How do you solve $v^{2} - 14 v = - 49$ $v^2 - 14v = -49$ ?

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Answer 1 · 2015-06-27T20:52:20

First add $49$ $49$ to both sides to get:

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

Now

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

So the solution is $v = 7$ $v=7$ .

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

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

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

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

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

$= (v - 7) (v - 7)$ $= (v-7)(v-7)$

$= {(v - 7)}^{2}$ $= (v-7)^2$

So this is zero when $v = 7$ $v=7$

Perfect square trinomials are of the form:

$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2 = (a+b)^2$

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

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.

How do you solve v2−14v=−49v^2 - 14v = -49?