How do you factor and solve #2x^2 - 3= 125#?

2 Answers
Anish M.
Apr 1, 2018

#x=+-8#

Explanation:

#2x^2-3=125#

Subtract 125 on both sides

#2x^2-128#=0

Divide both sides by 2

#x^2-64=0#

Using #a^2-b^2=(a+b)(a-b)#

#x^2-64=(x+8)(x-8)#

So #(x+8)(x-8)=0#

#x=+-8#

Sharkbasket
Apr 1, 2018

#2x^2-3=125# can be factored to:

#2(x-8)(x+8)=0#, and has the solution:

#color(red)(absx=8)#

Explanation:

Move all the terms to one side of the equation

#2x^2-3=125#

#2x^2-3-color(red)125=cancel125-cancelcolor(red)125#

#2x^2-128=0#

Now take out a factor of 2

#(color(red)2*x^2)-(color(red)2*64)=0#

#color(red)2(x^2-64)=0#

We now have a term in the parentheses that looks like

#(a^2-b^2)#

This is called a difference of squares

We can factor a difference of squares like this:

#(a^2-b^2)=(a-b)(a+b)#

Let's apply this to our expression

#2(x^2-color(red)64)=0#

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

#2(x-8)(x+8)=0#

This is the fully factored form.

By examining this equation, we can see that the solutions — the values of #x# that make the equation true — are

#x=8#

and

#x=-8#

or simply

#absx=8#

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