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How do you solve the equation #2/3x^2-4=12#?

Answer 1

Eli Assouline

The answer is #x=+-2sqrt6#.
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Solve:

#2/3x^2-4=12#

Add #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 each side's square root.

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

Simplify.

#x=+-sqrt24#

Simplify #sqrt24# using prime factorization.

#x=+-sqrt(2xx2xx2xx3)#

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

#x=+-2sqrt6#

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Answer 2

Abigail Allen

To solve the equation ( \frac{2}{3}x^2 - 4 = 12 ), you would first add 4 to both sides to isolate the term involving ( x^2 ), then divide both sides by ( \frac{2}{3} ), and finally take the square root of both sides to solve for ( x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.