Question Statement:-
If $a+b+c=0$ and $a,b,c$ are real, then prove that equation $(b-x)^2-4(a-x)(c-x)=0$ has real roots and the roots will not be equal unless $a=b=c$.
Solution with a standard approach:-
$(b-x)^2-4(a-x)(c-x)=0\implies -3x^2+2(2a+2c-b)x+(b^2-4ac)=0$
Now, as the roots of the quadratic equation need to be proved to be real, so we need to prove that $D\ge 0$
So,
$\begin{aligned} D &=4(2a+2c-b)^2-4(-3)(b^2-4ac)\\ &=16(a^2+b^2+c^2-ab-bc-ca)\\ &=8\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\ge 0\\ \end{aligned}$
As $D\ge0$, so the roots of the quadratic equation $(b-x)^2-4(a-x)(c-x)=0$ are real
For the roots to be equal $D=0$, which can only happen when $a=b=c$.
Attempt at a solution with a not so standard approach:-
On putting $x=0$ in the quadratic equation $(b-x)^2-4(a-x)(c-x)=0$ we get $b^2-4ac=0$. This leads us to conclude that the the equation the quadratic equation $ay^2+by+c=0$ has equal roots because $D=0$.
Now, as we already know that $a+b+c=0$, so this leads us to conclude that the quadratic equation $ay^2+by+c=0$ has equal roots which are both $1$.
From this we get that $b^2=4ac\implies b=\pm2\sqrt{ac} \qquad(ac\gt0)$.
Also, as the root is unity and $D=0$, hence $\dfrac{-b}{2a}=1\implies b^2=4a^2\implies 4ac=4a^2\implies 4a(c-a)=0\implies a=c,0$
Case 1:-
If $a=0$, then as $b^2-4ac=0\implies b^2=0\implies b=0$. Now, from the relation $a+b+c=0$, we get $a=b=c=0$
On putting $a=b=c=0$ in the quadratic equation $(b-x)^2-4(a-x)(c-x)=0$, we get $x^2=0$, which leaves us with two equal roots $0$
Case 2:-
If $a=c$, then form the relation $a+b+c=0$, we get ${b}=-2a=-2c$, which on putting in the quadratic equation $(b-x)^2-4(a-x)(c-x)=0$, we get $x=0,4a$. As, both the roots are not same so this is not the required condition.
My deal with the question:-
As can be seen in the "standard approach" solution, there was no need to use the relation $a+b+c=0$, so I thought that there must be a reason why the relation $a+b+c=0$ is given and hence tried finding an intuitive solution rather than a brute force one. I strongly feel that there must be a reason why the relation was given and hence attempted with a "not so standard approach".
I know my "not so standard approach solution" seems like a forced solution so please can you point out some mistakes or provide me with a "fluid solution".
We consider $f(x) = (b-x)^2 -4(a-x)(c-x)$
On taking derivative, we get $$f'(x) = -2(b-x) + 4(a-x) + 4(c-x)$$, which evaluates to $$f'(x) = -6x-6b = -6(x+b)$$
Hence, maximum value of $f(x)$ occurs at $x=-b$ ,which is
$$f(-b) = 4b^2 - 4(a+b)(c+b) = 4b^2 - 4ac = 4(b^2-ac)$$
Now, for $a+b+c = 0$ $$a^2-bc = b^2-ac = c^2-ab = k$$
$$\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2)) = 3k \ge 0$$
Hence, $k \ge 0$, and a quadratic with maximum value greater than 0 and coefficent of $x$ negative has at least one real root. Equality is trivial to check
Approach 2
Given $a+b+c=0$, hence the quadratic $ax^2+bx+c=0$ has a root $1$ . This mean that both the roots are real , which guarantees that $D \ge 0$
That means that $f(0) \ge 0$, and since $f(x)$ is a quadratic with leading coefficient negative, this would imply at least one real root