The Quadratic Formula: Additional Practice (Set 2)

A repeated root needs the discriminant to be $0$, so $b^2 - 4ac = 0$ with $a = 1$, $b = -8$.

$(-8)^2 - 4(1)c = 64 - 4c = 0 \quad\Rightarrow\quad c = 16$

Then $x^2 - 8x + 16 = (x - 4)^2$, whose only root is $x = 4$.

With $a = 4$, $b = -4$, $c = 1$, compute $b^2 - 4ac$.

$D = (-4)^2 - 4(4)(1) = 16 - 16 = 0$

A zero discriminant gives exactly one repeated real root, $x = \tfrac{1}{2}$, since $4x^2 - 4x + 1 = (2x - 1)^2$.

Read off $a = 1$, $b = 1$, $c = 1$, so $D = 1 - 4 = -3$, with $\sqrt{-3} = i\sqrt{3}$.

$x = \frac{-1 \pm i\sqrt{3}}{2}$

The negative discriminant makes the roots a complex conjugate pair.

Read off $a = 1$, $b = 2$, $c = 10$, so $D = 4 - 40 = -36$, with $\sqrt{-36} = 6i$.

$x = \frac{-2 \pm 6i}{2} = -1 \pm 3i$

Read off $a = 9$, $b = 6$, $c = 1$, and the discriminant is $36 - 36 = 0$.

$x = \frac{-6 \pm \sqrt{0}}{18} = \frac{-6}{18} = -\tfrac{1}{3}$

A repeated root, since $9x^2 + 6x + 1 = (3x + 1)^2$.

Read off $a = 3$, $b = 2$, $c = -8$, so $-4ac = +96$ and $D = 4 + 96 = 100$.

$x = \frac{-2 \pm \sqrt{100}}{6} = \frac{-2 \pm 10}{6}$

So $x = \frac{8}{6} = \tfrac{4}{3}$ or $x = \frac{-12}{6} = -2$.

Read off $a = 2$, $b = -4$, $c = -1$, so $-b = 4$ and $-4ac = +8$, giving $D = 16 + 8 = 24$, with $\sqrt{24} = 2\sqrt{6}$.

$x = \frac{4 \pm 2\sqrt{6}}{4} = \frac{2 \pm \sqrt{6}}{2}$

Reduce by the common factor $2$.

Read off $a = 1$, $b = -12$, $c = 40$, so $-b = 12$ and $D = 144 - 160 = -16$, with $\sqrt{-16} = 4i$.

$x = \frac{12 \pm 4i}{2} = 6 \pm 2i$

Read off $a = 3$, $b = -2$, $c = 1$, so $-b = 2$ and $D = 4 - 12 = -8$, with $\sqrt{-8} = 2i\sqrt{2}$.

$x = \frac{2 \pm 2i\sqrt{2}}{6} = \frac{2(1 \pm i\sqrt{2})}{6} = \frac{1 \pm i\sqrt{2}}{3}$

Read off $a = 4$, $b = 0$, $c = 9$, so the $-b$ term is $0$ and $D = 0 - 144 = -144$, with $\sqrt{-144} = 12i$.

$x = \frac{0 \pm 12i}{8} = \pm\tfrac{3}{2}i$

Pure imaginary roots, since there is no $x$ term and $c > 0$.

Read off $a = 6$, $b = 1$, $c = -1$, so $-4ac = +24$ and $D = 1 + 24 = 25$.

$x = \frac{-1 \pm \sqrt{25}}{12} = \frac{-1 \pm 5}{12}$

So $x = \frac{4}{12} = \tfrac{1}{3}$ or $x = \frac{-6}{12} = -\tfrac{1}{2}$.

First reach standard form: $x^2 - 2x + 3 = 0$, so $-b = 2$ and $D = 4 - 12 = -8$, with $\sqrt{-8} = 2i\sqrt{2}$.

$x = \frac{2 \pm 2i\sqrt{2}}{2} = 1 \pm i\sqrt{2}$