What is the sum of all possible solutions to this equations?

$|x+4|^2 -10|x+4|=24$

My attempt:

Since $(x+4)^2=|x+4|^2$, so I can ignore the absolute sign of the first term. So we only need to deal with the absolute sign of the second term. This become two cases:

1) $(x+4)^2-10(x+4)=24$

2) $(x+4)^2+10(x+4)=24$

But it turns out my answer is wrong. I am not sure where I did wrong. The correct way is to do a substitution that $u=|x+4|$, then it would become $u^2-10u=24$. But where did I do wrong? I didn't see it.


The first equation above is only valid when $x+4 \ge 0$, and we must disregard any solution that doesn't lie in this range, and similarly for the second equation and $x+4 < 0$.

$(1): (x - 1)^2 = 7^2 \iff x = 8 \text{ or } x = -6$, of which only the first solution is valid, and

$(2): (x + 9)^2 = 7^2 \iff x = -2 \text{ or } x = -16$, and only the second solution is valid.

Note that something similar happens for the equation in $u$. The solutions are $u = -2$ or $u = 12$, and the first case cannot happen because an absolute value cannot be negative. This gives you the same roots.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.