How do you solve #a^2 - 4a = 32#?

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Answer 1 · 2018-04-16T10:00:56

#a=8, a=-4#

Minus #32#

#-> a^2-4a-32#

List the factors of #32#:

#1# and #32#
#2# and #16#
#4# and #8#

As you can see, we can make #-4# from #8# and #4#

#-> (a-8)(a+4)#

Solve:

#a-8=0#

#a=8#

#a+4=0#

#a=-4#

Therefore #a=8, a=-4#

We can always check the answer:

#a=8#

#8^2-(4xx8)=64-32=32#

#a=-4#

#(-4)^2-(-4xx4)=16+16=32#

Hence #a=8, a=-4# is correct