NCERT Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2

NCERT Solutions Class 10 Maths
Chapter – 2 (Polynomials)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 2 Polynomials Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.

Chapter : 2 Polynomials

Exercise – 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8 (ii) 4s² – 4s + 1 (iii) 6x² – 3 – 7x
(iv) 4u² + 8u (v) t² – 15 (vi) 3x² – x – 4

Solutions –
(i) x²– 2x – 8
⇒x²– 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4)(x + 2)
x = 4 , x = -2 are the zeroes of the polynomial.
Thus, α = 4, β = -2
Sum of zeroes = – coefficient of x / coefficient of x²For x² – 2x – 8,
a = 1, b = – 2, c = – 8
α + β = – b / a
Here,
α + β = – 2 + 4 = 2
– b / a = – (- 2)/1 = 2
Hence, sum of the zeros α + β = – b/a is verified.
Now, Product of zeroes = constant term / coefficient of x²α × β = c / a
Here,
α × β = – 2 × 4 = – 8
c / a = – 8 / 1 = – 8
Hence, product of zeros α × β = c / a is verified.
Thus, x = 4, -2 are the zeroes of the polynomial.

(ii) 4s²– 4s + 1
⇒ 4s²– 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1)
s = 1/2, s = 1/2 are the zeroes of the polynomial.
Thus, α = 1/2 and β = 1/2
Now, let’s find the relationship between the zeroes and the coefficients.
For 4s² – 4s + 1,
a = 4, b = – 4 and c = 1
Sum of zeroes = – coefficient of s / coefficient of s²α + β = – b/a
Here,
α + β = 1/2 + 1/2 = 1
– b / a = – (- 4) / 4 = 1
Hence, α + β = – b / a, verified
Now, Product of zeroes = constant term / coefficient of s²α × β = c / a
Here,
α × β = 1/2 × 1/2 = 1/4
c / a = 1/4
Hence, α × β = c / a, verified.
Thus, s = 1/2, 1/2 are the zeroes of the polynomial.

(iii) 6x²– 3 – 7x
⇒ 6x²– 7x – 3
= 6x²– 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x – 3)
x = 3/2, x = – 1/3 are the zeroes of the polynomial.
Thus, α = 3/2 and β = -1/3
Now, let’s find the relationship between the zeroes and the coefficients:
For 6x² – 3 – 7x,
a = 6, b = – 7 and c = – 3
Sum of zeroes = – coefficient of x / coefficient of x²α + β = – b / a
Here,
α + β = 3/2 + (-1/3) = 7/6
– b / a = – (-7) / 6 = 7/6
Hence, α + β = – b/a, verified
Now, Product of zeroes = constant term / coefficient of x²α × β = c / a
Here,
α × β = 3/2 × (- 1/3) = -1/2
c / a = (- 3) / 6 = -1/2
Hence, α × β = c/a, verified.
Thus, x = 3/2, – 1/3 are the zeroes of the polynomial.

(iv) 4u²+ 8u
⇒ 4u(u + 2)
u = 0, u = – 2 are the zeroes of the polynomial
Thus, α = 0 and β = – 2
Now, let’s find the relationship between the zeroes and the coefficients
For 4u² + 8u,
a = 4, b = 8, c = 0
Sum of zeroes = – coefficient of u / coefficient of u²α + β = – b/a
Here,
α + β = 0 + (- 2) = – 2
– b / a = – (8) / 4 = – 2
Hence, α + β = – b / a, verified
Now, Product of zeroes = constant term / coefficient of u²α × β = c/a
Here,
α × β = 0 × (- 2) = 0
c / a = 0 / 4 = 0
Hence, α × β = c / a, verified.
Thus, u = 0, – 2 are the zeroes of the polynomial.

(v) t²– 15
⇒ t² = 15
⇒ t = ±√15
t = -√15, t = √15 are the zeroes of the polynomial.
Thus, α = -√15 and β = √15
Now, let’s find the relationship between the zeroes and the coefficients
For t² – 15,
a = 1, b = 0, c = -15
Sum of zeroes = – coefficient of t / coefficient of t²α + β = – b / a
Here,
α + β = -√15 + √15 = 0
– b / a = – 0 / 1 = 0
Hence, α + β = – b / a, verified
Now, Product of zeroes = constant term / coefficient of t²α × β = c / a
Here,
α × β = -√15 × √15 = -15
c / a = -15 / 1 = -15
Hence, α × β = c / a, verified.
Thus, t = -√15, √15 are the zeroes of the polynomial.

