How do you solve #x=sqrt(12x-32)#?

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Answer 1 · 2015-04-09T17:17:32

Square both sides of the given equation and then convert into a standard quadratic form and factor.

#x=sqrt(12x-32)#

#x^2 = 12x-32#

#x^2 -12x +32= 0#

#(x-4)(x-8) = 0#

#x=4#
or
#x=8#

Answer 2 · 2018-08-05T01:46:47

#x=4# and #x=8#

To get rid of the radical on the right, we can square both sides to get

#x^2=12x-32#

This is beginning to look like a quadratic, but to make it more obvious, we can subtract #12x# and add #32# to both sides to get

#x^2-12x+32=0#

To factor this, let's do a little thought experiment:

What two numbers sum up to #-12# (middle term) and have a product of the last term (#32#)?

After some trial and error, we arrive at #-8# and #-4#. This allows us to factor this as

#(x-8)(x-4)=0#

Setting both factors equal to zero, we get

#x=4# and #x=8#

Hope this helps!