How do you solve #-(3/4)x^2 + 2 = 0 #?

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Answer 1 · 2015-08-21T09:42:13

#x = +- (2sqrt(6))/3#

Start by isolating #x^2# on one side of the equation. You can do that by adding #-2# to both sides first

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

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

then multiply both sides by #-4/3# to get

#(-4/3) * (-3/4)x^2 = -2 * (-4/3)#

#x^2 = 8/3#

To solve for #x#, take the square root of both sides

#sqrt(x^2) = sqrt(8/3)#

#x = +- (2sqrt(2))/sqrt(3)#

You can simplify this by rationalizing the denominator

#x = +- (2sqrt(2) * sqrt(3))/(sqrt(3) * sqrt(3)) = color(green)(+- (2sqrt(6))/3)#