Question

2x^2 - 2√6x +3 =0 solve by factorization method​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{2x^2-2\sqrt{6}x+3=0}$

$\underline{\textbf{To find:}}$

$\textsf{Solution of the given quadratic equation}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{2x^2-2\sqrt{6}x+3=0}$

$\textsf{This can be written as,}$

$\mathsf{2x^2-\sqrt{6}\,x-\sqrt{6}\,x+3=0}$

$\mathsf{\sqrt{2}x(\sqrt{2}x-\sqrt{3})-\sqrt{3}(\sqrt{2}x-\sqrt{3})=0}$

$\mathsf{(\sqrt{2}x-\sqrt{3})(\sqrt{2}x-\sqrt{3})=0}$

$\implies\mathsf{(\sqrt{2}x-\sqrt{3})=0\;\;(or)\;\;(\sqrt{2}x-\sqrt{3})=0}$

$\implies\mathsf{\sqrt{2}x=\sqrt{3};\;(or)\;\;\sqrt{2}x=\sqrt{3}}$

$\implies\mathsf{x=\dfrac{\sqrt{3}}{\sqrt{2}};\;(or)\;\;x=\dfrac{\sqrt{3}}{\sqrt{2}}}$

$\therefore\mathsf{Solution\;is\;x=\dfrac{\sqrt{3}}{\sqrt{2}}}$

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