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

Chapter 4 - Quadratic Equations Exercise Ex. 4.1

Solution 1 (i)

Solution 1 (ii)

Solution 1 (iii)

Solution 1 (iv)

Solution 1 (v)

Solution 1(vi)

Solution 1 (vii)

Solution 1 (viii)

Solution 1 (ix)

Solution 1 (x)

Solution 1 (xi)

Solution 1 (xii)

Solution 1 (xiii)

Solution 1 (xiv)

Solution 1 (xv)

Solution 2 (i)

Solution 2 (ii)

Solution 2 (iii)

Solution 2 (iv)

Solution 2 (v)

= 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.

Solution 2 (vi)

Solution 2 (vii)

Solution 3 (i)

Solution 3 (ii)

Solution 3 (iii)

Solution 3 (iv)

Solution 4

Solution 5

Chapter 4 - Quadratic Equations Exercise Ex. 4.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Chapter 4 - Quadratic Equations Exercise Ex. 4.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

  

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

  

Solution 35

  

Solution 36

Solution 37

  

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

  

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

Solution 58

Solution 59

Solution 60

Solution 61

Solution 62

Chapter 4 - Quadratic Equations Exercise Ex. 4.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Given equation is x2 - 8x + 18 = 0

x2 - 2 × x × 4 + + 42 - 42 + 18 = 0

(x - 4)2 - 16 + 18 = 0

(x - 4)2 = 16 - 18

(x - 4)2 = -2

Taking square root on both the sides, we get 

Therefore, real roots does not exist.

Solution 7

Solution 8

Solution 9

Solution 10

Chapter 4 - Quadratic Equations Exercise Ex. 4.5

Solution 1 (i)

Solution 1 (ii)

Solution 1(iii)

Solution 1 (iv)

Solution 1(v)

Solution 1(vi)

Solution 1 (vii)

x2 + 2 × x × 5 + 52 = 10x - 6

x2 + 10x + 25 = 10x - 6

x2 + 31 = 0

Here, a = 1, b = 0 and c = 31

Therefore, the discriminant is

D = b2 - 4ac

 = 0 - 4 × 1 × 31

 = -124

Solution 2 (i)

Solution 2 (ii)

Solution 2 (iii)

Solution 2 (iv)

Solution 2 (v)

Solution 2 (vi)

Solution 2 (vii)

Solution 2 (viii)

Solution 2 (ix)

Solution 2(x)

Solution 2(xi)

Solution 2(xii)

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 3(v)

Chapter 4 - Quadratic Equations Exercise Ex. 4.6

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1 (vi)

Given quadratic equation is

Here,

Therefore, we have

As D = 0, roots of the given equation are real and equal.

Solution 2(i)

Solution 2(ii)

Solution 2(iii)

Solution 2(iv)

Solution 2(v)

Solution 2(vi)

Solution 2(vii)

Solution 2(viii)

Solution 2(ix)

Solution 2(x)

Solution 2(xi)

Solution 2(xii)

Solution 2(xiii)

Solution 2(xiv)

Solution 2(xv)

Solution 2(xvi)

Solution 2(xvii)

Solution 2(xviii)

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

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 3(iv)

Solution 3(v)

Solution 4(i)

Solution 4(ii)

Solution 4(iii)

Solution 4(iv)

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

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

Solution 5(iv)

Solution 5(v)

Solution 5(vi)

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

Solution 6 (i)

Solution 6 (ii)

Solution 7

Solution 8

Solution 9(i)

Solution 9 (ii)

Given quadratic equation is x2 + kx + 16 = 0

As it has equal roots, the discriminant will be 0.

Here, a = 1, b = k, c = 16

Therefore, D = k2 - 4(1)(16) = 0

i.e. k2 - 64 = 0

i.e. k = ± 8

When k = 8, the equation becomes x2 + 8x + 16 = 0

 or x2 - 8x + 16 = 0

As D = 0, roots of the given equation are real and equal.

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

  

Solution 15(i)

Solution 15(ii)

Solution 15(iii)

Solution 15(iv)

Solution 16(i)

Solution 16(ii)

Solution 16(iii)

Solution 16(iv)

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Chapter 4 - Quadratic Equations Exercise Ex. 4.7

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Let x be the natural number.

As per the question, we have

Therefore, x = 8 as x is a natural number.

Hence, the required natural number is 8.

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

  

Solution 40

Chapter 4 - Quadratic Equations Exercise Ex. 4.8

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

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.

Solution 11

Solution 12

Solution 13

Solution 14

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.

Solution 16

Let x km/hr be the speed of the stream.

Therefore, we have

Downstream speed = (9 + x) km/hr

Upstream speed = (9 - x) km/hr

Distance covered downstream = distance covered upstream

Total time taken = 3 hours 45 minutes =  hours

Therefore, x = 3 as the speed can't be negative.

Hence, the speed of the motor boat is 3 km/hr.

Chapter 4 - Quadratic Equations Exercise Ex. 4.9

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Chapter 4 - Quadratic Equations Exercise Ex. 4.10

Solution 1

Solution 2

Solution 3

Solution 4

Chapter 4 - Quadratic Equations Exercise Ex. 4.11

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5


Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Chapter 4 - Quadratic Equations Exercise Ex. 4.12

Solution 1

Solution 2

Solution 3


Solution 4

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.

  

Solution 6

Let the tap with smaller diameter takes x hours to completely fill the tank.

So, the other tap takes (x - 2) hours to fill the tank completely.

Total time taken to fill the tank   hours

As per the question, we have

When x = 5, then (x - 2) = 3

When   which can't be possible as the time becomes negative.

Hence, the smaller diameter tap fills in 5 hours and the larger diameter tap fills in 3 hours.

Chapter 4 - Quadratic Equations Exercise Ex. 4.13

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Chapter 4 - Quadratic Equations Exercise 4.82

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

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).

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

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

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).

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).

Solution 7

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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

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).

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).

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).

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).

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

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).

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).

Solution 15

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).

Chapter 4 - Quadratic Equations Exercise 4.84

Solution 16

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).

Solution 17

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

Solution 18

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).

Solution 19

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

Solution 20

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

Solution 21

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

Solution 22

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

Solution 23

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).

Solution 24

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

Solution 25

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).

Solution 26

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).

Solution 27

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

Solution 28

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

Solution 29

x+ ax + 3 = 0

product of roots = 3

One root is 1. Hence other root is 3.

So, the correct option is (a).

Chapter 4 - Quadratic Equations Exercise 4.85

Solution 30

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

Solution 31

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

Solution 32

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).