If the roots of $ax^2+2bx+c=0$ be real and different, then the roots of $(a+c)(ax^2+2bx+c)=2(ac-b^2)(x^2+1)$ will be imaginary and vice-versa .

Question

If the roots of $ax^2+2bx+c=0$ be real and different, then the roots of $(a+c)(ax^2+2bx+c)=2(ac-b^2)(x^2+1)$
will be non-real complex roots and vice-versa .

What I've tried :

I was thinking of finding the nature of discriminant of the equation to deduce things but I'm scared of expanding so many terms .

I too thought to use the idea of $ax^2 + 2bx + c = a(x-\alpha)(x-\beta)$ but found no use .

Can this problem have any better intuitive solution/approach than the standard one(posted by @Martin R in a comment) ?

Answer 1

We have to add an extra condition so that your statement becomes true. Let's examine the following case to see why your statement can be false.

If $a=0$, then $ax^2+2bx+c=0$ clearly cannot have two distinct real roots, but, for example, if $b=1$ and $c=0$, then $$ \begin{align*} (a+c)(ax^2+2bx+c)-2(ac-b^2)(x^2+1) &= 2(x^2+1), \end{align*} $$ which has two non-real roots.

If we exclude this case, then your statement is true. That is,

Let $a,b,c\in\mathbb{R}$. The following are equivalent:

1. $ax^2+2bx+c=0$ has two distinct real roots.
2. $a\neq 0$ and $(a+c)(ax^2+2bx+c)=2(ac-b^2)(x^2+1)$ has two non-real roots.

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Proof:

Suppose $ax^2+2bx+c=0$ has two distinct real roots. Then $a\neq 0$, or else the equation cannot have $2$ distinct roots. Now, we can then divide both sides of $ax^2+2bx+c=0$ by $a$ and obtain $$ x^2+\frac{2b}{a}x+\frac{c}{a}=0, $$ which also has two distinct real roots because division by $a$ does not change the roots. Let $p=b/a$ and $q=c/a$. Then $p^2-q>0$. In particular, we cannot have both $p=0$ and $q=1$. By expanding, we have $$ \begin{align*} &\quad \ \, (1+q)(x^2+2px+q)-2(q-p^2)(x^2+1) \\ &= (2p^2 - q + 1)x^2 + (2p(q + 1))x + 2p^2 + (q + 1)q - 2q. \end{align*} $$ The discriminant is $$ \begin{align*} &\ \ \quad 4(p^2(q + 1)^2 - (2p^2 + (q + 1)q - 2q)(2p^2 - q + 1)) \\ &= 4(-4p^4 - p^2q^2 + 6p^2q + q^3 - p^2 - 2q^2 + q) \\ &= -4(4p^2 + (q-1)^2)(p^2 - q) < 0 \end{align*} $$ where the last inequality is true because we cannot have both $p=0$ and $q=1$. This shows that $$ (1+q)(x^2+2px+q)=2(q-p^2)(x^2+1) $$ has two distinct real roots. Multiplying by $a^2$ (which is nonzero) on both sides, we have $$ (a+c)(ax^2+2bx+c)=2(ac-b^2)(x^2+1), $$ which shows that this equation also has two distinct real roots because multiplying by a nonzero number does not change the roots.

For the converse, suppose $a\neq 0$ and $$ (a+c)(ax^2+2bx+c)=2(ac-b^2)(x^2+1) $$ has two distinct real roots. Since $a\neq 0$, we can divide both sides by $a^2$ and obtain $$ (1+q)(x^2+2px+q)=2(q-p^2)(x^2+1) \tag{*} $$ which has two distinct real roots because division by a nonzero number does not change the roots. By the previous calculation, the discriminant is $$ -4(4p^2 + (q-1)^2)(p^2 - q), $$ which is negative because $(*)$ has two non-real roots. This implies that $p^2-q>0$. Multiplying by $a^2$ (which is nonzero), we have $b^2-ac>0$, which shows that $ax^2+2bx+c=0$ has two distinct real roots.