How do you solve #4x^2-16=28#?

1 Answer
Feb 13, 2017

Answer:

See the entire solution process below:

Explanation:

First, add #color(red)(16)# to each side of the equation to isolate the #x^2# term while keeping the equation balanced:

#4x^2 - 16 + color(red)(16) = 28 + color(red)(16)#

#4x^2 - 0 = 44#

#4x^2 = 44#

Next, divide each side of the equation by #color(red)(4)# to isolate #x^2# while keeping the equation balanced:

#(4x^2)/color(red)(4) = 44/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x^2)/cancel(color(red)(4)) = 11#

#x^2 = 11#

Now, take the square root of each side of the equation to solve for #x# while keeping the equation balanced. Remember, when taking the square root there is a positive and negative solution:

#sqrt(x^2) = +-sqrt(11)#

#x = +-sqrt(11) = +-3.317# rounded to the nearest thousandth.