How do you solve: $-2x^2 + 8 = 0#?

2 Answers
Mar 24, 2017

Answer:

See the entire solution process below:

Explanation:

First, subtract #color(red)(8)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#-2x^2 + 8 - color(red)(8) = 0 - color(red)(8)#

#-2x^2 + 0 = -8#

#-2x^2 = -8#

Next, divide each side of the equation by #color(red)(-2)# to isolate #x^2# while keeping the equation balanced:

#(-2x^2)/color(red)(-2) = (-8)/color(red)(-2)#

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

#x^2 = 4#

Now, take the square root of each side of the equation to solve for #x# while keeping the equation balanced. Remember, the square root of a number produces both a negative and positive result:

#sqrt(x^2) = sqrt(4)#

#x = +-2#

Mar 24, 2017

Answer:

#x = 2 ,-2#

Explanation:

#-2 x^2 + 8 =0#

subtract #-8# to both sides
#-2 x^2 + 8 - 8=0 -8#
#-2 x^2 = -8#

divide #-23# to both sides
#(-2 x^2) /-2 = -8/-2#
#x^2 = 4#

square root both sides
#sqrt (x^2) = sqrt 4#
#x = +-2#