Question

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

Answer 1

Hint:In this question sum of roots and their cubes are given. So by putting the formula of cube roots we find the value of the product of roots. After finding out the product and sum of roots we put in the standard form of a quadratic equation and find out the desirable quadratic equation.

Formula used:
We solve the question by using the formula:-
${{(m+n)}^{3}}={{m}^{3}}+{{n}^{3}}+3mn(m+n)$

Complete step by step Solution:
Given that sum of the roots of an equation is 2 and the sum of their cubes is 98
We have to find the quadratic equation.
First, we suppose that the roots of the quadratic equations are m and n
The sum of the roots = 2
That is m + n = 2
And the sum of their cubes :- ${{m}^{3}}+{{n}^{3}}=98$
We know the formula of ${{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$
By putting a = m and b = n, we get
${{(m+n)}^{3}}={{m}^{3}}+{{n}^{3}}+3mn(m+n)$
Now we put the values of m + n and $m{}^{3}+{{n}^{3}}$, we get
${{(2)}^{3}}=98+3mn(2)$
Solving the above equation, we get 8 = 98 + 6 mn
That is 6 mn = 8 – 98 = -90
And mn = -15
We know that for a quadratic equation , $a{{x}^{2}}+bx+c=0$
That is sum of its roots = $-\dfrac{b}{a}$= 2
As the value of a = 1
That is b = -2
And the product of its roots = $\dfrac{c}{a}=-15$
The value of c = -15
Then the quadratic equation is ${{x}^{2}}-2x-15=0$

Note:Students must know that when the sum of roots and the product of roots are given or we find out both values then we find out the quadratic equation with the help of ${{x}^{2}}$- (sum of roots)x + product of roots = 0