Question

Show that the roots of the equation
$(x-a)(x-b)(x-c)-f^2(x-a)-g^2(x-b)-h^2(x-c)+2fgh=0$ are all real.

Answer 1

From the given equation, we have
$(x-a)\left\{\right(x-b\left)\right(x-c)-f^2\}-\left\{g^2\right(x-b)-h^2(x-c)-2fgh\}=0$
Let $p,q$ be the roots of the quadratic
$(x-b)(x-c)-f^2=0$,
and suppose $p$ to be not less than $q$.

By solving the quadratic formula, we have
$2x=b+c\pm \sqrt{(b-c{)}^2+4f^2}....\left(1\right)$;
Now the value of the surd is greater than $b\sim c$, so that $p$ is greater than $b$ or $c$, and $q$ is less than $b$ or $c$.
In the given equation substitute for $x$ successively the values
$+\infty ,p,q,-\infty$;
The results are respectively
$+\infty ,-(g\sqrt{p-b}-h\sqrt{p-c}{)}^2,+(g\sqrt{b-q}-h\sqrt{c-q}{)}^2,-\infty$,
Since $(p-b)(p-c)=f^2=(b-q)(c-q)$.
Thus, the given equation has three real roots, one greater than $p$, one between $p$ and $q$, and one less than $q$.
If $p=q$, then from $\left(1\right)$ we have $(b-c{)}^2+4f^2=0$ and therefore $b=c,f=0$.
In this case the given equation becomes
$(x-b)\left\{\right(x-a\left)\right(x-b)-g^2-h^2\}=0$;
thus the roots are all real.
If $p$ is a root of the given equation, the above investigation fails; for it only shows that there is one root between $q$ and $+\infty$, namely $p$. But as before, there is a second real root less than $q$; hence the third root must also be real. Similarly if $q$ is a root of the given equation we can show that all the roots are real.