Question

Please do this question

Answer 1

$6 {x}^{2} - \sqrt{2} x - 2 \\ = 6 {x}^{2} - 3 \sqrt{2} x + 2 \sqrt{2} x - 2 \\ = 3x(2x - \sqrt{2} ) + \sqrt{2} (2x - \sqrt{2} ) \\ = (3 x + \sqrt{2} )(2x - \sqrt{2} )$

hope it helps.

Answer 2

Answer :-

By using quadratic formula / Sridhacharyas equation

which is = $\frac{ -b ±\sqrt{b^2 - 4ac}}{2a}$

Then by putting values above

= $\frac{ -(-\sqrt{2} ±\sqrt{\sqrt{2}^2 - 4 × 6 ×(- 2)}}{2 × 6}$

= $\frac{ \sqrt{2} ±\sqrt{ 2 + 48 }}{12}$

= $\frac{ \sqrt{2} ±\sqrt{ 50}}{12}$

= $\frac{ \sqrt{2} ± 5\sqrt{ 2 }}{12}$

= $\frac{ \sqrt{2} +5\sqrt{ 2 }}{12}$

= $\frac{ 6\sqrt{ 2 }}{12}$

= $\frac{ \sqrt{ 2 }}{2}$

And

= $\frac{ \sqrt{2} - 5\sqrt{ 2 }}{12}$

= - $\frac{ 4\sqrt{ 2 }}{12}$

= - $\frac{ \sqrt{ 2 }}{3}$

Then as x = - $\frac{ \sqrt{ 2 }}{3}$ and $\frac{ \sqrt{ 2 }}{2}$

therefore factorisation

= (x - $\frac{ \sqrt{ 2 }}{3}$ ) ×(x + $\frac{ \sqrt{ 2 }}{2}$ )

or

(3x - √2) × (2x + √2)