NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2

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) x²-2x-8

Solution:

Let, P(x) = x²-2x-8

Comparing P(x) with a quadratic equation ax²+bx+c

we have  = -2 and = -8.

P(x) = x²-2x-8

= x²-4x+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) 4s²-4s+1

Solution:

Let P(s) = 4s²-4s+1

Comparing P(s) with a quadratic equation as²+bs+c we have  = –  = -1 and =

P(s) = 4s²-4s+1

= (2s)²-2. 2s+1²

= (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) 6x²-3-7x

Solution:

Let, P(x) = 6x²-3-7x

Comparing P(x) with a quadratic equation ax²+bx+c we have = –  and = –

Now P(x) = 6x²-7x-3

= 6x²-9x+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) 4u²+8u

Solution:

Let, P(u) = 4u²+8u

Comparing P(u) with a quadratic equation au²+bu+c we have = = 2 and = = 0

P(u) = 4u²+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) t²-15

Solution:

Let P(t) = t²-15

Comparing P(t) with a quadratic equation at²+bt+c we have = 0 and = -15

Therefore, the value of P(t) will be zero if t²-15 = 0 i.e., t²= 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) 3x²– x-4

Solution:

Let P(x) = 3x²-x-4

Comparing P(x) with a quadratic equation ax²+bx+c we have = – and = –

P(x) = 3x²-x-4

= 3x²-4x+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 ax²+bx+c and its zeros be α, β.

Then α+β = = –  

And αβ =  -1 =

Then the quadratic polynomial is given by

  x² –(α+β)x+ αβ = x²– 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

x² – (α+β)x + αβ = x²– √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

x² – ( α+β)x + αβ = x²– 0.x + √5 = x²+ √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

x² – ( α+β)x + αβ = x²– 1.x + 1 = x²– 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

x² – ( α+β)x + αβ = x²– .x + = x²+ 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

x² –( α+β)x+ αβ = x²– 4.x + 1 = x²– 4x +1

Download NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2

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