(vi) 3x² – x – 4
⇒ 3x² – x – 4 = 0
⇒ 3x² – 4x + 3x – 4 = 0
⇒ x (3x – 4) + 1(3x – 4) = 0
⇒ (x + 1)(3x – 4) = 0
x = – 1, x = 4/3 are the zeroes of the polynomial.
Thus, α = – 1 and β = 4/3
Now, let’s find the relationship between the zeroes and the coefficients
For 3x² – x – 4,
a = 3, b = – 1, c = – 4
Sum of zeroes = – coefficient of x / coefficient of x²α + β = – b / a
Here,
α + β = – 1 + 4/3 = 1/3
– b / a = – (-1) / 3 = 1/3
Hence, α + β = – b / a, verified
Now, Product of zeroes = constant term / coefficient of x²α × β = c/a
Here,
α × β = – 1 × (4/3) = – 4/3
c / a = – 4/3
Hence, α × β = c / a, verified.
Thus, x = – 1, 4/3 are the zeroes of the polynomial.

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.
(i) $\mathbf{\frac{1}{4}}$ , – 1 (ii) √2, $\mathbf{\frac{1}{3}}$ (iii) 0, √5

(iv) 1, 1 (v) – $\mathbf{\frac{1}{4}}$ , $\mathbf{\frac{1}{4}}$ (vi) 4, 1

Solution –

(i) $\mathbf{\frac{1}{4}}$ , – 1
Sum of zeroes = α+β = 1/4
Product of zeroes = α β = -1
We know that the general equation of a quadratic polynomial is:
x² – (sum of roots) x + (product of roots)
x²–(α + β)x + αβ = 0
x² – (1/4)x + (- 1)
x² – (1/4)x – 1
4x²–x-4 = 0
Thus, 4x²–x–4 is the quadratic polynomial.

(ii) √2, $\mathbf{\frac{1}{3}}$
Sum of zeroes = α + β =√2
Product of zeroes = α β = 1/3
We know that the general equation of a quadratic polynomial is:
x² – (sum of roots) x + (product of roots)
x²– (α + β)x + αβ = 0
x² –(√2)x + (1/3) = 0
3x²– 3√2x + 1 = 0
Thus, 3x²– 3√2x + 1 is the quadratic polynomial.

(iii) 0, √5
Sum of zeroes = α + β = 0
Product of zeroes = α β = √5
We know that the general equation of a quadratic polynomial is:
x² – (sum of roots) x + (product of roots)
x²– (α + β)x + αβ = 0
x²– (0)x + √5= 0
Thus, x²+√5 is the quadratic polynomial.

(iv) 1, 1
Sum of zeroes = α+β = 1
Product of zeroes = α β = 1
We know that the general equation of a quadratic polynomial is:
x² – (sum of roots) x + (product of roots)
x²– (α + β)x + αβ = 0
x²– x + 1 = 0
Thus, x²– x + 1 is the quadratic polynomial.

(v) – $\mathbf{\frac{1}{4}}$ , $\mathbf{\frac{1}{4}}$
Sum of zeroes = α+β = -1/4
Product of zeroes = α β = 1/4
We know that the general equation of a quadratic polynomial is:
x² – (sum of roots) x + (product of roots)
x²– (α + β)x + αβ = 0
x²– (-1/4)x + (1/4) = 0
4x²+x+1 = 0
Thus, 4x²+ x + 1 is the quadratic polynomial.

(vi) 4, 1
Sum of zeroes = α+β =4
Product of zeroes = αβ = 1
We know that the general equation of a quadratic polynomial is:
x² – (sum of roots) x + (product of roots)
x²– (α + β)x + αβ = 0
x²– 4x + 1 = 0
Thus, x²– 4x + 1 is the quadratic polynomial.

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NCERT Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2