Question

find the sum of zeroes and product of zeroes x2+1/6x-2=0​

Answer 1

Answer:

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Answer 2

Given:-

$\sf{Polynomial = x^2 + \dfrac{1x}{6} - 2 = 0 }$

To find:-

• Sum of zeroes
• Product of zeroes

Solution:-

$\sf{x^2 + \dfrac{1x}{6} - 2 = 0}$

By taking LCM,

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

$\sf{\implies 6x^2 + x - 12 = 0\times 6}$

$\sf{\implies 6x^2 + x - 12 = 0}$

By splitting the middle term,

= $\sf{6x^2 + 9x - 8x - 12 = 0}$

$\sf{\implies 3x(2x + 3) - 4(2x+3) = 0}$

$\sf{\implies (2x+3)(3x-4) = 0}$

Either,

$\sf{2x+3=0}$

$\sf{\implies 2x = -3}$

$\sf{\implies x = \dfrac{-3}{2}}$

Or,

$\sf{3x-4=0}$

$\sf{\implies 3x = 4}$

$\sf{\implies x = \dfrac{4}{3}}$

Now,

$\sf{Let \:\alpha = \dfrac{-3}{2} \:and\:\beta = \dfrac{4}{3}}$

Therefore,

$\sf{Sum \:of\: zeroes = \alpha + \beta}$

= $\sf{\dfrac{-3}{2}+\dfrac{4}{3}}$

= $\sf{\dfrac{-9+8}{6}}$

= $\sf{\dfrac{-1}{6}}$

$\sf{\underline{\therefore Sum\:of\:zeroes} = \dfrac{-1}{6}}$

$\sf{Product\:of\:zeroes = \alpha\times \beta}$

= $\sf{\dfrac{-3}{2}\times \dfrac{4}{3}}$

= $\sf{-2}$

$\sf{\underline{\therefore Product\:of\:zeroes} = -2}$

Verification:-

$\sf{Sum\:of\:zeroes = \dfrac{-Coefficient\:of\:x}{Coefficient\:of\:x^2}}$

= $\sf{\dfrac{-1}{6} = \dfrac{-1}{\dfrac{6}{1}}}$

= $\sf{\dfrac{-1}{6} = \dfrac{-1}{6}\:\:[Verified]}$

$\sf{Product\:of\:zeroes = \dfrac{Constant\:term}{Coefficient\:of\:x^2}}$

= $\sf{-2 = \dfrac{-2}{1}}$

= $\sf{-2=-2\:\:[Verified]}$