Question 1148967
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The equation is satisfied if...<br>
(1) (x^2+7x+12) = 0  (any number to the 0 power is 1)<br>
(2) (x^2 +3x+1) = 1  (1 raised to any power is 1)<br>
(3) (x^2+3x+1) = -1 AND (x^2+7x+12) is even  (-1 raised to an even power is 1)<br>
(1) {{{x^2+7x+12 = (x+3)(x+4) = 0}}}
x = -3 and x = -4 are solutions.<br>
(2) {{{x^2+3x+1 = 1}}}
{{{x^2+3x = 0}}}
{{{x(x+3) = 0}}}
x = -3 and x = 0 are solutions (x = -3 is already known to be a solution).<br>
(3) {{{x^2+3x+1 = -1}}}
{{{x^2+3x+2 = 0}}}
{{{(x+1)(x+2) = 0}}}
x = -1 and x = -2 are solutions IF they make x^2+7x+12 = (x+3)(x+4) even.<br>
Note that, for both x = -1 and x = -2, (x+3)(x+4) are consecutive integers.  One of them has to be even and one odd; so their product is even.<br<
Therefore, both x = -1 and x = -2 are solutions.<br>
ANSWER: The solutions are 0, -1 -2, -3, and -4.<br>