roots comma space then space straight k space equals
left parenthesis straight a right parenthesis space fraction numerator negative 2 over denominator 3 end fraction comma space 1
open parentheses straight b close parentheses space 2 over 3 comma space minus 1
open parentheses straight c close parentheses space 3 over 2 comma space 1 third
open parentheses straight d close parentheses space fraction numerator negative 3 over denominator 2 end fraction comma space fraction numerator negative 1 over denominator 3 end fraction end style

Solution 6

begin mathsize 12px style For space equation comma space straight x squared space plus space straight k open parentheses 4 straight x space plus space straight k space minus space 1 close parentheses space plus space 2 space equals space 0
space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight x squared space plus space 4 kx space plus space straight k open parentheses straight k space minus space 1 close parentheses space plus space 2 space equals space 0
Condition space for space equal space roots space is comma
open parentheses 4 straight k close parentheses squared space minus space 4 space open parentheses straight k open parentheses straight k space minus space 1 close parentheses space plus space 2 close parentheses space equals space 0
16 straight k squared space minus space 4 open parentheses straight k squared space minus space straight k space plus space 2 close parentheses space equals space 0
4 straight k squared space minus space open parentheses straight k squared space minus space straight k space plus space 2 close parentheses space equals space 0
4 straight k squared space minus space straight k squared space plus space straight k space minus space 2 space equals space 0
3 straight k squared space plus space straight k space minus 2 space equals space 0
3 straight k squared space plus space 3 straight k space minus space 2 straight k space minus space 2 space equals space 0
3 straight k space open parentheses straight k space plus space 1 close parentheses space minus space 2 open parentheses straight k space plus space 1 close parentheses space equals space 0
open parentheses 3 straight k space minus space 2 close parentheses space open parentheses straight k space plus space 1 close parentheses space equals space 0
straight k space equals space minus 1 space or space straight k space equals space 2 over 3
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 7

begin mathsize 12px style If space the space sum space and space product space of space the space roots space of space the space equation space kx squared space plus space 6 straight x space plus 4 straight k space equals space 0 space are space equal comma space then space straight k space equals
open parentheses straight a close parentheses space fraction numerator negative 3 over denominator 2 end fraction
open parentheses straight b close parentheses space 3 over 2
open parentheses straight c close parentheses space 2 over 3
open parentheses straight d close parentheses space fraction numerator negative 2 over denominator 3 end fraction end style

Solution 7

begin mathsize 12px style For space any space quadratic space equation space ax squared space plus space bx space plus space straight c space equals space 0
sum space of space roots space equals space fraction numerator negative straight b over denominator straight a end fraction space and space product space of space roots space equals space straight c over straight a
For space equation comma space kx squared space plus space 6 straight x space plus space 4 straight k space equals space 0 comma space given space sum space and space product space of space roots space of space this space equation space are space equal comma
Hence comma
fraction numerator negative open parentheses 6 close parentheses over denominator straight k end fraction space equals space fraction numerator 4 straight k over denominator straight k end fraction
4 straight k space equals space minus 6
straight k space equals space fraction numerator negative 3 over denominator 2 end fraction
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 8

