Find the quadratic equation from given relatioship the quadratic equation whose roots are a and b where $a^2 +b^2=5$ and
$3(a^5+b^5)=11(a^3+b^3)$
What I Tried
$a^2 +b^2=5$  $(a+b)^2-2ab=5$  $(\text{sum of roots})^2 -2(\text{products of roots})=5$

$3(a^5+b^5)=11(a^3+b^3)$  $a^3(3a^2-11)=b^3(11-3b^2)$

Hint: If $a+b = 0$, then you have $a = \pm \sqrt{\frac{5}{2}}, b = \mp \sqrt{\frac{5}{2}}$. From this you can find the equation. If $a + b \neq 0\implies 3(a^4-a^3b+a^2b^2-ab^3+b^4)=11(a^2-ab+b^2)= 11(5-ab)$. Using $a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2 = 25 - 2(ab)^2, a^3b + ab^3 = ab(a^2+b^2) = 5ab$. Thus you do have a quadratic equation in $ab$, and once you solve for $ab$, then you can solve for $a,b$ and you are done !
