How do you solve #v- 3= \sqrt { - 2v + 21}#?

1 Answer
George C.
Dec 13, 2016

#v=6#

Explanation:

Given:

#v-3 = sqrt(-2v+21)#

First square both sides, noting that this may introduce spurious solutions:

#v^2-6v+9 = -2v+21#

Next add #2v-21# to both sides to get:

#v^2-4v-12 = 0#

We can solve this by finding a pair of factors of #12# with difference #4#. The pair #6, 2# works, so we find:

#0 = v^2-4v-12 = (v-6)(v+2)#

So #v=6# or #v=-2#

The value #v=-2# is not a solution of the original problem, since it results in #v-3 = -5 < 0#, which is the negative square root of #25# not the positive one required.

The value #v=6# is a solution of the original equation, since we find:

#6-3 = 3 = sqrt(9) = sqrt(-12+21) = sqrt(-2(6)+21)#

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