If sin α and cos α are the roots of the equation ax+ bx + c = 0, then b2

(a) a- 2ac

(b) a+ 2ac

(c) a- ac

(d) a+ ac

Solution 8

begin mathsize 12px style ax squared space plus space bx space plus space straight c space equals space 0
sin space straight alpha space plus space cos space straight alpha space equals space fraction numerator negative straight b over denominator straight a end fraction space space..... open parentheses 1 close parentheses space space space space and
sin space straight alpha space cosα space equals space straight c over straight a space space space space space space space...... open parentheses 2 close parentheses
On space squaring space we space get comma
open parentheses sin space straight alpha space plus space cos space straight alpha close parentheses squared space equals space straight b squared over straight a squared
sin squared straight alpha space plus space cos squared straight alpha space plus space 2 space sinα space cosα space equals space straight b squared over straight a squared space space space space space space space space space space space space.... open parentheses 3 close parentheses
We space know space sin squared space straight alpha space plus space cos squared space straight alpha space equals space 1 space space space space space space space..... open parentheses 4 close parentheses
from space open parentheses 2 close parentheses comma space open parentheses 3 close parentheses comma space open parentheses 4 close parentheses
1 space plus space 2 open parentheses straight c over straight a close parentheses space equals space straight b squared over straight a squared
straight b squared over straight a squared space equals space fraction numerator straight a space plus space 2 straight c over denominator straight a end fraction
straight b squared space equals straight a to the power of 2 space end exponent plus space 2 ac
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 9

If 2 is a root of the equation x+ ax + 12 = 0 and the quadratic equation x + ax + q = 0 has equal roots, then q =

(a) 12

(b) 8

(c) 20

(d) 16

Solution 9

Given, 2 is a root of equation x+ ax + 12 = 0

so (2)+ a(2) + 12 = 0

4 + 2a + 12 = 0

a = -8     .....(1)

Given x+ ax + q = 0 has equal roots so

a- 4q = 0   ......(2)

from (1) & (2)

(-8)- 4q = 0

4q = 64

q = 16

So, the correct option is (d).

Question 10

If the sum of the roots of the equation x2 - (k + 6)x + 2(2k - 1) = 0 is equal to half of their product, then k =

(a) 6

(b) 7

(c) 1

(d) 5

Solution 10

begin mathsize 12px style For space any space quadratic space equation space straight x squared space minus space left parenthesis straight k space plus space 6 right parenthesis straight x space plus space 2 left parenthesis 2 straight k space minus space 1 right parenthesis space equals space 0
open parentheses sum space of space roots close parentheses space equals space fraction numerator negative open curly brackets negative open parentheses straight k space plus space 6 close parentheses close curly brackets over denominator 1 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight k space plus space 6 space space space space space space space......... open parentheses 1 close parentheses
open parentheses product space of space roots close parentheses space equals space fraction numerator 2 open parentheses 2 straight k space minus space 1 close parentheses over denominator 1 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 open parentheses 2 straight k space minus space 1 close parentheses space space space space space space space space space space space space space space...... open parentheses 2 close parentheses
Given space that space the space sum space of space roots space of space equation space equals space 1 half open parentheses product space of space roots close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
straight k space plus space 6 space equals space 1 half open parentheses 2 open parentheses 2 straight k space minus space 1 close parentheses close parentheses
straight k space plus space 6 space equals space 2 straight k space minus space 1
box enclose 7 space equals space straight k end enclose
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 11

If a and b are roots of the equation x2 + ax + b = 0, then a + b =

(a) 1

(b) 2

(c) -2

(d) -1

Solution 11

If a and b are roots of the equation x+ ax + b = 0

Then, sum of roots = -a

a + b = -a

2a + b = 0          .......(1)

product of roots = b

ab = b

ab - b = 0

b(a - 1) = 0         ........(2)

from (1) and (2)

-2a(a - 1) = 0...(From (1), we have b = -2a)

a(a - 1) = 0

a = 0 or a = 1

if a = 0 rightwards double arrow b = 0

if a = 1 rightwards double arrow b = -2

Now a and b can't be zero at same time, so correct solution is

a = 1 and b = -2

a + b = -1

So, the correct option is (d).

Question 12

A quadratic equation whose one root is 2 and the sum of whose roots is zero, is

(a) x+ 4 = 0

(b) x- 4 = 0

(c) 4x- 1 = 0

(d) x- 2 = 0

Solution 12

Given sum of roots is zero and one root is 2.

So the other root must be -2

so any quadratic equation having root 2 and -2 is

(x - 2) (x - (-2)) = 0

(x - 2) (x + 2) = 0

x- 4 = 0

So, the correct option is (b).

