Question

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

Answer 1

Answer:

X² + 1/6x - 2 = 0

Writing 12 on both sides to make 1/6 as integer from fraction.

(X² + 1/6 - 2)12 = 0x12

X² + 2x - 24 = 0. factors of 24 are 1x24, 2x12, 3x8, 4x6

Taking +6x and -4x

X² + 6x - 4x - 24

x(x + 6) -4 (x + 6)

(x + 6)(x - 4)

Hence the of the polynomial are -6 and +4

product of zeroes = -24

Step-by-step explanation:

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Answer 2

$\sf\underline{\green{\:\:\: Given:-\:\:\:}}$

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

$\sf\underline{\green{\:\:\:To \:Find:-\:\:\:}}$

• Sum of zeroes

• Product of zeroes

$\sf\underline{\green{\:\:\: Solution:-\:\:\:}}$

⠀$\underline{\:\textsf{ We have :}}$⠀⠀⠀⠀⠀

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

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

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

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

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

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

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

$\sf Hence :-$

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

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

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

$\sf OR,$

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

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

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

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⠀⠀⠀⠀⠀⠀⠀

$\sf Now$

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

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

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

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

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

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

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

$\leadsto \sf{-2}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

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

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

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