Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its coefficients.

Question : Find the zeroes of the quadratic polynomials and verify a relationship between zeroes and its coefficients.

(i)  \(x^2 – 2x – 8\)

(ii)  \(4s^2 – 4s + 1\)

(iii)  \(6x^2 – 3 – 7x\)

(iv)  \(4u^2 + 8u\)

(v)  \(t^2 – 15\)

(vi)  \(3x^2 – x – 4\)

Solution :

(i)  \(x^2 – 2x – 8\) = \(x^2 – 4x + 2x – 8\) = x(x – 4) + 2(x – 4)  = (x – 4)(x + 2)

So, the value of \(x^2 – 2x – 8\) is zero when x – 4 = 0 or x + 2 = 0. i.e. when x = 4 or x = -2.

So, the zeroes of \(x^2 – 2x – 8\) are 4, -2.

Sum of the zeroes = 4 – 2 = 2 = \(-(-2)\over 1\) = \(-coefficient of x\over coefficient of x^2\) = 2

Product of the zeroes = 4(-2) = -8 = \(-8\over 1\) = \(constant term\over coefficient of x^2\) = -8

(ii)  \(4s^2 – 4s + 1\) = \(4s^2 – 2s -2s + 1\) = \((2s – 1)^2\)

So, the value of \(4s^2 – 4s + 1\) is zero when 2s – 1 = 0 or s = \(1\over 2\)

So, the zeroes of \(x^2 – 2x – 8x\) are \(1\over 2\), \(1\over 2\)

Sum of the zeroes = \(1\over 2\) + \(1\over 2\) = 1 = \(-(-4)\over 4\) = \(-coefficient of s\over coefficient of s^2\) = 1

Product of the zeroes = (\(1\over 2\))(\(1\over 2\)) = \(1\over 4\) = \(1\over 4\) = \(constant term\over coefficient of s^2\) = \(1\over 4\)

(iii) \(6x^2 – 3 – 7x\) = \(6x^2 – 7x – 3\) = \(6x^2 – 9x – 2x – 3\)

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

So, the value of \(6x^2 – 3 – 7x\) is zero when the value of (3x + 1)(2x – 3) = 0 i.e.

when 3x + 1 = 0 or 2x – 3 = 0  i.e. when x = \(-1\over 3\) or x = \(3\over 2\).

So, the zeroes of \(6x^2 – 3 – 7x\) are \(-1\over 3\) and \(3\over 2\).

Sum of the zeroes = \(-1\over 3\) + \(3\over 2\) = \(7\over 6\) = \(-(-7)\over 6\) = \(-coefficient of x\over coefficient of x^2\) = \(7\over 6\)

Product of the zeroes = (\(-1\over 3\))(\(3\over 2\)) = \(-3\over 6\) = (\(-1\over 3\))(\(3\over 2\)) = \(constant term\over coefficient of x^2\) = \(-3\over 6\)

(iv) We have : \(4u^2 + 8u\) = 4u(u + 2)

The value of \(4u^2 + 8u\) is zero when the value of 4u(u + 2) = 0 i.e. when u = 0 or u + 2 = 0 i.e. u = 0 or u = -2.

So, the zeroes of \(4u^2 + 8u\) are 0, -2.

Sum of the zeroes = 0 + (- 2) = -2 = \(-(-8)\over 4\) = \(-coefficient of u\over coefficient of u^2\) = –2

Product of the zeroes = (0)(-2) = 0 = \(0\over 4\) = \(constant term\over coefficient of u^2\) = 0

(v)  We have, \(t^2 – 15\) = \((t – \sqrt{15})\)\((t + \sqrt{15})\)

The value of \(t^2 – 15\) is zero when the value of \((t – \sqrt{15})\)\((t + \sqrt{15})\) is 0, i.e. when \((t – \sqrt{15})\) = 0 or \((t + \sqrt{15})\) = 0

i.e. when t = \(\sqrt{15}\), \(-\sqrt{15}\).

So, the zeroes of \(t^2 – 15\) are  \(\sqrt{15}\), \(-\sqrt{15}\).

Sum of the zeroes = (\(\sqrt{15}\)) +  (\(-\sqrt{15}\)) = 0 = \(-(-0)\over 1\) = \(-coefficient of t\over coefficient of t^2\) = 0

Product of the zeroes = (\(\sqrt{15}\))(\(-\sqrt{15}\)) = -15 = \(-15\over 1\) = \(constant term\over coefficient of t^2\) = -15

(vi)  We have : \(3x^2 – x – 4\) = \(3x^2 – 3x – 4x – 4\) = 3x(x +1) – 4(x + 1) = (x + 1)(3x – 4)

The value of \(3x^2 – x – 4\) is zero when the value of (x + 1)(3x – 4) is 0, i.e. when x + 1 = 0 or 3x – 4 = 0, i.e. when x = -1 or x = \(4\over 3\)

So, the zeroes of \(3x^2 – x – 4\) are -1, \(4\over 3\).

Sum of the zeroes = -1 + \(4\over 3\) = \(-3 + 4\over 3\) = \(1\over 3\) = \(-(-1)\over 3\) =  \(-coefficient of x\over coefficient of x^2\) = \(1\over 3\).

Product of the zeroes = (-1)(\(4\over 3\)) = \(constant term\over coefficient of x^2\) = \(-4\over 3\)

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