Question 13

If one root of the equation ax+ bx + c = 0 is three times the other, then b2 : ac =

(a) 3 : 1

(b) 3 : 16

(c) 16 : 3

(d) 16 : 1

Solution 13

begin mathsize 12px style If space straight alpha comma space straight beta space are space roots space of space equation space ax squared space plus space bx space plus space straight c space equals space 0
Then comma space straight alpha space plus space straight beta space equals space fraction numerator negative straight b over denominator straight a end fraction space space space space space space space space space...... open parentheses 1 close parentheses
and space αβ space equals space straight c over straight a space space space space space space...... open parentheses 2 close parentheses
It space is space given space that space one space root space is space three space times space the space other space one
Let space straight alpha space equals space 3 straight beta space space space space space space space space space....... open parentheses 3 close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 3 close parentheses
4 straight beta space equals space fraction numerator negative straight b over denominator straight a end fraction
straight beta space equals space fraction numerator negative straight b over denominator 4 straight a end fraction space space space space space space space space space space space...... open parentheses 4 close parentheses
from space open parentheses 2 close parentheses space & space open parentheses 3 close parentheses
3 straight beta squared space equals space straight c over straight a space space space space space space space space space.......... open parentheses 5 close parentheses
from space open parentheses 4 close parentheses space & space open parentheses 5 close parentheses
3 open parentheses fraction numerator negative straight b over denominator 4 straight a end fraction close parentheses squared space equals space straight c over straight a
fraction numerator 3 straight b squared over denominator 16 straight a squared end fraction equals straight c over straight a
straight b squared over ac equals space 16 over 3
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 14

If one root of the equation 2x+ kx + 4 = 0 is 2, then the other root is

(a) 6

(b) -6

(c) -1

(d) 1

Solution 14

begin mathsize 12px style We space know space that comma space for space straight a space quadratic space equation space ax squared space plus space bx space plus space straight c space equals space 0
product space of space roots space equals space straight c over straight a space space space space space space space space space space space space space........ open parentheses 1 close parentheses
According space to space the space question comma space equation space is space 2 straight x squared space plus space kx space plus space 4 space equals space 0
from space open parentheses 1 close parentheses comma
product space of space roots space equals 4 over 2
αβ space equals space 2 space space space space space space space space space space....... open parentheses 2 close parentheses
given space one space root space is space 2. space Hence space from space open parentheses 2 close parentheses space other space root space is space 1.
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Chapter 4 - Quadratic Equations Exercise 4.85

Question 1

If one root of the equation 4x- 2x + (λ - 4) = 0 is a reciprocal of the other, then λ =

(a) 8

(b) -8

(c) 4

(d) -4

Solution 1

begin mathsize 12px style Let space apostrophe straight a apostrophe space be space straight a space space root space of space given space equation.
Then space according space to space the space question comma space other space root space is space 1 over straight a.
4 straight x squared space minus space 2 straight x space plus space open parentheses straight lambda space minus space 4 close parentheses space equals space 0
product space of space roots space equals space fraction numerator straight lambda space minus space 4 over denominator 4 end fraction
straight a space cross times space 1 over straight a space equals space fraction numerator straight lambda space minus space 4 over denominator 4 end fraction
1 space equals fraction numerator space straight lambda space minus space 4 over denominator 4 end fraction
straight lambda space minus space 4 space equals space 4
straight lambda space equals space 8
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 2

begin mathsize 12px style If space straight y space equals space 1 space is space straight a space common space root space of space the space equations space ay squared plus space ay space plus space 3 space equals space 0 space and space straight y squared space plus space straight y space plus straight b space equals space 0 comma space then space ab space equals
open parentheses straight a close parentheses space 3
open parentheses straight b close parentheses space fraction numerator negative 7 over denominator 2 end fraction
open parentheses straight c close parentheses space 6
open parentheses straight d close parentheses space minus 3 end style

