How do you solve #x^2 + 6x - 7 = 0#?

1 Answer
Jul 7, 2015

Answer:

Notice #x=1# is a root since the sum of the coefficients is #0#.
Then find the other root #x=-7#

Explanation:

Let #f(x) = x^2+6x-7#

First notice that the sum of the coefficients is zero: #1 + 6 - 7 = 0#
so #f(1) = 0#, #1# is a root and #(x-1)# is a factor of #f(x)#.

In order to get the #x^2# leading term and #-7# constant term, the other factor must be #(x+7)#, giving the other root as #x = -7#

#x^2+6x-7 = (x-1)(x+7)#