I'm having trouble doing the following question, was wondering if anyone was able to lend a hand, would be greatly appreciated as i'm not too sure where to start or how to go about this.
The general quartic equation is
$$ ax^4 + bx^3 + cx^2 + dx + f = 0, $$
where $a,b,c,d$ and $f$ are real numbers and $a>0$. We suppose the coefficients obey the extra condition that
$$ d = \frac{b(4ac-b^2)}{8a^2}. $$
(c) Expand and simplify $$ \left( \sqrt a\,x^2 + \frac{b}{2\sqrt a}x + \frac{4ac-b^2}{8a\sqrt a} \right)^2.$$
Start with
$$ \left( \sqrt a\,x^2 + \frac{b}{2\sqrt a}x + \frac{4ac-b^2}{8a\sqrt a} \right)^2.$$
and expand. Then, the coefficient on $x^4$ is
$(\sqrt a)(\sqrt a)=a$
The coefficient on $x^3$ is
$(\sqrt a)\left(\frac{b}{2\sqrt a}\right)+(\sqrt a)\left(\frac{b}{2\sqrt a}\right)=b$
The coefficient on $x^2$ is
$(\sqrt a)\left(\frac{4ac-b^2}{8a\sqrt a}\right)+(\sqrt a)\left(\frac{4ac-b^2}{8a\sqrt a}\right)+\left(\frac{b}{2\sqrt a}\right)\left(\frac{b}{2\sqrt a}\right)=c$
The coefficient on $x^1$ is
$\left(\frac{b}{2\sqrt a}\right)\left(\frac{4ac-b^2}{8a\sqrt a}\right)+\left(\frac{b}{2\sqrt a}\right)\left(\frac{4ac-b^2}{8a\sqrt a}\right)=\frac{b(4ac-b^2)}{8a^2}=d$
by assumption.
The coefficient on $x^0$ is
$\left( \frac{4ac-b^2}{8a\sqrt a} \right)\left( \frac{4ac-b^2}{8a\sqrt a} \right)=a\left(\frac{d}{b}\right)^2$
which does not match, in general, the coefficient $f$. If we impose the second condition that $f=a\left(\frac{d}{b}\right)^2$, then the quartic can be factored as suggested.