Question

26. 3x^2 - 2√6x+2=0​

Answer 1

$\huge\star{\purple{\underline{\mathfrak{Answer :-}}}}$

$x = (\sqrt {\dfrac {2}{3})} \:and\; (\sqrt {\dfrac {2}{3}})$

$\huge\star{\purple{\underline{\mathfrak{Explanation :-}}}}$

Let solve this quadratic equation by quadratic formula for Sridharacharya formula :-

$\dfrac {-b \pm \sqrt {{b}^{2} - 4ac}}{2a}$

Putting in the values :-

$\leadsto {\dfrac {-(-2\sqrt {6})\pm \sqrt {{(-2\sqrt {6}})^{2}-4 (3)(2)}}{2 (3)}}$

$\leadsto {\dfrac {2\sqrt {6}\pm \sqrt {(4)(6)-24}}{6}}$

$\leadsto {\dfrac {2\sqrt {6} \pm \sqrt {24-24}}{6}}$

$\leadsto {\dfrac {2\sqrt {6} \pm \sqrt {0}}{6}}$

$\leadsto {\dfrac {\cancel {2}\sqrt {6}}{\cancel {6}\:\:3}}$

$\leadsto{\dfrac {\sqrt {6}}{3}}$

Simplifying further,

$\leadsto{\dfrac {\sqrt {2}\times\cancel{\sqrt{3}}}{\sqrt {3}\times\cancel{\sqrt{3}}}}$

$\leadsto {\dfrac {\sqrt {2}}{\sqrt {3}}}$

So, the roots are :- $x = (\sqrt {\dfrac {2}{3})} \:and \:(\sqrt {\dfrac {2}{3}})$

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NOTE :-

The discriminant here that is $\sqrt {{b}^{2}-4ac}$ is equal to 0.

Hence, the roots are real and equal.

_____________________

Answer 2

$\purple{ \sf \large \underline{ \underline{ \: Solution : \: \: \: }}}$

Given ,

The quadratic equation is

3x² - 2√6x + 2 = 0

By prime factorisation method ,

$\sf \hookrightarrow 3 {x}^{2} - 2 \sqrt{6} + 2 = 0 \\ \\ \sf \hookrightarrow3 {x}^{2} - \sqrt{6}x - \sqrt{6}x + 2 \\ \\ \sf \hookrightarrow \sqrt{3} x( \sqrt{3} x - \sqrt{2} ) - \sqrt{2} ( \sqrt{3} x - \sqrt{2} ) \\ \\ \sf \hookrightarrow (\sqrt{3}x - \sqrt{2} )( \sqrt{3}x - \sqrt{2} ) \\ \\ \sf \hookrightarrow x = \frac{ \sqrt{2} }{ \sqrt{3} } \\ \\ \sf \hookrightarrow x = \frac{ \sqrt{6} }{3}$

Hence , the roots of given quadratic equation is √6/3 and √6/3