NCERT solution Class 10 polynomials

NCERT Solutions

Class 10 Maths

Polynomials

NCERT Questions

Ex. 2.1 Q1
Ex. 2.2 Q1
Ex. 2.2 Q2

Ex. 2.2 Q1

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

(i) x² – 2x – 8

= x² – 4x + 2x - 8

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

= (x - 4)(x + 2)

The value of x² – 2x – 8 is zero if x + 2 = 0 or x – 4 = 0

So, x = -2, 4

Therefore zeroes of x² – 2x – 8 are -2 and 4

Now, Sum of zeroes = -2 + 4 = 2 = -(-2)/1 = -(Coefficient of x)/ (Coefficient of x²)

Product of zeroes = (-2) * 4 = -8 = -8/1 = Constant term/ (Coefficient of x²)

(ii) 4s² – 4s + 1

= 4s² – 2s – 2s + 1

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

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

The value of 4s² – 4s + 1 is zero if 2s - 1 = 0

So, s = 1/2, 1/2

Therefore zeroes of 4s² – 4s + 1 are 1/2 and 1/2

Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Coefficient of x)/ (Coefficient of x²)

Product of zeroes = 1/2 * 1/2 = 1/4 = Constant term/ (Coefficient of x²)

(iii) 6x² – 3 – 7x

= 6x² – 7x - 3

= 6x² – 9x + 2x - 3

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

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

The value of 6x² – 7x – 2 is zero if 2x - 3 = 0 or 3x + 1 = 0

So, x = -1/3, 3/2

Therefore zeroes of 6x² – 7x – 3 are -1/3 and 3/2

Sum of zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6 = -(-7)/6 = -(Coefficient of x)/ (Coefficient of x²)

Product of zeroes = (-1/3) * 3/2 = -1/2 = -3/6 = Constant term/ (Coefficient of x²)

(iv) 4u² + 8u

= 4u(u + 2)

The value of 4u² + 8u is zero if 4u = 0 or u + 2 = 0

=> u = 0, -2

Therefore, the zeroes of 4u² + 8u are 0 and -2.

Now, Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of x)/ (Coefficient of x²)

Product of zeroes = 0 * (-2) = 0 = 0/4 = Constant term/ (Coefficient of x²)

(v) t² – 15

= t² – (√15)²

= (t - √15)(t + √15)

The value of t² – 15 is zero if t - √15 or t + √15 = 0

=> t = √15, -√15

Therefore, the zeroes of t² – 15 are √15 and -√15.

Now, Sum of zeroes = -√15 + √15 = 0 = -(0)/1 = -(Coefficient of x)/ (Coefficient of x²)

Product of zeroes = (-√15) * √15 = -15 = -15/1 = Constant term/ (Coefficient of x²)

(vi) 3x² – x – 4

= 3x² – 4x + 3x - 4

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

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

The value of 6x² – 7x – 2 is zero if 3x - 4 = 0 or x + 1 = 0

So, x = 4/3, -1

Therefore zeroes of 3x² – x – 4 are 4/3 and -1

Sum of zeroes = 4/3 + (-1) = (4 - 3)/3 = 1/3 = -(-1)/3 = -(Coefficient of x)/ (Coefficient of x²)

Product of zeroes = 4/3 * 1 = -4/3 = Constant term/ (Coefficient of x²)

Video solution:

Polynomials Class 10 Maths NCERT Chapter 2 NCERT Solutions

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