How do you solve #5x^2 - 125 = 0#?

2 Answers
Mar 13, 2018

Answer:

Start by factorizing.

Explanation:

Start by factorizing

#5(x^2 - 25) = 0#

then you can divide both sides by 5 and since #0/5 =0# you get the expression:

#x^2 - 25 = 0#

rearrange so that 25 is on the right hand side

#x^2 =25#

#x=± sqrt25#

#x=± 5#

Mar 13, 2018

Answer:

#x = +- 5#

Explanation:

To make progress, an attempt should be made to have just #x# on one side of the equation to see what it is equal to (on the other side of the equation).

By inspection, both #5# and #-125# are divisible by #5#. Zero is also divisible by #5# in the sense #0/5 = 0#

So, dividing both sides of the equation by #5# (also called "dividing through by #5#)

#5x^2 - 125 = 0#

implies

#(5x^2)/5 - (125)/5 = 0/5#

that is

#x^2 - 25 = 0#

Now #25# may be added to both sides to give

#x^2 - 25 + 25 = 0 + 25#

that is

#x^2 = 25#

You will recognise #25# as a perfect square so finding a solution should be easy but take care! Remember square numbers have two roots, a positive one and a negative one. So

#x^2 = 25#

implies

#sqrt(x^2) = sqrt(25)#

that is

#x = 5#
or
#x = -5#