2
$\begingroup$

I am trying to solve $a^2+b^2$ and $a^3$ given $a+b = 2, ab=4$. I have the key with the answers $a^2+b^2=-4$ and $a^3=-8$ but am wondering which steps to take to get to that answer.

My understanding of the problem so far is that from this system of equations it is not possible to get integer solutions for $a$ and $b$ (which is not asked for). I have not really gotten further in solving it than this realization.

$\endgroup$
  • 1
    $\begingroup$ $(a+b)^2=a^2+b^2+2ab$ so $(a+b)^2-2ab=a^2+b^2$ and you know the value of both $(a+b)^2$ and $ab$, geez i wonder what could you do... $\endgroup$ – cirpis Jul 21 '14 at 12:47
  • $\begingroup$ x^2-(a+b)x+ab=0 has the roots a and b. $\endgroup$ – Bumblebee Jul 21 '14 at 13:32
9
$\begingroup$

HINT:

$a^2+b^2=(a+b)^2-2(ab)$

and

$a^3+b^3=(a+b)^3-3ab(a+b)$

EDIT: To further elongate the second equation

$a^3+b^3=-16$

Now, substitute b=4/a and solve.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ in your second hint, how do you arrive to $a^3$? $\endgroup$ – cirpis Jul 21 '14 at 12:51
  • $\begingroup$ After you get the sum, which I think will be -16, you need to back substitute b=4/a and solve. :) $\endgroup$ – MonK Jul 21 '14 at 12:55
  • $\begingroup$ the sum is -16 i think $\endgroup$ – cirpis Jul 21 '14 at 13:02
  • $\begingroup$ Yeah! sorry about that, got mixed up with my work. it is -16 $\endgroup$ – MonK Jul 21 '14 at 13:02
3
$\begingroup$

Hint: $$ (a+b)^2 = a^2 + 2ab + b^2 $$

Hint:

Two equations in two unknowns can be enough to determine two unknowns. Pick an equation and try to express one unknown by another and use the other equation.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

if all else fails, remember that you can solve for $a$ and $b$ explicitly. I get roots $1+\sqrt{3}i$ and $1-\sqrt{3}i$, and $-8 = (1+\sqrt{3}i)^3 = (1-\sqrt{3}i)^3$

| cite | improve this answer | |
$\endgroup$

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.