How do you factor and solve 2x^2 - 3= 1252x23=125?

2 Answers
Apr 1, 2018

x=+-8x=±8

Explanation:

2x^2-3=1252x23=125

Subtract 125 on both sides

2x^2-1282x2128=0

Divide both sides by 2

x^2-64=0x264=0

Using a^2-b^2=(a+b)(a-b)a2b2=(a+b)(ab)

x^2-64=(x+8)(x-8)x264=(x+8)(x8)

So (x+8)(x-8)=0(x+8)(x8)=0

x=+-8x=±8

Apr 1, 2018

2x^2-3=1252x23=125 can be factored to:

2(x-8)(x+8)=02(x8)(x+8)=0, and has the solution:

color(red)(absx=8)|x|=8

Explanation:

Move all the terms to one side of the equation

2x^2-3=1252x23=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