Roots of polynomial equations 4

[Erfan]

The product of two of the roots of the equation ax^4 + bx^3 + cx^2 + dx + e = 0 is equal to the product of the other two roots. Prove that a*d^2 = b^2 * e

 

[anemone]

The product of two of the roots of the equation ax^4 + bx^3 + cx^2 + dx + e = 0 is equal to the product of the other two roots. Prove that a*d^2 = b^2 * e

If we let $p, q, r$ and $s$ to represents the 4 roots of the equation $ax^4 + bx^3 + cx^2 + dx + e = 0$, we're told that $pq=rs$.

We can also apply Vieta's formulas to get the following relations between a, b, e and d:

$p+q+r+s=-\frac{b}{a}$---(1)

$pqr+pqs+prs+qrs=-\frac{d}{a}$---(2)

$pqrs=\frac{e}{a}$---(3)


Simplifying (2) yields

$pq(r+s)+rs(p+q)=-\frac{d}{a}$

$rs(r+s)+rs(p+q)=-\frac{d}{a}$ since $pq=rs$

$rs(r+s+p+q)=-\frac{d}{a}$

$rs(-\frac{b}{a})=-\frac{d}{a}$ since $p+q+r+s=-\frac{b}{a}$---(1)

$rs=-\frac{d}{b}$

Substituting $pq=rs$ and $rs=-\frac{d}{b}$ into equation (3) gives

$pqrs=\frac{e}{a}$

$(rs)rs=\frac{e}{a}\rightarrow\;\;(rs)^2=\frac{e}{a}$.

Hence, $(-\frac{d}{b})^2=\frac{e}{a}$ and this gives $ad^2=eb^2$. So we're done.