The Quadratic Formula and the Discriminant: Practice

With $a = 1$, $b = -6$, $c = 7$, the discriminant is $36 - 28 = 8$.

$x = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}$

Simplifying $\sqrt{8} = 2\sqrt{2}$ and cancelling the factor of $2$ gives the finished form $3 \pm \sqrt{2}$.

With $a = 1$, $b = 2$, $c = -4$, the discriminant is $4 + 16 = 20$.

$x = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5}$

Simplify $\sqrt{20} = 2\sqrt{5}$, then cancel the $2$ shared by both numerator terms and the denominator.

Compute the discriminant with $a = 2$, $b = -4$, $c = 5$.

$\Delta = (-4)^2 - 4(2)(5) = 16 - 40 = -24 < 0$

A negative discriminant means no real solutions; the parabola stays clear of the $x$-axis.

Exactly one real solution is the case $\Delta = 0$. With $a = 1$, $b = k$, $c = 9$:

$\Delta = k^2 - 36 = 0 \ \Rightarrow \ k^2 = 36 \ \Rightarrow \ k = \pm 6$

Both values give a double root: $k = 6$ gives $(x + 3)^2 = 0$ and $k = -6$ gives $(x - 3)^2 = 0$.

With $a = 2$, $b = 5$, $c = 2$, the discriminant is $25 - 16 = 9$.

$x = \frac{-5 \pm \sqrt{9}}{4} = \frac{-5 \pm 3}{4}$

That gives $x = \tfrac{-2}{4} = -\tfrac{1}{2}$ and $x = \tfrac{-8}{4} = -2$.

With $a = 5$, $b = -2$, $c = -3$, substitute with parentheses.

$\Delta = (-2)^2 - 4(5)(-3) = 4 + 60 = 64$

The $-4ac$ term is $-4(5)(-3) = +60$, so the discriminant is $64$, a positive perfect square.

The vertex height is $-\tfrac{\Delta}{4a}$. With $\Delta < 0$ and $a > 0$, this is a negative divided by a positive, negated.

$-\frac{\Delta}{4a} > 0 \quad (\text{since } \Delta < 0,\ a > 0)$

So the vertex sits above the axis, and an upward parabola with its lowest point above the axis never meets it, matching $\Delta < 0$.

The roots straddle the axis $x = -\tfrac{b}{2a} = \tfrac{8}{2} = 4$, each the same distance from it, so their average is $4$ and their sum is twice that.

$\text{sum} = 2\left(-\frac{b}{2a}\right) = -\frac{b}{a} = \frac{8}{1} = 8$

The $\pm$ parts cancel when the two roots are added, leaving $-\tfrac{b}{a} = 8$.

With $a = 1$, $b = 4$, $c = 1$, the discriminant is $16 - 4 = 12$.

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

Simplify $\sqrt{12} = 2\sqrt{3}$ and cancel the shared factor of $2$.

With $a = 3$, $b = -2$, $c = -1$, the discriminant is $4 + 12 = 16$.

$x = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6}$

That gives $x = \tfrac{6}{6} = 1$ and $x = \tfrac{-2}{6} = -\tfrac{1}{3}$.

For integer coefficients, factoring over the rationals happens exactly when the discriminant is a perfect square.

$30 \ \text{is not a perfect square} \ \Rightarrow \ \sqrt{30} \ \text{irrational}$

So the roots $\tfrac{-b \pm \sqrt{30}}{2a}$ are irrational, and the quadratic does not factor over the rationals, even though $\Delta > 0$ gives two real roots.

The discriminant is $\Delta = 9 + 40 = 49$, a perfect square, so integer factoring succeeds and reads off the roots instantly.

$x^2 - 3x - 10 = (x - 5)(x + 2) = 0 \ \Rightarrow \ x = 5 \ \text{ or } \ x = -2$

The formula would also work but is slower here, and it is certainly not the only option, so factoring is the fastest correct first step.