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 \(b = 0\).