Relation between Roots and Coefficients Question

Question

If the roots of the equation $x^3+px^2+qx+r=0$ are consecutive terms of a geometric series, prove that $q^3 = p^3r$.

Show that this condition is satisfied for the equation $8x^3-100x^2+250x-125=0$ and solve this equation.

Answer 1

The roots can be written as $a, ab, ab^2$. Then $(x-a)(x-ab)(x-ab^2) = x^3 + px^2 + qx +r$. Multiplying the left hand side and comparing yields $p = -a(1+b+b^2), q = a^2b(1+b+b^2), r = -a^3b^3$. This shows that $q^3 = p^3r$.

Since $r = -a^3b^3$, we get $ab = -\sqrt[3]{r}$ which is one of the three roots. Thus, if $q^3 = p^3r$, then it is worth to check whether $-\sqrt[3]{r}$ is a root.

Note, however, that $q^3 = p^3r$ is not sufficient in order that the roots have the above form. Take for example $q = r = 0$ and $p \ne 0$. Then the polynomial has a zero of order two at $0$ and a zero of order one at $-p$.

For the special equation we have $(100/8)^3 125/8 = (5^2 2^2/2^3)^3 5^3/2^3 = 5^9/2^6 =5^9 2^3/2^9 = (5^3 2/2^3)^3 = (250/8)^3$.

This means that we should consider $-\sqrt[3]{r} = 5/2$. In fact, this is a root. Polynomial divison produces a quadratric equation with solutions $5(2 - \sqrt{3})/2$ and $5(2 + \sqrt{3})/2$. Hence we get $a = 5(2 - \sqrt{3})/2$ and $b = 2 + \sqrt{3}$.