How do you solve #x^2 - 21 = 3(5 - x^2)#?

1 Answer
Mar 28, 2018

Answer:

#x=+- 3#

Explanation:

We want to expand the brackets firstly:

#3 xx 5=15#

#3xx-x^2=-3x^2#

This, therefore:

#x^2-21=3(5-x^2) -> x^2-21=15-3x^2#

As we want to isolate the #x# to one side, we do not want #-3x^2#, and therefore we do the opposite which is to #+3x^2#. Notice that these cancel out. Also, remember what we do to one side we must do to another.

#x^2-21=15-3x^2 -> 4x^2-21=15#

We do not want #-21#, as we would like to get #x# on its own, and therefore do the opposite to #-21# which is to #+21#, notice the #-21+21# cancels out. Also, remember what you do to one side you MUST do to another.

#4x^2-21=15 -> 4x^2=36#

As we would like the value of #x#, first we realise we have #4# lots of #x# and divide both sides by #4#

#4x^2=36 -> x^2=9#

As we would like #x#, we need to #sqrt# as that it the opposite to #^2#, notice the #^2# cancels out. Also, remember to do this to both sides. We must also remember to take both square roots as they both satisfy the original equation

#x^2=9 -> x=+-sqrt9=+-3#

#therefore# #x=+-3#