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

Question

Hazel Anglin · Accepted Answer

#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)#

Avery Alexander · Answer

undefined To solve the equation -(3/4)x^2 + 2 = 0:

1. Subtract 2 from both sides to isolate the term -(3/4)x^2:
   -(3/4)x^2 = -2

2. Divide both sides by -(3/4) to solve for x^2:
   x^2 = (-2) / (-(3/4))

3. Simplify the expression on the right side:
   x^2 = (2) / (3/4)
   x^2 = 2 * (4/3)

4. Further simplify the expression:
   x^2 = 8/3

5. Take the square root of both sides to solve for x:
   x = ± √(8/3)

6. Simplify the square root:
   x = ± √(8/3)
   x = ± √(8) / √(3)
   x = ± √(8) / √(3) * (√(3) / √(3))
   x = ± (√(8 * 3) / 3)
   x = ± (√24 / 3)
   x = ± (√(4 * 6) / 3)
   x = ± (√4 * √6 / 3)
   x = ± (2√6 / 3)

Therefore, the solutions to the equation -(3/4)x^2 + 2 = 0 are:
x = (2√6) / 3 and x = -(2√6) / 3.