**Download NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2**

**1. Find the zeros of the quadratic polynomials and verify the relationships between the zeros and the coefficients.**

**(i) x ^{2}-2x-8**

**Solution: **

Let, P(x) = x^{2}-2x-8

Comparing P(x) with a quadratic equation ax^{2}+bx+c

we have = -2 and = -8.

P(x) = x^{2}-2x-8

= x^{2}-4x+2x-8

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

= (x-4) (x+2)

Therefore, the value of P(x) will be zero if x-4=0 or x+2=0 i.e., when x=4 or x=-2

Hence the zeros of P(x) are 4, -2

Sum of zeros = 4+(-2) = 2 = – = –

Product of the zeros = 4(-2) = -8 =

**(ii) 4s ^{2}-4s+1**

**Solution: **

Let P(s) = 4s^{2}-4s+1

Comparing P(s) with a quadratic equation as^{2}+bs+c we have = – = -1 and =

P(s) = 4s^{2}-4s+1

= (2s)^{2}-2. 2s+1^{2}

= (2s-1)^{2}

Therefore, the value of P(s) will be zero if (2s-1)^{2} = 0 i.e., if s = ,or

Hence the zeros of P(s) are ,

Sum of zeros = + = = = 1 = –

Product of the zeros = . = = .

**(iii) 6x ^{2}-3-7x**

**Solution: **

Let, P(x) = 6x^{2}-3-7x

Comparing P(x) with a quadratic equation ax^{2}+bx+c we have = – and = –

Now P(x) = 6x^{2}-7x-3

= 6x^{2}-9x+2x-3

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

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

Therefore the value of P(x) will be zero if 2x-3=0 or 3x+1=0

Then, 2x = 3

x =

Or, 3x = -1

x = –

Hence the zeros of P(x) are , –

Sum of zeros = + = – = = = –

Product of the zeros = . = – =

**(iv) 4u ^{2}+8u**

**Solution:**

Let, P(u) = 4u^{2}+8u

Comparing P(u) with a quadratic equation au^{2}+bu+c we have = = 2 and = = 0

P(u) = 4u^{2}+8u

= u(4u+8)

Therefore, the value of P(u) will be zero

Then, u = 0

Or, 4u+8 = 0

4u = -8

u = -2

Hence the zeros of P(s) are 0, -2

Now sum of zeros = 0+(-2) =-2 =-

And product of the zeros = 0. 2= 0 =

**(v) t ^{2}-15**

**Solution: **

Let P(t) = t^{2}-15

Comparing P(t) with a quadratic equation at^{2}+bt+c we have = 0 and = -15

Therefore, the value of P(t) will be zero if t^{2}-15 = 0 i.e., t^{2}= 15 so t = ±√15

Hence the zeros of P(s) are √15, – √15

Sum of zeros = √15 – √15 = 0 = –

Product of the zeros = √15(-√15) = -15 =

**(vi) 3x ^{2}– x-4**

**Solution: **

Let P(x) = 3x^{2}-x-4

Comparing P(x) with a quadratic equation ax^{2}+bx+c we have = – and = –

P(x) = 3x^{2}-x-4

= 3x^{2}-4x+3x-4

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

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

Therefore the value of P(x) will be zero if 3x-4=0 or x+1=0

Then, 3x = 4

x =

Or, x = -1

Hence the zeros of P(x) are , – 1

Sum of zeros = – 1= = = –

Product of the zeros = (-1) = – =

**2. Find the quadratic polynomials each with the given numbers as the sum and products of its zeros respectively.**

**(i) , -1**

Solution: Let the quadratic polynomial be ax^{2}+bx+c and its zeros be α, β.

Then α+β = = –

And αβ = -1 =

Then the quadratic polynomial is given by

x^{2 }–(α+β)x+ αβ = x^{2}– x – 1

**(ii) √2, **

Solution: Let α, β be the two zeros of the quadratic polynomial.

Then α+β = √2 and αβ =

Hence the required quadratic polynomial is given by

x^{2 }– (α+β)x + αβ = x^{2}– √2x +

**(iii) 0, √5**

Solution: Let α, β be the two zeros of the quadratic polynomial.

Then α+β = 0 and αβ = √5

Hence the required quadratic polynomial is given by

x^{2 }– ( α+β)x + αβ = x^{2}– 0.x + √5 = x^{2}+ √5

**(iv) 1, 1**

Solution: Let α, β be the two zeros of the quadratic polynomial.

Then α+β = 1 and αβ = 1

Hence the required quadratic polynomial is given by

x^{2 }– ( α+β)x + αβ = x^{2}– 1.x + 1 = x^{2}– x + 1

**(v) – , **

Solution: Let α, β be the two zeros of the quadratic polynomial.

Then α+β = – and αβ =

Hence the required quadratic polynomial is given by

x^{2 }– ( α+β)x + αβ = x^{2}– .x + = x^{2}+ x +

**(vi) 4, 1**

Solution: Let α, β be the two zeros of the quadratic polynomial.

Then α+β = 4 and αβ = 1

Hence the required quadratic polynomial is given by

x^{2 }–( α+β)x+ αβ = x^{2}– 4.x + 1 = x^{2}– 4x +1

**Download NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2**