Question

If the sum of the roots of an equation is $2$ and sum of their cubes is $98$, then find the equation.

Answer 1

Let the roots be $a$ and $b$.

It is given that the sum of the roots of an equation is $2$ and the sum of their cubes is $98$, therefore, we have:

$a+b=2$, and

$a^3+b^3=98\Rightarrow (a+b{)}^3-3ab(a+b)=98(\because (a+b{)}^3=a^3+b^3+3ab(a+b\left)\right)\Rightarrow (2{)}^3-3ab(2)=98(\because a+b=2)\Rightarrow 8-6ab=98\Rightarrow -6ab=98-8\Rightarrow -6ab=90\Rightarrow ab=-\frac{90}{6}\Rightarrow ab=-15$

Therefore, the sum of the equation is $a+b=2$ and the product is $ab=-15$ and thus the equation is:

$x^2-(a+b)x+ab=0\Rightarrow x^2-2x-15=0$

Hence, the equation is $x^2-2x-15=0$.