Question

Find
the zeroes of polynomials x2 + 1/6x -2 and verify the relationship between zeroes and coefficient ​

Answer 1

Answer:

Hope this helps uhh!!❤

Answer 2

$\text{Zeros of polynomial } : x = \frac{4}{3} \text{ and } x = \frac{-3}{2}$

Solution:

Given polynomial equation is:

$x^2 + \frac{1}{6}x -2 = 0\\\\6x^2 +x - 12 = 0$

We have to find the zeros of polynomial

$6x^2+x+12 = 0\\\\Split\ the\ middle\ term\\\\6x^2-8x+9x+12=0\\\\\mathrm{Break\:the\:expression\:into\:groups}\\\\\left(6x^2-8x\right)+\left(9x-12\right) = 0\\\\\mathrm{Factor\:out\:}2x\mathrm{\:from\:}6x^2-8x\mathrm{:\quad }2x\left(3x-4\right)\\\\\mathrm{Factor\:out\:}3\mathrm{\:from\:}9x-12\mathrm{:\quad }3\left(3x-4\right)\\\\2x\left(3x-4\right)+3\left(3x-4\right) = 0\\\\\mathrm{Factor\:out\:common\:term\:}3x-4\\\\\left(3x-4\right)\left(2x+3\right) = 0$

$Therefore \\\\3x-4 = 0\\\\3x = 4 \\\\x = \frac{4}{3}$

$Also\\\\2x+3=0\\\\2x=-3\\\\x = \frac{-3}{2}$

Thus,

$\text{Zeros of polynomial } : x = \frac{4}{3} \text{ and } x = \frac{-3}{2}$

$Sum\ of\ zeros = \frac{4}{3} + \frac{-3}{2} \\\\Sum\ of\ zeros = \frac{-1}{6}\\\\Product\ of\ zeros = \frac{4}{3} \times \frac{-3}{2} =-2$

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