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

2 Answers
Apr 19, 2017

Answer:

#x^2=7# or #x=sqrt7#

Explanation:

Divide both sides by 4 (first step):

#x^2=28/4#
#x^2=7#

Take square root (of both sides):
#x=sqrt7#

This is your answer: #x=sqrt7#

Apr 19, 2017

Answer:

#x=+-sqrt7#

Explanation:

#color(blue)"Isolate " x^2" by dividing both sides by 4"#

#(cancel(4) x^2)/cancel(4)=28/4#

#rArrx^2=7#

#color(blue)"take the square root of both sides"# remembering that the square root of a number can have a positive/negative value.

#sqrt(x^2)=color(red)(+-)sqrt7#

#rArrx=color(red)(+-)sqrt7#