5
$\begingroup$

If $a^4+64=0$ and $a^2 \ne-4a-8$, what is the value of $a^2-4a$?

I tried to add $-16a^2$ to both sides and doing some algebra but it didnt help much.

$\endgroup$
5
  • 1
    $\begingroup$ If $a^4=-64$, then $a^2=\pm8i$. What is $(\pm8i)!$ supposed to mean? The extension per $\Gamma$ function? $\endgroup$ – Hagen von Eitzen Jul 26 '13 at 21:55
  • $\begingroup$ This is a high school question,I think it is supposed to be simple. $\endgroup$ – guest Jul 26 '13 at 21:58
  • $\begingroup$ By "$!=$", do you mean $\ne$? $\endgroup$ – 6005 Jul 26 '13 at 21:58
  • $\begingroup$ @HagenvonEitzen: I believe $!=$ is supposed to be $\neq$. OP, please confirm. $\endgroup$ – Ross Millikan Jul 26 '13 at 21:58
  • $\begingroup$ Yes it means "not equal".Sorry for that. $\endgroup$ – guest Jul 26 '13 at 21:59
10
$\begingroup$

Note that $a^4+64=(a^2+4a+8)(a^2-4a+8)$. So if one of the items on the right is not $0$, then $\dots$

$\endgroup$
8
  • $\begingroup$ How could you seperate it like that? Yes the answer is -8,thank you. $\endgroup$ – guest Jul 26 '13 at 22:01
  • $\begingroup$ Great!Thanks for the quick solution. $\endgroup$ – guest Jul 26 '13 at 22:04
  • 1
    $\begingroup$ You are welcome. You can check the factorization by multiplying out. You can maybe discover it by $a^4+64=(a^2+8)^2 -(4a)^2$, difference of squares. Your adding $16a^2$ to both sides was a great idea, would have worked. $\endgroup$ – André Nicolas Jul 26 '13 at 22:05
  • $\begingroup$ Super @Andre Nicolas . Cannot be quite without congratulating on the answer . $\endgroup$ – Harish Kayarohanam Jul 26 '13 at 22:07
  • $\begingroup$ How can I mark the question as solved? $\endgroup$ – guest Jul 26 '13 at 22:08
4
$\begingroup$

Here's how to find the factorization, it is very similar to completing the square, we have $$\begin{align*}a^4 + 64 &= a^4 + 16a^2 + 64 - 16a^2\\ &= (a^2 + 8)^2 - 16a^2 \\ &= (a^2 - 4a + 8)(a^2 + 4a + 8)\end{align*}$$

See also http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity

$\endgroup$
1
  • $\begingroup$ Thanks for all answers and comments.I spent more than 20 minutes on this question.I think I should sleep :) $\endgroup$ – guest Jul 26 '13 at 22:10

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.