Question

If the equation $x^4+a{x}^3+b{x}^2+cx+d=0$ has three equal roots, show that each of them is equal to $\frac{6c-ab}{3a^2-8b}$.

Answer 1

Three roots of the given equation are same hence, the roots can be assumed as $\alpha ,\alpha ,\alpha ,\beta$

Here, $S_1=3\alpha +\beta =-a;S_2=3\alpha (\alpha +\beta )=b;S_3={\alpha }^2(\alpha +3\beta )=-c;S_4={\alpha }^3\beta =d$

We need to evaluate the value of $\frac{6c-ab}{3a^2-8b}$

$6c-ab=\alpha (3{\alpha }^2-6\alpha \beta +3{\beta }^2)$

$3a^2-8b=3{\alpha }^2-6\alpha \beta +3{\beta }^2$

$\therefore \frac{6c-ab}{3a^2-8b}=\frac{\alpha (3{\alpha }^2-6\alpha \beta +3{\beta }^2)}{3{\alpha }^2-6\alpha \beta +3{\beta }^2}=\alpha$, which is the common root of the given equation