Solution 2

begin mathsize 12px style If space straight y space equals space 1 space is space straight a space root space of space equation space ay squared space plus space ay space plus space 3 space equals space 0
Then comma
straight a open parentheses 1 close parentheses squared space plus space straight a open parentheses 1 close parentheses space plus space 3 space equals space 0
2 straight a space plus space 3 space equals space 0
straight a space equals space fraction numerator negative 3 over denominator 2 end fraction space space space space space space space space space space space space space........ open parentheses 1 close parentheses
Also space straight y space equals space 1 space is space straight a space root space of space equation space straight y squared space plus space straight y space plus space straight b space equals space 0
Then comma
open parentheses 1 close parentheses squared space plus space 1 space plus space straight b space equals space 0
box enclose straight b space equals space minus 2 end enclose space space space space space space space space space space space space space space space space...... open parentheses 2 close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
ab space equals space open parentheses fraction numerator negative 3 over denominator 2 end fraction close parentheses open parentheses negative 2 close parentheses
space space space space space space equals space 3
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 3

begin mathsize 12px style The space values space of space straight k space for space which space the space quadratic space equation space 16 straight x squared space plus space 4 kx space plus space 9 space equals space 0 space has space real space and space equal space root space are
open parentheses straight a close parentheses space 6 comma space fraction numerator negative 1 over denominator 6 end fraction
open parentheses straight b close parentheses space 36 comma space minus 36
left parenthesis straight c right parenthesis space 6 comma space minus 6
open parentheses straight d close parentheses space 3 over 4 comma space fraction numerator negative 3 over denominator 4 end fraction end style

Solution 3

Any quadratic equation, ax+ bx + c = 0 has real and equal roots if b- 4ac = 0

For the question, equation is 16x+ 4kx + 9 = 0,

(4k)- 4 × 16 × 9 = 0

16k- 36 × 16 = 0

k2 - 36 = 0

k= 36

k = ± 6

So, the correct option is (c).

Chapter 4 - Quadratic Equations Exercise Ex. 4.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

solutions of the equation 2x2 - x + 9 = x2 + 4x + 3 , x = 2 and x = 3

Solution 20

= 2x- x + 9 – x+ 4x + 3

= x2 - 5x + 6 = 0

Here, LHS = x2 - 5x + 6 and RHS = 0

Substituting x = 2 and x = 3

= x2 - 5x + 6

= (2)2 – 5(2) + 6

=10-10

=0

= RHS

= x2 - 5x + 6

= (3)2 – 5(3) + 6

= 9 - 15 + 6

=15 - 15

=0

= RHS

x = 2 and x = 3 both are the solutions of the given quadratic equation.

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 4 - Quadratic Equations Exercise Ex. 4.10

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Solution 4

Chapter 4 - Quadratic Equations Exercise Ex. 4.11

Question 1

Solution 1

Question 2

Solution 2

Question 3

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm2. Find the sides of the squares.

Solution 3

Question 4

Solution 4

Question 5

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution 5


Question 6

Solution 6

Question 7

Sum of the areas of two squares is 640 m2. If the difference of their perimeter is 64 m, find the sides of the two squares.

Solution 7

Question 8

Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of two squares.

Solution 8

Question 9

The area of a rectangular plot is 528 m2. The length of the plot (in meters) is one meter more than twice its breadth. Find the length and the breadth of the plot.

Solution 9

Question 10

In the centre of a rectangular lawn of dimension 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2. Find the length and breadth of the pond.

Solution 10

Chapter 4 - Quadratic Equations Exercise Ex. 4.12

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3


Question 4

Solution 4

Question 5

To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Solution 5

Let us assume that the larger pipe takes 'x' hours to fill the pool.

So, as per the question, the smaller pipe takes 'x + 10' hours to fill the same pool.

