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

Jun 27, 2015

First add $49$ to both sides to get:

${v}^{2} - 14 v + 49 = 0$

Now

${v}^{2} - 14 v + 49 = {\left(v - 7\right)}^{2}$

So the solution is $v = 7$.

#### Explanation:

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

${v}^{2} - 14 v + 49 = 0$

Notice that $49 = {7}^{2}$ and $14 = 2 \cdot 7$, so this is a perfect square trinomial:

${v}^{2} - 14 v + 49$

$= {v}^{2} - \left(2 \cdot v \cdot 7\right) + {7}^{2}$

$= \left(v - 7\right) \left(v - 7\right)$

$= {\left(v - 7\right)}^{2}$

So this is zero when $v = 7$

Perfect square trinomials are of the form:

${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

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.