Question

If alpha and beta are roots of the equation x²-px+q=0 and alpha>0, beta >0 then find the value of alpha raise to the power 5 and beta raise to the power 5.​

Answer 1

Hint.

$(x+y)^2=x^2+y^2+2xy$

$(x+y)^3=x^3+y^3+3xy(x+y)$

$(x^2+y^2)(x^3+y^3)=x^5+y^5+x^2y^2(x+y)$

In this type of question, we find the values stepwise, and approach to the conclusion. We are given to use the identity of polynomials.

Solution.

Let's recall the sum and product of solutions.

$x^2-px+q=0$ and $x=\alpha ,\beta$

$\implies \begin{cases} & \alpha +\beta =p \\ & \alpha \beta =q \end{cases}$

Squaring both sides of $\alpha +\beta =p$ we get $\alpha ^2+\beta ^2+2\alpha \beta =p^2$, so $\alpha ^2+\beta ^2=p^2-2q$. $...(1)$

Cubing both sides of $\alpha +\beta =p$ we get $\alpha ^3+\beta ^3+3\alpha \beta (\alpha +\beta )=p^3$, so $\alpha ^3+\beta ^3=p^3-3pq$. $...(2)$

Multiplying both equations, we have $\alpha ^5+\beta ^5+\alpha ^2\beta ^2(\alpha +\beta )=(p^2-2q)(p^3-3pq)$, so $\alpha ^5+\beta ^5+\alpha ^2\beta ^2(\alpha +\beta )=p^5-3p^3q-2p^3q+6pq^2$.

Simplifying further, we arrive at the conclusion that $\alpha ^5+\beta ^5=(p^5-5p^3q+6pq^2)-pq^2=\boxed{p^5-5p^3q+5pq^2}$.

Answer 2

Given :-

If alpha and beta are roots of the equation x²-px+q=0 and alpha>0, beta >0

To Find :-

$\sf \alpha ^5+\beta ^5$

Solution :-

At first here

Sum of zeroes = α + β = -b/a

= -(-p)/1

= p/1

= p

Product of zeroes = αβ = c/a

= q/1

= q

Now

$\sf (\alpha +\beta ) = p$

$\sf (\alpha ^2 + \beta ^2) = p^2$

$\sf \alpha ^2+\beta ^2+2\alpha \beta =p^2$

$\sf (\alpha +\beta )^2 + 2\alpha \beta = p^2$

$\sf (p)^2 - 2(q)$

$\sf p^2-2q = p^2$

$\sf p^2-2q = \alpha ^2+\beta ^2$

Now

$\sf (\alpha +\beta )=p$

$\sf (\alpha +\beta )^3=p^3$

$\sf (\alpha+\beta) ^3+3\alpha \beta (\alpha +\beta )=p^3$

$\sf (p)^3 + 3(q)(p)=p^3$

$\sf p^3 + 3qp= \alpha ^3+\beta ^3$

$\sf \alpha ^5+\beta ^5 = (p^2-2q)(p^3 +3pq)$

$\sf \alpha ^5+\beta ^5 = p^5 - 2qp^3 -6pq^2$