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

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

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