How do you solve $2x^2 = 16$ ?

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Answer 1 · 2015-08-02T00:32:54

$x=2sqrt2, -2sqrt 2$

$2x^2=16$

Divide both sides of the equation by $2$ .

$x^2=16/2$ =

$x^2=8$

Take the square root of both sides.

$x=+-sqrt8$

$x=+-sqrt(2xx2xx2)$

$x=+-sqrt(2^2xx2)$

Apply the rule $sqrt (a^2)=a$ .

$x=+-2sqrt2$

Solve for $x$ .

$x=2sqrt 2$

$x=-2sqrt2$

How do you solve 2x^2 = 162x^2 = 16?