How do you solve 16x^2 - 81 = 0 by factoring?

2 Answers
Apr 30, 2018

x=-9/4,9/4

Explanation:

Use the rule for difference of squares.
16x^2-81=0
(4x-9)(4x+9)=0

This equation will be true if either (4x-9) or (4x+9) is 0.
4x+9=0
4x=-9
x=-9/4
Or
4x-9=0
4x=9
x=9/4

x=-9/4,9/4

Aug 10, 2018

x=pm9/4

Explanation:

Recall that this is a difference of squares which factors as

bar ul(|color(white)(2/2)a^2-b^2=(a+b)(a-b)color(white)(2/2)|

Both of our terms are perfect squares, where our a=4x and b=9. This allows us to factor this as

(4x+9)(4x-9)=0

We can set both factors equal to zero to get

4x+9=0=>4x=-9=>x=-9/4 and

4x-9=0=>4x=9=>x=9/4

Hope this helps!