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

1 Answer
May 22, 2018

Answer:

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

Explanation:

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#