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Class 10 SELINA Solutions Maths Chapter 5 - Quadratic Equations

Quadratic Equations Exercise Ex. 5(A)

Solution 1(i)

(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.

Solution 2(i)

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. 

Solution 2(ii)

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. 

Solution 3

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

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

Solution 4

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,

Solution 5

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,

Solution 1(ii)

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.

 

Solution 1(iii)

(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. 

Solution 1(iv)

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. 

Solution 1(v)

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.

 

Solution 1(vi)

(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.

 

Quadratic Equations Exercise Ex. 5(B)

Solution 1


 


 

Solution 2(i)

Solution 2(ii)

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 

Solution 3

Solution 4

Solution 5

Quadratic Equations Exercise Ex. 5(C)

Solution 21

Solution 22(i)

Solution 22(ii)

Solution 23

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

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

Solution 24

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

Solution 25

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

Put 2x+3 = y

Solution 26

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.

Solution 27

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.

Solution 28

7x2+mx - 3=0

Given x = is the solution of the given equation.

Put given value of x in the given equation

Solution 29

Solution 30

Solution 31

Solution 33

Solution 34

 

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 32

 


or x = -(a + b)

Quadratic Equations Exercise Ex. 5(D)

Solution 1

 

Solution 2


Solution 3(i)

Solution 3(ii)



Solution 3(iv)



Solution 4

Solution 5


Solution 6


Solution 7

 

Solution 8

Solution 9

Consider the given equation:

 

 

Solution 10

Solution 11

x2 - 3(x + 3) = 0

  

Solution 3(iii)

4x2 - 5x - 3 = 0

Here, a = 4, b = -5 and c = -3

  

Solution 3(v)

Quadratic Equations Exercise Ex. 5(E)

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

Consider the given equation:

 

 

Quadratic Equations Exercise 5(F)

Solution 3

Given quadratic equation is x2 + (3-2a) x - 6a = 0

Now substitute x = -3

So, (-3)2 + (3-2a)(-3) - 6a = 0

0=0

Hence x = -3 satisfies the equation, so -3 is a root.

Now,

x2 + (3-2a) x - 6a = 0

x2 + 3x-2ax - 6a = 0

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

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

x=-3, 2a

Hence, the other root is 2a.

Quadratic Equations Exercise Ex. 5(F)

Solution 1(iv)

2x - 3 =

Squaring on both the sides, we get

(2x - 3)2 = 2x2 - 2x + 21

4x2 - 12x + 9 = 2x2 - 2x + 21

2x2 - 10x - 12 = 0

 x2 - 5x - 6 = 0 …..Dividing equation by 2

 x2 - 6x + x - 6 = 0

 x(x - 6) + 1(x - 6) = 0

 (x - 6)(x + 1) = 0

 (x - 6) = 0 or (x + 1) = 0

 x = 6 or x = -1

Solution 1(i)

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

  

Solution 1(ii)

Given:

  

Solution 1(iii)

Given:

  or   

Solution 2

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 .

Solution 3

Given quadratic equation is   …. (i)

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

Hence, the other root is 2a.

Solution 4

Given   i.e

So, the given quadratic equation becomes

Hence, the values of x are   and .

Solution 5

Given quadratic equation is  ….. (i)

Also, given  and

  and

So, the equation (i) becomes

Hence, the solution of given quadratic equation are   and .

Solution 6

Given quadratic equation is

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

  and

Hence,  .

Solution 7

Given quadratic equation is

Using quadratic formula,

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

 

Solution 8

Given quadratic equation is

(i) When  the equation   has no roots

(ii) When  the roots of   are

 or

Solution 9

Given quadratic equation is

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

  or

Solution 10

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 =

Solution 11

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 .

Solution 12

(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.

Solution 13

(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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