Review question

# Can we show $(b^2 - 2ac)x^2 + 4(a + c)x = 8$ always has real roots? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8812

## Solution

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

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.

Find the conditions that the roots are equal.

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 $b = 0$.