Chapter 5 : Quadratic Equations - Selina Solutions for Class 10 Maths ICSE

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Chapter 5 - Quadratic Equations Excercise Ex. 5(A)

Question 1

x2 + 5x - 5 = (x - 3)2

Solution 1

x2 + 5x - 5 = (x - 3)2

x2 + 5x - 5 = x2 - 6x + 9

11x - 14 =0; which is not of the form ax2 + bx + c = 0.

Given equation is not a quadratic equation. 

Question 2

7x3 - 2x2 + 10 = (2x - 5)2

Solution 2

7x3 - 2x2 + 10 = (2x - 5)2

7x3 - 2x2 + 10 = 4x2 - 20x + 25

7x3 - 6x2 + 20x - 15 = 0; which is not of the form ax2 + bx + c = 0.

Given equation is not a quadratic equation.

 

Question 3

(x - 1)2 + (x + 2)2 + 3(x +1) = 0

Solution 3

(x - 1)2 + (x + 2)2 + 3(x +1) = 0

x2 - 2x + 1 + x2 + 4x + 4 + 3x + 3 = 0

2x2 + 5x + 8 = 0; which is of the form ax2 + bx + c = 0.

Given equation is a quadratic equation.

 

Question 4

Find which of the following equations are quadratic:

(3x - 1)2 = 5(x + 8)

Solution 4

(3x - 1)2 = 5(x + 8)

(9x2 - 6x + 1) = 5x + 40

9x2 - 11x - 39 =0; which is of the form ax2 + bx + c = 0.

Given equation is a quadratic equation.

Question 5

5x2 - 8x = -3(7 - 2x)

Solution 5

5x2 - 8x = -3(7 - 2x)

5x2 - 8x = 6x - 21

5x2 - 14x + 21 =0; which is of the form ax2 + bx + c = 0.

Given equation is a quadratic equation.

 

Question 6

(x - 4)(3x + 1) = (3x - 1)(x +2)

Solution 6

(x - 4)(3x + 1) = (3x - 1)(x +2)

3x2 + x - 12x - 4 = 3x2 + 6x - x - 2

16x + 2 =0; which is not of the form ax2 + bx + c = 0.

Given equation is not a quadratic equation. 

Question 7

Is x = 5 a solution of the quadratic equation x2 - 2x - 15 = 0?

Solution 7

x2 - 2x - 15 = 0

For x = 5 to be solution of the given quadratic equation it should satisfy the equation.

So, substituting x = 5 in the given equation, we get

L.H.S = (5)2 - 2(5) - 15

 = 25 - 10 - 15

 = 0

 = R.H.S

Hence, x = 5 is a solution of the quadratic equation x2 - 2x - 15 = 0. 

Question 8

Is x = -3 a solution of the quadratic equation 2x2 - 7x + 9 = 0?

Solution 8

2x2 - 7x + 9 = 0

For x = -3 to be solution of the given quadratic equation it should satisfy the equation

So, substituting x = 5 in the given equation, we get

L.H.S=2(-3)2 - 7(-3) + 9

 = 18 + 21 + 9    

 = 48 

  R.H.S

Hence, x = -3 is not a solution of the quadratic equation 2x2 - 7x + 9 = 0. 

Question 9

If   is a solution of equation 3x2 + mx + 2 = 0, find the value of m.

Solution 9

For x =   to be solution of the given quadratic equation it should satisfy the equation

So, substituting x =   in the given equation, we get

Question 10

  and 1 are the solutions of equation mx2 + nx + 6 = 0. Find the values of m and n.

Solution 10

For x =   and x = 1 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x =   and x = 1 in the given equation, we get

Solving equations (1) and (2) simultaneously,

Question 11

If 3 and -3 are the solutions of equation ax2 + bx - 9 = 0. Find the values of a and b.

Solution 11

For x = 3 and x = -3 to be solutions of the given quadratic equation it should satisfy the equation

So, substituting x = 3 and x = -3 in the given equation, we get

Solving equations (1) and (2) simultaneously,

Chapter 5 - Quadratic Equations Excercise Ex. 5(B)

Question 1

Without solving, comment upon the nature of roots of each of the following equations :

(i)7x2 - 9x +2 =0 (ii)6x2 - 13x +4 =0

 

(iii)25x2 - 10x +1=0 (iv)

 

(v)x2 - ax - b2 =0 (vi)2x2 +8x +9=0

Solution 1


 


 

Question 2

Find the value of p, if the following quadratic equation has equal roots : 4x2 - (p - 2)x + 1 = 0

Solution 2

Question 3

Find the value of 'p', if the following quadratic equations have equal roots :

x2 + (p - 3)x + p = 0

Solution 3

x2 + (p - 3)x + p = 0

 Here, a = 1, b = (p - 3), c = p

 Since, the roots are equal,

b2- 4ac = 0

(p - 3)2- 4(1)(p) = 0

p2 + 9 - 6p - 4p = 0

p2- 10p + 9 = 0

p2-9p - p + 9 = 0

p(p - 9) - 1(p - 9) = 0

(p -9)(p - 1) = 0

p - 9 = 0 or p - 1 = 0

p = 9 or p = 1 

Question 4

The equation 3x2 - 12x + (n - 5)=0 has equal roots. Find the value of n.

