How do you solve $10 + \frac{1}{3} x^{2} = 18$ $10+1/3x^2=18$ ?

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Answer 1 · 2016-07-04T22:15:53

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

$10 + \frac{1}{3} x^{2} = 18$ $10 + 1/3x^2 = 18$

Subtract 10 from both sides

$\frac{1}{3} x^{2} = 8$ $1/3x^2 = 8$

Multiply both sides by 3

$x^{2} = 24$ $x^2 = 24$

Take square root of both sides:

$x = \pm \sqrt{24} = \pm \sqrt{4} \sqrt{6} = \pm 2 \sqrt{6}$ $x = +-sqrt(24) = +-sqrt(4)sqrt(6) = +-2sqrt(6)$

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