Exercise 2.1

Question 1. find the zeros of the following quadratic polynomial and verify the relationship between the zeros and coefficients.

(i)x²-2x-8

Solution: x² -2x-8 = x² - 4x + 2x -8

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

=(x+2) (x-4)

So, zeoes are -2 and 4 because

Value x²-2x-8 is zero, when

x+2=0

x=-2

Or, x-4=0

x=4

Verification :

Zeros are -2 and 4.

α+β=coefficient of x / coefficient of x²

-2+4= -(-2)

2=2 verified

α.β = constant Term/Coefficient of x²

(-2)(4)=-8/1=-8

-8=-8 verified.

(ii) 4s² - 4s+1

Solution : 4s² - 2s-2s+1

=2s(2s-1)-1(2s-1)

=(2s-1)(2s-1)

So,zeroes are 1/2 and 1/2,because the value of 4s²-4s+1 is zero,when

2x-1=0

=2x=1

=s=1/2

Or, 2x-1= 0

S=1/2

Verification:

α=1/2, β=1/2

α+β= Coefficient of x /Coefficient of x²

1/2+1/2 = -4/4

1=-(-1)=1 verified.

αβ= constant term/Coefficient of x²

(1/2) (1/2) =1/4

1/4=1/4 verified.

(iii) 6x² -3 -7x

=6x²-9x+2x-3

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

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

So,zeroes are -1/3 and 3/2

Because the value of 6x²-7x-3 is zero,when

3x+1=0

X=-1/3

2x-3=0

x=3/2

Verification :

α=-1/3,β=3/2

Therefore,α+β= -1/2+3/2 =-2+9/6=7/6

=-(-7)/6= coefficientof x / coefficient of x²

And, αβ = constant term/coefficient of x²

= -1/3 X 3/2 =-1/2 = -3/6 Verified.

(iv)4u²+8u=4u(u+2)

Solution: zeroes are 0 amd -2 ,so value of 4u²+8u is zero,when

4u=0

u=0

Or, u+2=0

=u=-2

verifiction : α=0,β=-2

Therefore,α+β =0+(-2) =-2=-8/4

=coefficientof u/coefficient of u²

And αβ =0(-2)=0/4

= constant term/Coefficient of u² verified.

(v) t²-15=t²-(√15)²

Solution :=(t-√15)(t+√15)

So,zeroes are √15 -√15,

Because value of t²-15 is zero when

t-√15=0

t=√15

Or, t+√15=0

t= - √15

Verification : α=√15, β=√15

Therefore, α+β=-√15+√15

=0=-0/1

= coefficient of t/coefficient of t² verified.

And. αβ=(-√15)(√15)

=-15=-15/1

=constant term / coefficient of t²

Verified.

(vi)3x²-x-4

Solution : 3x²-4x+3x-4

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

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

So,zeroes are 4/3 and -1.

Because vale of 3x²-x-4 is zero,

When, 3x-4=0

x=4/3

Or, x+1=0

x=-1

Verification : α=4/3, β=-1

Therefore, α+β =4/3+(-1) =1/3=-(-1)/3

=-coefficient of x²/coeffient of x² verified.

αβ=4/3(-1)=-4/3

=constant term/coefficient of x² verified.

Question 2: to find a quadratic polynomial each with the given numbers as the sum and product of its zeros respectively.

(i) 1/4, -1

Solution : let the zeroes of polynomial be α and β.

Then,α+β=1/4 and αβ =-1

Therfore, required polynomial is given,

x²-(α+β)x+αβ=x²-1/4x +(-1)

=x²-1/4x-1

=4x²-x-4

(ii) √2,1/3

Solution : let the zeroes of polynomial be α and β.

Then,α+β=√2 and αβ=1/3

Therefore, required polynomial is:

x²-(α+β)x+αβ=x²-√2x+1/3

=3x²-3√2x+1

(iii) 0,√15

Solution:let the zeroes of polynomial be α and β.

Then,α+β=0 and αβ=√15

Therefore,required polynomial

=x²-(α+β)x +αβ

=x²-0(α+β)x+αβ

=x²-0xx+√5

=x²+√5

(iv)1,1

Solution:let the zeroes of polynomial be α and β.

Then,α+β=1 and αβ=1.

Therefore, required polynomial

=x²-(α+β)x+αβ

=x²-x+1

(v)-1/4,1/4

Solution : let the zeroes of polynomial be α and β.

Then, α+=β =-1/4 and αβ=1/4

Therefore, required polynomial

=x²-(α+β)x+αβ

=x²-(-1/4)+1/4

=4x²+x+1=0

(iv)4,1

Solution :let the zeroes of polynomial be α and β.

Then, α+β=4 and αβ=1.

Therefore,required polynomial

= x²-(α+β)x + αβ

=x²-4x+1

We hope CBSE/MP Board Class 10th Maths Chapter 2 "Polynomials" Exercise 2.1 will help you.

Written By - Abhishek Dohare