Solution 4

Question 5

Find the value of m, if the following equation has equal roots : (m - 2)x2 - (5+m)x +16 =0

Solution 5

Question 6

Find the value of k for which the equation 3x2- 6x + k = 0 has distinct and real roots.

Solution 6

Chapter 5 - Quadratic Equations Excercise Ex. 5(C)

Question 1

Solve :

Solution 1

Question 2

Solve :

Solution 2

Question 3

Solve :

Solution 3

Question 4

Solve :

Solution 4

Question 5

Solve :

Solution 5

Question 6

Solve :

Solution 6

Question 7

Solve :

Solution 7

Question 8

Solve :

Solution 8

Question 9

Solve :

Solution 9

Question 10

Solve :

Solution 10

Question 11

Solve :

Solution 11

Question 12

Solve :

Solution 12

Question 13

Solve :

Solution 13

Question 14

Solve :

Solution 14

Question 15

Solve :

Solution 15

Question 16

2x2 - 9x + 10 = 0, When

(i) x N

(ii) x Q

Solution 16

Question 17

Solve :

Solution 17

Question 18

Solve :

Solution 18

Question 19

Solve :

Solution 19

Question 20

Solve :

Solution 20

Question 21

Find the quadratic equation, whose solution set is :

(i) (ii)

Solution 21

Question 22

 

Solution 22

Question 23

 

Solution 23

Question 24

Find the value of x, if a + 1=0 and x2 + ax - 6 =0.

Solution 24

If a+1=0, then a = -1

Put this value in the given equation x2 + ax - 6 =0

Question 25

Find the value of x, if a + 7=0; b + 10=0 and 12x2 = ax - b.

Solution 25

If a + 7 =0, then a = -7

and b + 10 =0, then b = - 10

 

Put these values of a and b in the given equation

Question 26

Use the substitution y= 2x +3 to solve for x, if 4(2x+3)2 - (2x+3) - 14 =0.

Solution 26

4(2x+3)2 - (2x+3) - 14 =0

Put 2x+3 = y

Question 27

Without solving the quadratic equation 6x2 - x - 2=0, find whether x equals 2 over 3 is a solution of this equation or not.

Solution 27

Consider the equation, 6x2 - x - 2=0

Put x equals 2 over 3 in L.H.S.

Since L.H.S.= R.H.S., then x equals 2 over 3 is a solution of the given equation.

Question 28

Determine whether x = -1 is a root of the equation x2 - 3x +2=0

or not.

Solution 28

x2 - 3x +2=0

Put x = -1 in L.H.S.

L.H.S. = (-1)2 - 3(-1) +2

= 1 +3 +2=6 R.H.S.

Then x = -1 is not the solution of the given equation.

Question 29

If x = is a solution of the quadratic equation 7x2+mx - 3=0;

Find the value of m.

Solution 29

7x2+mx - 3=0

Given x = is the solution of the given equation.

Put given value of x in the given equation

Question 30

If x = -3 and x = are solutions of quadratic equation mx2 + 7x + n = 0, find the values of m and n.

Solution 30

Question 31

If quadratic equation x2 - (m + 1) x + 6=0 has one root as x =3;

find the value of m and the root of the equation.

Solution 31

Question 32

Given that 2 is a root of the equation 3x2 - p(x + 1) = 0 and that the equation px2 - qx + 9 = 0 has equal roots, find the values of p and q.

Solution 32

Question 33

Solution 33


or x = -(a + b)

Question 34

Solution 34

Question 35

If -1 and 3 are the roots of x2+px+q=0
then find the values of p and q

Solution 35

 

Chapter 5 - Quadratic Equations Excercise Ex. 5(D)

Question 1

Solve each of the following equations using the formula :

 

(i)x2 - 6x =27 (ii)x2 - 10x +21=0

 

(iii)x2 +6x - 10 =0 (iv)x2 +2x - 6=0

 

(v)3x2+ 2x - 1=0 (vi)2x2 + 7x +5 =0

 

(vii) (viii)

(ix) (x)

(xi) (xii)

(xiii) (xiv)

Solution 1

 

Question 2

Solve each of the following equations for x and give, in each case, your answer correct to one decimal place :

(i)x2 - 8x+5=0

(ii)5x2 +10x - 3 =0

Solution 2


Question 3

x2 - 5x - 10 = 0

Solution 3

Question 4

Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :

x2 - 3x - 9 =0

Solution 4



Question 5

Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :

Solution 5



Question 6

Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :

(i)2x2 - 10x +5=0

Solution 6

Question 7

Solve each of the following equations for x and give, in each case, your answer correct to 3 decimal places :

(i)3x2 - 12x - 1 =0

(ii)x2 - 16 x +6= 0

(iii)2x2 + 11x + 4= 0

Solution 7

Question 8

Solve:

(i)x4 - 2x2 - 3 =0

(ii)x4 - 10x2 +9 =0

Solution 8


Question 9

Solve :

(i)(x2 - x)2 + 5(x2 - x)+ 4=0

(ii)(x2 - 3x)2 - 16(x2 - 3x) - 36 =0

Solution 9


Question 10

Solve :

(i)

(ii)

(iii)

Solution 10

 

Question 11

Solve the equation . Write your answer correct to two decimal places.

