Question

√2x^2+ √3 (1 - 2√2)x-6=0​

Answer 1

We have,

$\sf{ \sqrt{2}x {}^{2} + \sqrt{3}(1 - 2 \sqrt{2})x - 6 = 0 } \\ \\ \implies \: \sf{ \sqrt{2}x {}^{2} + \sqrt{3}x - 2 \sqrt{6}x - 6 = 0 } \\ \\ \implies \: \sf{x( \sqrt{2}x + \sqrt{3}) - 2\sqrt{3}( \sqrt{2}x + \sqrt{3}) = 0 } \\ \\ \implies \: \boxed{ \sf{(x - 2 \sqrt{3}) \: and \: ( \sqrt{2}x + \sqrt{3}) }}$

On equating them with zero,we get

x - 2√3 = 0 and, √2x + √3 = 0

→ x = 2√3 or - √3/√2

Answer 2

$\huge\bf{Answer:-}$

• $\begin{lgathered}\bf{ \sqrt{2}x {}^{2} + \sqrt{3}(1 - 2 \sqrt{2})x - 6 = 0 } \\\\ => \: \bf{ \sqrt{2}x {}^{2} + \sqrt{3}x - 2 \sqrt{6}x - 6 = 0 } \\ \\ => \: \bf{x( \sqrt{2}x + \sqrt{3}) - 2\sqrt{3}( \sqrt{2}x + \sqrt{3}) = 0 } \\ \\ => \: \bf{ \bf{(x - 2 \sqrt{3}) \: and \: ( \sqrt{2}x + \sqrt{3}) }}\end{lgathered}$

[x - 2√3 = 0 ],[√2x + √3 = 0] So....[{x = 2√3 or - √3/√2}]