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
= x – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
The value of x2 – 2x – 8 is zero if (x + 2) = 0 and (x – 4) = 0
x = -2 or x = 4
Sum of zeroes = (-2 + 4) = 2 = - coefficient of x
coefficient of x2
Product of zeroes = (-2) × 4 = -8 = 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 and (2s-1 ) = 0
s = 1/2 , 1/2
Sum of zeroes = (1/2 + 1/2) = 1 - coefficient of x
coefficient of x2
Product of zeroes =1/2 × 1/2 = 1/4 = constant term
coefficient of x2
(iii) 6x2 –7x – 3
= 6x – 9x + 2x – 3
= 3x (2x – 3) + 1(2x – 3)
= (3x + 1) (2x – 3)
The value of 6x2 –7x – 3 is zero, if (3x + 1) = 0 and (2x – 3) = 0
X = -1 /3 , 3/2
Sum of zeroes = ( -1/3 + 3/2) = 7/6 = - coefficient of x
coefficient of x2
Product of zeroes = -1/3 × 3/2 = -3/2 = constant term
coefficient of x2
(iv) 4u2+8u
4u(u+2)
The value of 4u2+8u is zero, if 4u = 0 and (u+2) =0
u = 0, - 2
Sum of zeroes = ( 0+ (-2)) = -2 = - coefficient of x
coefficient of x2
Product of zeroes = (-2) × 0 = 0 = constant term
coefficient of x2
(v)
(vi)
3x2–x–4
3x – 4x + 3x – 4
= x (3x – 4) + 1 (3x – 4)
The value of 3x – x + 4 is zero, if (3x – 4) = 0 and (x + 1) = 0
Sum of zeroes = [4/3 + ( -1)] = 1/3 = - coefficient of x
coefficient of x2
Product of zeroes = (-1) × 4/3 = -4/3 = constant term
coefficient of x2
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Welcome to the NCERT Solutions for Class 10 Mathematics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroe....
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