Solution 11

Question 12

Solve the following equation and give your answer correct to 3 significant figures:

Solution 12

Consider the given equation:

 

 

Question 13

Solve for x using the quadratic formula. Write your answer correct to two significant figures.

(x - 1)2 - 3x + 4 = 0

Solution 13

Question 14

Solve the quadratic equation x2 - 3(x + 3) = 0; Give your answer correct to two significant figures.

Solution 14

x2 - 3(x + 3) = 0

  

Chapter 5 - Quadratic Equations Excercise Ex. 5(E)

Question 1

Solve: 

Solution 1

Question 2

Solve: (2x+3)2=81

Solution 2

Question 3


Solution 3

Question 4

Solution 4

Question 5

begin mathsize 11px style straight x space plus space 4 over straight x space equals space minus 4 semicolon space straight x space not equal to 0 end style

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

Solve each of the following equations, giving answer upto two decimal places.(i)x2 - 5x -10=0(ii) 3x2 - x - 7 =0

Solution 12


Question 13


Solution 13

Question 14

Solve :

(i)x2 - 11x - 12 =0; when x N

 

(ii)x2 - 4x - 12 =0; when x I

 

(iii)2x2 - 9x + 10 =0; when x Q

Solution 14



Question 15


Solution 15

Question 16


Solution 16

Question 17

Solution 17

 


Question 18

Solution 18



Question 19


Solution 19

Question 20

Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots.

Solution 20

Consider the given equation:

 

 

Chapter 5 - Quadratic Equations Excercise Ex. 5(F)

Question 1

Solve : (x+5)(x-5)=24 

Solution 1

Given: (x+5)(x-5)=24

  

Question 2

Solve :   

Solution 2

Given:

  

Question 3

Solve :   

Solution 3

Given:

  or   

Question 4

One root of the quadratic equation   is  . Find the value of m. Also, find the other root of the equation. 

Solution 4

Given quadratic equation is   …. (i)

One of the roots of (i) is  , so it satisfies (i)

So, the equation (i) becomes

Hence, the other root is .

Question 5

One root of the quadratic equation   is -3, find its other root. 

Solution 5

Given quadratic equation is   …. (i)

One of the roots of (i) is -3, so it satisfies (i)

Hence, the other root is 2a.

Question 6

If  and  ;find the values of x. 

Solution 6

Given   i.e

So, the given quadratic equation becomes

Hence, the values of x are   and .

Question 7

Find the solution of the equation ; if  and  . 

Solution 7

Given quadratic equation is  ….. (i)

Also, given  and

  and

So, the equation (i) becomes

Hence, the solution of given quadratic equation are   and .

Question 8

If m and n are roots of the equation  where x ≠ 0 and x ≠ 2; find m × n. 

Solution 8

Given quadratic equation is

Since, m and n are roots of the equation, we have

  and

Hence,  .

Question 9

Solve, using formula :

  

Solution 9

Given quadratic equation is

Using quadratic formula,

x = a + 1 or x = -a - 2 = -(a + 2)

 

Question 10

Solve the quadratic equation

(i) When  (integers)

(ii) When  (rational numbers) 

Solution 10

Given quadratic equation is

(i) When  the equation   has no roots

(ii) When  the roots of   are

 or

Question 11

Find the value of m for which the equation   has real and equal roots. 

Solution 11

Given quadratic equation is

The quadratic equation has real and equal roots if its discriminant is zero.

  or

Question 12

Find the values of m for which equation   has equal roots. Also, find the roots of the given equation. 

Solution 12

Given quadratic equation is  …. (i)

The quadratic equation has equal roots if its discriminant is zero

When  , equation (i) becomes

When  , equation (i) becomes

 x =

Question 13

Find the value of k for which equation   has real roots. 

Solution 13

Given quadratic equation is  …. (i)

The quadratic equation has real roots if its discriminant is greater than or equal to zero

Hence, the given quadratic equation has real roots for .

Question 14

Find, using quadratic formula, the roots of the following quadratic equations, if they exist

(i)

(ii)   

Solution 14

(i) Given quadratic equation is

D = b2 - 4ac =  = 25 - 24 = 1

Since D > 0, the roots of the given quadratic equation are real and distinct.

Using quadratic formula, we have

  or

(ii) Given quadratic equation is

D = b2 - 4ac =  = 16 - 20 = - 4

Since D < 0, the roots of the given quadratic equation does not exist.

Question 15

Solve :

(i)   and x > 0.

(ii)   and x < 0. 

Solution 15

(i) Given quadratic equation is

  or

But as x > 0, so x can't be negative.

Hence, x = 6.

(ii) Given quadratic equation is

  or

But as x < 0, so x can't be positive.

Hence,

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