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$\sqrt{2} x (3 \sqrt{2} x - 6) = 0$ $sqrt2 x(3sqrt2 x-6)=0$

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Answer 1 · 2018-03-01T00:00:58

$x = 0, x = \sqrt{2}$ $x=0, x=sqrt(2)$

Apply the Distributive Property, $a (b \pm c) = a b \pm a c$ $a(b+-c)=ab+-ac$

$\sqrt{2} x (3 \sqrt{2} x - 6) = \sqrt{2} x \cdot 3 \sqrt{2} x - \sqrt{2} x \cdot 6$ $sqrt(2)x(3sqrt(2)x-6)=sqrt(2)x*3sqrt(2)x-sqrt(2)x*6$

$3 {\sqrt{2}}^{2} x^{2} - 6 \sqrt{2} x = 0$ $3sqrt(2)^2x^2-6sqrt(2)x=0$

$3 (2) x^{2} - 6 \sqrt{2} x = 0$ $3(2)x^2-6sqrt(2)x=0$ (Because ${\sqrt{a}}^{2} = a$ $sqrt(a)^2=a$ )

$6 x^{2} - 6 \sqrt{2} x = 0$ $6x^2-6sqrt(2)x=0$

Both of our terms include an instance of $6 x$ $6x$ , so we can factor out $6 x :$ $6x:$

$6 x (x - \sqrt{2}) = 0$ $6x(x-sqrt(2))=0$

We now have two equations two solve for $x :$ $x:$

$6 x = 0$ $6x=0$ and $x - \sqrt{2} = 0$ $x-sqrt(2)=0$

$6 x = 0$ $6x=0$

$(cancel6x)/cancel6=0/6$

$x=0$

So, $x=0$ is one of our answers.

$x-sqrt(2)=0$

$xcancel(-sqrt(2)+sqrt(2))=0+sqrt(2)$

$x=sqrt(2)$

So, $x=sqrt(2)$ is our other answer.