How do you solve $x^2+14x + 49= 19$ ?

Question

2057 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-22T07:55:03

$x^2+14x+49=19$ has the two solutions $x=-7+-\sqrt19$
(i.e. $x_1=-7-\sqrt19$ and $x_2=-7+\sqrt19$ )

The first I notice is that $14=2*7$ , and $49=7^2$

Please also remember that $(x+a)^2=x^2+2a+a^2$
Therefore $x^2+14x+49=(x+7)^2=19$

It follows, therefore, that
$x+7=+-\sqrt19$

We, therefore, have two values of x here:
$x=-7+-\sqrt19$

How do you solve x^2+14x + 49= 19x^2+14x + 49= 19?