Gujarat Board GSEB Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2

Question 1.
Find the zeroes of the following polynomial and verify the relationship between the zeroes and the coefficients

1. x² – 2x – 8
2. 4s² – 4s + 1
3. 6x³ – 3 – 7x
4. 4u² + 8u
5. t² – 15
6. 3x² – x – 4

Solution:
1. p(x) = x² – 2x – 8
= x² – 4x + 2x – 8
[By splitting middle term]
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
For zeroes of p(x),
p(x) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0
or x + 2 = 0
⇒ x = 4
or x = -2
⇒ x = 4, – 2.
Zeroes of p(x) are 4 and -2.
⇒ α = 4, β = -2

Verification of relationship between zeroes and coefficient:
Sum of the zeroes = 4 + (-2) = 2
⇒ α + β = 2
and (frac{-b}{a}=frac{-(text { coefficient of } x)}{text { coefficient of } x^{2}})
= (frac{-(-2)}{1}) = 2
⇒ α + β = (frac{-b}{a})
Product of the zeroes
αβ = 4 × (-2) = -8
and (frac{c}{a}=frac{text { constant term }}{text { coefficient of } x^{2}})
= (frac{-8}{1}) = -8
⇒ αβ = (frac{c}{a})

2. p(s) = 4s² – 4s + 1
= 4s² – 2s – 2s + 1
[By splitting middle term]
= 2s(2s – 1) – 1 (2s – 1)
(2s – 1) (2s – 1)
For zeroes of p(s),
P(s) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1 = 0
or 2s – 1 = 0

⇒ s = (frac{1}{2})
or s = (frac{1}{2})
⇒ s = (frac{1}{2}), (frac{1}{2})
⇒ zeroes of p(s) are (frac{1}{2}), (frac{1}{2})
⇒ α = (frac{1}{2}), β = (frac{1}{2})
Verification of relationship between zeroes and coefficients:
Sum of the zeroes = (frac{1}{2}) + (frac{1}{2}) = 1
⇒ α + β = 1

3. p(x) = 6x² – 3 – 7x
p(x) = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
For zeroes p(x) = 0
⇒ (2x – 3) (3x + 1) = 0
⇒ 2x – 3 = 0
or 3x + 1 = 0

⇒ x = (frac{3}{2})
or x = (frac{-1}{3})
x = (frac{3}{2}), (frac{-1}{3})
⇒ α = (frac{3}{2}), β = (frac{-1}{3})
Now, verification of relationship between zeroes and coefficients.
Sum of the zeroes

Also, product of the zeroes;

4. p(u) = 4u² + 8 u
p(u) = 4u (u + 2)
For zeroes, p(u) = 0
⇒ 4 u(u + 2) = 0
⇒ u(u + 2) = 0
⇒ u = 0
or u + 2 = 0
⇒ u = 0
⇒ u = -2
So, zeroes of p(u) are 0 and -2.
⇒ α = 0, β = -2
Verification of relationship between zeroes and coefficients:
Sum of the zeroes,
α + β = 0 + (-2) = -2

5. t² – 15
P(t) = t² – 15
For zeroes p(t) = 0
t² – 15 = 0
t² – ((sqrt { 15 } ))² = 0
(t – (sqrt { 15 } )) (t + (sqrt { 15 } )) = o
⇒ t – (sqrt { 15 } ) = 0
or t + (sqrt { 15 } ) = 0
⇒ t = (sqrt { 15 } )
t = –(sqrt { 15 } )
So, zeroes of p(t) are (sqrt { 15 } ) and –(sqrt { 15 } )
⇒ α = (sqrt { 15 } ), β = –(sqrt { 15 } )
Verification of relationship between zeroes
and coefficients:
Sum of the zeroes,
α + β = (sqrt { 15 } ) + (-(sqrt { 15 } )) = 0

6. p(x) = 3x² – x – 4
p(x) = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4) (x + 1)
For zeroes, p(x) = 0
⇒ (3x – 4)(x + 1) = 0
⇒ 3x – 4 = 0
or x + 1 = 0
⇒ x = (frac{4}{3})
or x = -1
⇒ Zeroes of p(x) are (frac{4}{3}) and -1.
⇒ α = (frac{4}{3}), β = -1
Verification of relationship between zeroes and coefficients:
Now sum of the zeroes,

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

1. (frac{1}{4}), -1
2. (sqrt { 2 } ), (frac{1}{3})
3. 0, (sqrt { 5 } )
4. 1, 1
5. (frac{-1}{4}), (frac{1}{4})
6. 4, 1

Solution:
1. A quadratic polynomial, when the sum and product of its zeroes are given, is
p(x) = K [x² – (sum of the zeroes) x + Product of zeroes]
where K is constant.
Now sum of the zeroes = (frac{1}{4})
Product of the zeroes = -1
∴ Required quadratic polynomial is given

or K[4x² – x – 4]

2. (sqrt { 2 } ), (frac{1}{3})
Sum of the zeroes, S = (sqrt { 2 } )
Product of the zeroes, P = (frac{1}{3})
∴ Required polynomial is
p(x) = K[x² – Sx + P]

3. 0, (sqrt { 5 } )
Sum of the zeroes S = 0
Product of the zeroes, P = (sqrt { 5 } )
∴ Required polynomial is
p(x) = K[x² – Sx + P]
= x² – 0x + (sqrt { 5 } )
p(x) = x² + (sqrt { 5 } )

4. 1, 1
Sum of the polynomial, S = 1
Product of the polynomial, P = 1
∴ Required polynomial is
p(x) = K[x² – Sx + P]
= x² – 1x + 1
p(x) = x² – x + 1

5. (frac{-1}{4}), (frac{1}{4})
Sum of the zeroes, S = (frac{-1}{4})
Product of the zeroes, P = (frac{1}{4})
∴ Required polynomial is
p(x) = K[x² – Sx + P]

6. 4, 1
Sum of the zeroes, S = 4
Product of the zeroes, P = 1
∴ Required polynomial is
p(x) = K[x² – Sx + P]
p(x) = x² – 4x + 1