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 $D\ge 0$


$\begin{aligned} D &=4(2a+2c-b)^2-4(-3)(b^2-4ac)\\ &=16(a^2+b^2+c^2-ab-bc-ca)\\ &=8\left[(a-b)^2+(b-c)^2+(c-a)^2\right]\ge 0\\ \end{aligned}$

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

  • 1
    $\begingroup$ I was hoping for a solution that would use the fact that the determinant of $ax^2+bx+c$ is $b^2-4ac$, because that is what we get when we put $x=0$ in $(b-x)^2-4(a-x)(c-x)$ and we are also given $a+b+c=0$ which means that $ax^2+bx+c$ has one of the roots as $1$ hence both the roots would be rational. Anyways, I still liked how you used $a+b+c=0$ to get at $a^2-bc = b^2-ac = c^2-ab$ $\endgroup$
    – user350331
    Nov 14 '16 at 10:01
  • $\begingroup$ Approach 2 should be better $\endgroup$ Nov 14 '16 at 10:24
  • 1
    $\begingroup$ I just came up with the way to complete your solution (Approach 2). For $f(x)=(b-x)^2-4(a-x)(c-x)$ to have real roots $$f(x)=0\implies (x-b)^2=4(x-a)(x-c)$$. As we have already proved that the roots are real so we just need to find the condition for which the roots are equal. $$\\$$ Now, as $(x-b)^2$ touched the $x-$ axis at $x=b$ and $4(x-a)(x-c)$ at $x=a,c$, so for equal roots the graph of both the quadratic expressions $(b-x)^2$ and $4(x-a)(x-c)$should touch each other that would only happen when $a=b=c$ $\endgroup$
    – user350331
    Nov 14 '16 at 11:43
  • $\begingroup$ Why would that be? Isn't the question asking for equal roots of $(b-x)^2-4(a-x)(c-x)$ $\endgroup$
    – user350331
    Nov 14 '16 at 12:40
  • 1
    $\begingroup$ That gives $a=b=c=0$ $\endgroup$ Nov 14 '16 at 16:50

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