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

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Answer 1 · 2018-04-30T08:48:07

#x=-9/4,9/4#

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#

Answer 2 · 2018-08-10T02:38:49

#x=pm9/4#

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!