Question

Show that the equation $x^4+p{x}^3+q{x}^2+rx+s=0$ may be solved as a quadratic if $r^2=p^{2}s$.

Answer 1

Let $r^2=p^{2}r;s=\frac{r^2}{p^2}$

We have

$x^4+p{x}^3+q{x}^2+rx+s$

$=x^4+p{x}^3+q{x}^2+rx+\frac{r^2}{p^2}$

$={(x^2+\frac{p}{2}x+\frac{r}{p})}^2-(\frac{p^2}{4}+\frac{2r}{p}-r)x^2=0$

${(x^2+\frac{p}{2}x+\frac{r}{p})}^2=a^2{x}^2$ (say)

thus $x^2+\frac{px}{2}x+\frac{r}{p}=\pm ax$

Thus the given equation can be solved as a quadratic equation.