Question

1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
(i) x2 - 2x -- 8
(ii) 452-4s +1
(ii) 6x2-3-7x​

Answer 1

(i) x2 – 2x – 8 = (x - 4) (x + 2) The value of x2 – 2x – 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2 Therefore, the zeroes of x2 – 2x – 8 are 4 and -2.Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x2 Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x2 (ii) 4s2 – 4s + 1 = (2s-1)2 The value of 4s2 - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2 Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2 Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s2

(iii) 6x2 – 3 – 7x = 6x2 – 7x – 3 = (3x + 1) (2x - 3) The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2 Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.Sum of zeroes = -1/3 + 3/2 = 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 = 4u2 + 8u + 0 = 4u(u + 2) The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2 Therefore, the zeroes of 4u2 + 8u are 0 and - 2. Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2 Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of (v) t2 – 15 = t2 - 0.t - 15 = (t - √15) (t + √15) The value of t2 - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15 Therefore, the zeroes of t2 - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2 Product of zeroes

Answer 2

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(i) x2 - 2x - 8 = (x - 4) (x + 2) The value of x2 - 2x

-8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2 Therefore, the zeroes of x2 - 2x - 8 are 4 and -2.Sum of zeroes = 4 + (-2) = 2 = -(-2)/ 1 = -(Coefficient of x)/Coefficient of x2 Product of zeroes = 4 * (-2) = -8 = -8/1 = Constant term/ Coefficient of x2

(ii) 4s2 - 4s + 1 = (25-1)2 The value of 4s2 - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2 Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s) Coefficient of s2 Product of zeroes = 1/2 x 1/2 = 1/4 = Constant term/Coefficient of s2

(iii) 6x2-3-7x = 6x2 - 7x - 3 = (3x + 1) (2x -3)

The value of 6x2 3 -7x is zero when 3x + 1 = O or 2x - 3 = 0, i.e., X = -1/3 or x = 3/2 Therefore, the zeroes of 6x2 3 7x are -1/3 and 3/2.Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of(i) x2 - 2x - 8 = (x - 4) (x + 2) The value of x2 - 2x)/Coefficient of x2 Product of zeroes = -1/3 x 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2

(iv) 4u2 + 8u = 4u2 + 8 + 0 = 4u(u + 2) The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = -2 Therefore, the zeroes of 4u2 + 8u are 0 and - 2. Sum of zeroes = 0 + (-2) = -2 = -(8)/ 4 = -(Coefficient of u)/Coefficient of u2 Product of zeroes = 0 Coefficient of (-2) = 0 = 0/4 = Constant term/

(v) t2- 15 = t2 0.t 15

= (t - V15) (t + V15) The value of t2 - 15 is zero when t-V15 = 0 or t + 15 = 0, i.e., when t = v15 or t = -V15 Therefore, the zeroes of t2 - 15 are 15 and -15.Sum of zeroes = v15 +

- 15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2 Product of zeroes = (v15) (-V15) = -15 = -15/1 = Constant term/Coefficient of u2.

-4/3

(vi) 3x2 x 4 = (3x 4) (x + 1) The value of 3x2-x-4 is zero when 3x -4 = 0 and x + 1= 0,i.e., when x = 4/3 or x = -1 Therefore, the zeroes of 3x2 - X-4 are 4/3 and -1.Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/ Coefficient of x2 Product of zeroes = 4/3 x (-1) = Constant term/Coefficient of x2.