Question

Find the condition that $x^3-p{x}^2+qx-r=0$ may here
(1) two roots equal but of opposite sign;
(2) the roots in geometrical progression.

Answer 1

Given, $x^3-p{x}^2+qx-r=0$

Case 1: Roots are $\alpha ,-\alpha ,\beta$

Sum of the roots, $S_1=\alpha -\alpha +\beta =p⟹\beta =p$

Sum of the roots taken 2 at a time, $S_2=-{\alpha }^2+\alpha \beta -\alpha \beta =q⟹-{\alpha }^2=q$

Product of the roots, $S_3 = -\alpha^2\beta = r

From the above equations we can see that $S_1\cdot S_3=S_2⟹pq=r$

Case 2: Roots are in geometric progression. Let the roots be $\alpha ,\beta ,\gamma$

$\therefore \frac{\alpha }{\beta }=\frac{\beta }{\gamma }$

$⟹{\beta }^2=\alpha \gamma ⟹{\beta }^3=\alpha \beta \gamma =r$

Since, $\beta$ is the roots of the given equation, we have

${\beta }^3-p{\beta }^2+q\beta -r=0⟹r–p\cdot r^{2/3}+q\cdot r^{1/3}–r=0$

$⟹p^{3}r=q^3\left[\text{taking cubes on both sides and simplifying}\right]$