Question

xx
$6x ^{2} - x - 2 = 0$
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Answer 1

Solution

Given Equation :

• f(x) = 6x² - x - 2 = 0

☞ 6x² + 3x- 4x - 2 = 0

☞ 3x(2x+1) - 2(2x+1) = 0

☞ (3x- 2)(2x+1) = 0

Zeros of Equation :

f(x) = 3x- 2 0r f(x) = 2x+1

☞ 3x - 2 = 0 0r ☞ 2x + 1 = 0

☞ 3x = 2 0r. ☞ 2x =- 1

☞ x = 2/3 0r ☞ x = -½

Therefore, -½ and 2/3 are roots of this Quadratic Equation.

Answer 2

$\huge{\blue{\underline{\mathfrak{\underline{Solution:}}}}}$

$\underline{\textsf{Given Equation}} \ratio \\ \sf f(x) = 6x^2 - x - 2 = 0 \\ \sf \implies 6x^2 + 3x - 4x - 2 = 0 \\ \sf \implies 3x(2x + 1) - 2(2x + 1) = 0 \\ \sf \implies (3x - 2)(2x + 1) = 0 \\ \\ \underline{\textsf{Zero of the Equation}} \ratio\\ \sf f(x) = 3x - 2 = 0 \\ \sf \implies 3x = 2 \\ \sf \implies x = \dfrac{2}{3} \\ \\ \boxed{\sf OR} \\ \\ \sf f(x) = 2x + 1 = 0 \\ \sf \implies 2x = -1 \\ \sf \implies x = \dfrac{-1}{2} \\ \\ \therefore \: \sf \dfrac{2}{3} \: and \: \dfrac{-1}{2} \textsf{ are roots of this Equation.}$