How do you solve #(x - 7)^{2} - 3( x - 7) - 28= 0#?

1 Answer
Nghi N.
Sep 15, 2016

3 and 14

Explanation:

Call (x - 7) = X, we get a quadratic equation in X:
#X^2 - 3X - 28 = 0#.
Roots have opposite signs because ac < 0.
Compose factor pairs of (c = - 28) --> (-2, 14)(-4, 7). This sum is
(7 - 4 = 3 = -b). Then, the 2 real roots are: X = -4 and X = 7
Next, solve for x:
a. X = (x - 7) = - 4 --> #x = 7 - 4 = 3#
b. X = (x - 7) = 7 --> #x = 7 + 7 = 14#.

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