How do you solve the equation $\frac{2}{3} x^{2} - 4 = 12$ $2/3x^2-4=12$ ?

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Answer 1 · 2017-06-12T03:11:18

The answer is $x = \pm 2 \sqrt{6}$ $x=+-2sqrt6$ .
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Solve:

$\frac{2}{3} x^{2} - 4 = 12$ $2/3x^2-4=12$

Add $4$ $4$ to both sides.

$2/3x^2-color(red)cancel(color(black)(4))+color(red)cancel(color(black)(4))=12+4$

Simplify.

$2/3x^2=16$

Multiply both sides by $3$ .

$color(red)cancel(color(black)(3))xx2/color(red)cancel(color(black)(3))x^2=16xx3$

Simplify.

$2x^2=48$

Divide both sides by $2$ .

$color(red)cancel(color(black)(2))/color(red)cancel(color(black)(2))x^2=48/2$

Simplify.

$x^2=24$

Take the square root of both sides.

$sqrt(x^2)=+-sqrt24$

Simplify.

$x=+-sqrt24$

Simplify $sqrt24$ using prime factorization.

$x=+-sqrt(2xx2xx2xx3)$

$x=+-sqrt(2^2xx2xx3)$

$x=+-2sqrt6$

How do you solve the equation 2/3x^2-4=1223x2−4=122/3x^2-4=12?