Question

Find the roots of the quadratic equation
$3x { }^{2} - 2 \sqrt{6} x + 2 = 0$

Answer 1

$3x^2 - 2 \sqrt{6}x + 2 = 0 \\ \\ 3x^2 - 1\sqrt6x - 1\sqrt6 + 2 = 0 \\ \\ \sqrt3x ( \sqrt3x - \sqrt2) - \sqrt2 (\sqrt3x - \sqrt2x) = 0 \\ \\ (\sqrt3x - \sqrt2)(\sqrt3x - \sqrt2) = 0 \\ \\ x = \frac{\sqrt2}{\sqrt3}\: and \: \frac{\sqrt2}{\sqrt3}$

Answer 2

hey !!!

3x² - 2√6x + 2 =0

3x ²-√6 x -√6x + 2 { by using spliting method]

3x² - √2×3x - √2×3 +2

√3×√3x² - √2×3 - √2×3x +√2×√2

√3x(√3x - √2) -√2( √3x -√2)

(√3x-√2 ) ( √3x -√2) { taking common }

√3x = √2

x = √2/√3

hope it helps !!!

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