How do you solve #v^2 - 14v = -49#?

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Answer 1 · 2015-06-27T20:52:20

First add #49# to both sides to get:

#v^2-14v+49 = 0#

Now

#v^2-14v+49 = (v-7)^2#

So the solution is #v=7#.

Having added #49# to both sides of the equation we have

#v^2-14v+49 = 0#

Notice that #49 = 7^2# and #14 = 2 * 7#, so this is a perfect square trinomial:

#v^2-14v+49#

#= v^2-(2*v*7)+7^2#

#= (v-7)(v-7)#

#= (v-7)^2#

So this is zero when #v=7#

Perfect square trinomials are of the form:

#a^2+2ab+b^2 = (a+b)^2#

In our case #a=v# and #b = -7#

If you see a quadratic with first and last terms being square, check the middle term to see if it's a perfect square trinomial.