Question

${x}^{2} - 4 \sqrt{2} x + 6 = 0$

Answer 1

Answer:

$\boxed{x = \sqrt{2} , \bf{or}\: 3\sqrt{2}}$

Step-by-step explanation:

Observe that:

$6 = \sqrt{2}\times\sqrt{2}\times3$

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$x^2 - 4\sqrt{2} x + 6 = 0$

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

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

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

Two cases arise :-

$\implies x = \sqrt{2}$

$\implies x = 3 \sqrt{2}$

Hope it helps you

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