Can we show $(b^2 - 2ac)x^2 + 4(a + c)x = 8$ always has real roots?

Can we show $(b^2 - 2ac)x^2 + 4(a + c)x = 8$ always has real roots?

Question

Prove that the roots of $(b^2 - 2ac)x^2 + 4(a + c)x = 8$ are real if $a$, $b$, $c$ are real. Find the conditions that the roots are equal.

Suggestion

Would the discriminant be helpful here?

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The curve here is $y=(b^2 - 2ac)x^2 + 4(a + c)x -8$.

What can we say about the curve if the roots of the equation are real?

What would the curve look like if the roots of the equation are equal?

Solution

The roots of a quadratic equation are real if the discriminant is greater than or equal to zero.

We will assume throughout this question that $b^2-2ac$, the coefficient of $x^2$, is non-zero. If it were zero, we would not have a quadratic equation any longer!

The quadratic equation here is

$$(b^2 - 2ac)x^2 + 4(a + c)x - 8=0,$$

and so its discriminant (denoted by

$\Delta$

) is

$$\begin{align*} \Delta &= 16(a + c)^2 - 4(b^2 - 2ac)(-8) \\ &= 16(a + c)^2 + 32(b^2 - 2ac). \end{align*}$$

There is a common factor of $16$, which it would be helpful to take out. By expanding and simplifying, we then see that

$$\begin{align*} \Delta &= 16\bigl((a+c)^2+2(b^2-2ac)\bigr) \\ &= 16(a^2 + 2ac + c^2 + 2b^2 - 4ac) \\ &= 16(a^2 - 2ac + c^2 + 2b^2) \\ &= 16\bigl((a - c)^2 + 2b^2\bigr), \end{align*}$$

which is nonnegative for all real $a$, $b$, and $c$, as it is a sum of squares, and a square cannot be negative.

The two roots are equal when the discriminant, $\Delta$, is zero. This occurs precisely when both of the squares $(a-c)^2$ and $b^2$ are zero, that is, when $a = c$ and $b = 0$.