Question

if p, q be the roots of 2x^2-4x-1=0 find the value of p^2/q+q^2/p​

Answer 1

$\large\underline{\text{Question}}$

If $p$ and $q$ are the roots of $2x^{2}-4x-1=0$, find the value of $\dfrac{p^{2}}{q}+\dfrac{q^{2}}{p}$.

$\large\underline{\text{Important Facts}}$

$\red{\bigstar}$ Vieta's formula.

$\rightarrow$By Vieta's formula(relation between roots and coefficients), the zeros, $\alpha$ and $\beta$ of the quadratic equation $ax^{2}+bx+c=0$ has relations as

$\implies\begin{cases} & \alpha+\beta =-\dfrac{b}{a}\\ & \alpha\beta=\dfrac{c}{a} \end{cases}$

$\large\underline{\text{Solution}}$

By Vieta's formula

$\implies\begin{cases} & p+q=2 \\ & pq=-\dfrac{1}{2} \end{cases}$

We need the value of $p^{3}+q^{3}$.

$\implies p^{3}+q^{3}=(p+q)(p^{2}-pq+q^{2})$

The second factor can be written as

$\implies p^{2}-pq+q^{2}=(p+q)^{2}-3pq$

So

$\implies p^{2}-pq+q^{2}=2^{2}+\dfrac{3}{2}$

$\implies p^{2}-pq+q^{2}=\dfrac{11}{2}$

Hence

$\implies p^{3}+q^{3}=2\times\dfrac{11}{2}$

$\implies p^{3}+q^{3}=11$

The value needed to be found is

$\implies\dfrac{p^{3}+q^{3}}{pq}=\dfrac{11}{-\dfrac{1}{2}}=-22$

$\large\underline{\text{Answer}}$

The answer is $-22$.