The Quadratic Formula and the Discriminant: Additional Practice (Set 2)

With $a = 3$, $b = 2$, $c = -2$, the discriminant is $4 + 24 = 28$.

$x = \frac{-2 \pm \sqrt{28}}{6} = \frac{-2 \pm 2\sqrt{7}}{6} = \frac{-1 \pm \sqrt{7}}{3}$

Simplify $\sqrt{28} = 2\sqrt{7}$, then cancel the factor of $2$ shared by $-2$, $2\sqrt{7}$, and $6$.

With $a = 6$, $b = 7$, $c = -3$, the discriminant is $49 + 72 = 121 = 11^2$.

$x = \frac{-7 \pm \sqrt{121}}{12} = \frac{-7 \pm 11}{12}, \qquad x = \tfrac{1}{3} \ \text{ or } \ x = -\tfrac{3}{2}$

A perfect-square discriminant with integer coefficients gives rational roots, so it factors over the rationals as $(3x - 1)(2x + 3)$.

First $\Delta = 8^2 - 4(2)(3) = 64 - 24 = 40$. Then the vertex height with $a = 2$ is

$y = -\frac{\Delta}{4a} = -\frac{40}{8} = -5$

Since $a > 0$ the parabola opens upward, and its vertex at $y = -5$ lies below the axis, consistent with two real roots.

With $a = 2$, $b = 6$, $c = 3$, the discriminant is $36 - 24 = 12$.

$x = \frac{-6 \pm \sqrt{12}}{4} = \frac{-6 \pm 2\sqrt{3}}{4} = \frac{-3 \pm \sqrt{3}}{2}$

Simplify $\sqrt{12} = 2\sqrt{3}$, then cancel the factor of $2$ from $-6$, $2\sqrt{3}$, and $4$.

With $a = 5$, $b = -3$, $c = -1$, the discriminant is $9 + 20 = 29$.

$x = \frac{3 \pm \sqrt{29}}{10}$

Since $29$ is prime the radical will not simplify, and the denominator is $2a = 10$, so the finished form is $\tfrac{3 \pm \sqrt{29}}{10}$.

Expand and move everything to one side: $x^2 + x - 56 = 0$, with discriminant $1 + 224 = 225 = 15^2$.

$x = \frac{-1 \pm \sqrt{225}}{2} = \frac{-1 \pm 15}{2}, \qquad x = 7 \ \text{ or } \ x = -8$

The positive value is $x = 7$, giving the pair $7$ and $8$ with product $56$.

A repeated root means $\Delta = b^2 - 36 = 0$, so $b = \pm 6$, and the root is $-\tfrac{b}{2a} = -\tfrac{b}{2}$.

$b = 6 \Rightarrow x = -3, \qquad b = -6 \Rightarrow x = 3$

So the repeated root is $3$ or $-3$ (the two ways the vertex can touch the axis).

Multiply by $-1$ to get $2x^2 - 4x - 1 = 0$, with $a = 2$, $b = -4$, $c = -1$ and discriminant $16 + 8 = 24$.

$x = \frac{4 \pm \sqrt{24}}{4} = \frac{4 \pm 2\sqrt{6}}{4} = \frac{2 \pm \sqrt{6}}{2}$

Simplify $\sqrt{24} = 2\sqrt{6}$, then cancel the factor of $2$.

One real solution means $\Delta = 0$, with $a = 1$, $b = -6$.

$\Delta = 36 - 4c = 0 \ \Rightarrow \ c = 9$

Then $x^2 - 6x + 9 = (x - 3)^2 = 0$, a double root at $x = 3$.

A repeated root means $\Delta = 0$, with $a = 1$, $b = p$, $c = p$.

$\Delta = p^2 - 4p = p(p - 4) = 0 \ \Rightarrow \ p = 0 \ \text{ or } \ p = 4$

At $p = 0$ the equation is $x^2 = 0$, and at $p = 4$ it is $(x + 2)^2 = 0$, each with a double root.

With $a = 4$, $b = -4$, $c = -1$, the discriminant is $16 + 16 = 32$.

$x = \frac{4 \pm \sqrt{32}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2}$

Simplify $\sqrt{32} = 4\sqrt{2}$, then cancel the factor of $4$ common to $4$, $4\sqrt{2}$, and $8$.

With $a = 1$, $b = 10$, $c = 18$, the discriminant is $100 - 72 = 28$.

$x = \frac{-10 \pm \sqrt{28}}{2} = \frac{-10 \pm 2\sqrt{7}}{2} = -5 \pm \sqrt{7}$

Simplify $\sqrt{28} = 2\sqrt{7}$, then cancel the factor of $2$.