  

Chapter 4 - Quadratic Equations Exercise Ex. 4.13

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of his marks, would have been 180. Find his marks in the two subjects.

Solution 9

Question 10

Solution 10

Question 11

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.

Solution 11

Question 12

At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than   minutes. Find t.

Solution 12

Chapter 4 - Quadratic Equations Exercise Ex. 4.2

Question 1

The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.

Solution 1

Question 2

John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.

Solution 2

Question 3

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.

Solution 3

Question 4

The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.

Solution 4

Question 5

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.

Solution 5

Question 6

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.

Solution 6

Chapter 4 - Quadratic Equations Exercise Ex. 4.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solve the following quadratic equation by factorisation:

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the following quadratic equation by factorisation:

Solution 15

Question 16

Solution 16

Question 17

9x2 - 6b2x - (a4 - b4) = 0

Solution 17

  

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solve the following quadratic equation by factorisation:

2x2 + ax - a2 = 0

Solution 20

Question 21

Solve the following quadratic equation by factorisation:

Solution 21

Question 22

Solve the following quadratic equations by factorization:

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solve the following quadratic equation by factorisation:

Solution 29

Question 30

Solve the following quadratic equation by factorisation:

Solution 30

Question 31

Solve the following quadratic equation by factorisation:

Solution 31

Question 32

Solve the following quadratic equation by factorisation:

Solution 32

Question 33

Solve the following quadratic equation by factorisation:

Solution 33

Question 34

  

Solution 34

  

Question 35

  

Solution 35

  

Question 36

Solution 36

Question 37

  

Solution 37

  

Question 38

Solve the following quadratic equation by factorisation:

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solve the following quadratic equations by factorization:

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solve the following quadratic equations by factorization:

  

Solution 47

  

Question 48

Solve the following quadratic equations by factorization:

Solution 48

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solve the following quadratic equation by factorisation:

Solution 58

Question 59

Solve the following quadratic equation by factorisation:

Solution 59

Question 60

Solve the following quadratic equation by factorisation:

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Chapter 4 - Quadratic Equations Exercise Ex. 4.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 4 - Quadratic Equations Exercise Ex. 4.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

3x2 - 5x + 2 = 0

Solution 18

Question 19

Solution 19

Question 20

Solve for x:

Solution 20

Question 21

Solution 21

Question 22

Solve for x:

Solution 22

Question 23

Solve for x:

 

Solution 23

Chapter 4 - Quadratic Equations Exercise Ex. 4.6

Question 1

Determine the nature of the roots of the following quadratic equations:

2x2 - 3x + 5 = 0

Solution 1

Question 2

Determine the nature of the roots of the following quadratic equations:

2x2 - 6x + 3 = 0

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Find the values of k for which the roots are real and equal in each of the following equations:

4x2 - 2 (k + 1)x + (k + 1) = 0 

Solution 23

4x2 - 2 (k + 1)x + (k + 1) = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = -2(k + 1), c = k + 1 According to the question, roots are real and equal. Hence, b2 - 4ac = 0

Question 24

Solution 24

Question 25

In the following, determine the set of values of k for which the given quadratic equation has real roots:

 

2x2 + x + k = 0

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Find the values of k for which the following equations have real and equal roots:

x2 + k(2x + k - 1) + 2 = 0 

Solution 32

x2 + k(2x + k - 1) + 2 = 0 x2 + 2kx + k(k - 1) + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 2k, c = k(k - 1) + 2 According to the question, roots are real and equal. Hence, b2 - 4ac = 0

Question 33

Find the values of k for which the roots are real and equal in each of the following equations:

2x2 + kx + 3 = 0

Solution 33

Question 34

Find the values of k for which the roots are real and equal in each of the following equations:

kx (x - 2) + 6 = 0

Solution 34

Question 35

Find the values of k for which the roots are real and equal in each of the following equations:

x2 - 4kx + k = 0

Solution 35

Question 36

Find the value of k for which the roots are real and equal in the following equation:

Solution 36

Question 37

Find the value of p for which the roots are real and equal in the following equation:

px(x - 3) + 9 = 0

Solution 37

Question 38

Find the values of k for which the following equations have real roots.

4x2 + kx + 3 = 0 

Solution 38

4x2 + kx + 3 = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = k, c = 3 According to the question, roots are real and equal. Hence, b2 - 4ac = 0

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Find the values of k for which the quadratic equation (3k + 1)x2 + 2(k + 1)x + 1 = 0 has equal roots. Also, find the roots.

Solution 43

Question 44

Find the values of p for which the quadratic equation (2p + 1)x2 - (7p + 2)x + (7p - 3) = 0 has equal roots. Also, find these roots.

Solution 44

Question 45

 If -5 is a root of the quadratic equation, 2x2 + px - 15 = 0, and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k.

Solution 45

Question 46

 If 2 is a root of the quadratic equation 3x2 + px - 8 = 0 and the quadratic equation 4x2 - 2px + k = 0 has equal roots, find the value of k.

Solution 46

Question 47

If 1 is root of the quadratic equation 3x2 + ax - 2 = 0 and the quadratic equation a(x2 + 6x) - b = 0 has equal roots, find the value of b.

Solution 47

Question 48

Find the value of p for which the quadratic equation (p + 1)x2 - 6(p + 1)x + 3(p + 9) = 0, p -1 has equal roots. Hence, find the roots of equation.

Solution 48

  

Question 49

Solution 49

Question 50

Solution 50

Question 51

Solution 51

Question 52

Solution 52

Question 53

Solution 53

Question 54

Solution 54

Question 55

Solution 55

Question 56

Solution 56

Question 57

Solution 57

Question 58

Solution 58

Question 59

Solution 59

Question 60

Solution 60

Question 61

Solution 61

Question 62

Solution 62

Question 63

Solution 63

Question 64

Solution 64

Question 65

Solution 65

Chapter 4 - Quadratic Equations Exercise Ex. 4.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The sum of a number and its square is 63/4. Find the numbers.

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers.

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Find two consecutives odd positive integers, sum of whose squares is 970.

Solution 35

Question 36

The difference of two natural numbers is 3 and the difference of their reciprocal is  . Find the numbers.

Solution 36

Question 37

The sum of the squares of two consecutive odd numbers is 394. Find the numbers.

Solution 37

Question 38

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.

Solution 38

Question 39

The sum of the squares of two consecutive even numbers is 340. Find the numbers.

Solution 39

Question 40

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is . Find the original fraction.

Solution 40

  

Question 41

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Solution 41

Question 42

A natural number when increased by 12 equals 160 times its reciprocal. Find the number.

Solution 42

Chapter 4 - Quadratic Equations Exercise Ex. 4.8

Question 1

Solution 1

Question 2

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed, If it takes 3 hours to complete total journey, what is its original average speed?

Solution 8

Question 9

Solution 9

Question 10

Solution 10



Concept Insight: Use the relation s =d/t to crack this question and remember here distance is constant so speed and time will vary inversely.

Question 11

Solution 11

Question 12

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

Solution 12

Question 13

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.

Solution 13

Question 14

A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream than to return downstream to the same spot. Find the speed of the stream.

Solution 14

Question 15

A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.

Solution 15

Let the speed of a car be x km/hr. According to the question, time is  hr. Distance = Speed × Time 2592 =    x = 72 km/hr Hence, the time taken by a car to cover a distance of 2592 km is 36 hrs.

Chapter 4 - Quadratic Equations Exercise Ex. 4.9

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?

Solution 8

Question 9

At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

Solution 9

Chapter 8 - Quadratic Equations Exercise Ex. 8.6

Question 1

Find the value of p for which the roots are real and equal in the following equation:

4x2 + px + 3 = 0

Solution 1

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