The Quadratic Formula and the Discriminant: Practice

A repeated root is the case $\Delta = 0$. With $a = 1$, $b = -k$, $c = 4$:

$\Delta = (-k)^2 - 4(1)(4) = k^2 - 16 = 0 \ \Rightarrow \ k = \pm 4$

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

No real solutions means $\Delta < 0$. With $a = 1$, $b = k$, $c = k + 3$:

$\Delta = k^2 - 4(k + 3) = k^2 - 4k - 12 = (k - 6)(k + 2)$

The product $(k - 6)(k + 2)$ is negative exactly between its roots, so $\Delta < 0$ for $-2 < k < 6$.

One solution means $\Delta = 0$, so $b^2 - 4(1)(25) = 0$, giving $b^2 = 100$ and, since $b > 0$, $b = 10$.

$x = -\frac{b}{2a} = -\frac{10}{2} = -5$

The double root sits on the axis at $x = -5$; indeed $x^2 + 10x + 25 = (x + 5)^2$.

Since $a < 0$ the parabola opens downward, and $\Delta > 0$ guarantees two distinct real roots, hence two crossings.

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

The vertex sits above the axis and the arms open downward, so the curve dips down through the axis at two points.

The two numbers are roots of $x^2 - 6x + 7 = 0$, whose discriminant is $36 - 28 = 8 > 0$, so real numbers do exist.

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

Check: the sum $(3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$ and the product $(3)^2 - (\sqrt{2})^2 = 9 - 2 = 7$.

Tangent at $x = 3$ means a double root at $3$, so the quadratic is $(x - 3)^2$.

$(x - 3)^2 = x^2 - 6x + 9$

Matching coefficients gives $b = -6$ and $c = 9$. (Check: $\Delta = 36 - 36 = 0$, confirming tangency.)

With $a = 1$, $b = -1$, $c = -1$, the discriminant is $1 + 4 = 5$.

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

The positive root $\tfrac{1 + \sqrt{5}}{2} \approx 1.618$ is the golden ratio; both roots straddle the axis $x = \tfrac{1}{2}$.

Since $\Delta = 45 > 0$, there are two distinct real roots, but $45$ is not a perfect square, so $\sqrt{45}$ is irrational.

$x = \frac{-b \pm \sqrt{45}}{2a}, \qquad \sqrt{45} = 3\sqrt{5} \ \text{ irrational}$

With integer coefficients, an irrational root means no rational factorization, though the quadratic still factors over the reals.

Two distinct real roots needs $\Delta > 0$, with $a = 1$, $b = 2$.

$\Delta = 4 - 4c > 0 \ \Rightarrow \ 4c < 4 \ \Rightarrow \ c < 1$

At $c = 1$ the discriminant is $0$ (a double root), and for $c > 1$ it is negative (no real roots).

With rational coefficients, the irrational partner $2 - \sqrt{3}$ must also be a root, so the quadratic is a multiple of $(x - (2 + \sqrt{3}))(x - (2 - \sqrt{3})) = x^2 - 4x + 1$.

$\Delta = (-4)^2 - 4(1)(1) = 12, \quad \text{not a perfect square}$

An irrational root forces a non-perfect-square discriminant, so the equation does not factor over the rationals, even though it factors over the reals.

Set $-5t^2 + 20t = 15$, then move everything to one side and divide by $-5$.

$-5t^2 + 20t - 15 = 0 \ \Rightarrow \ t^2 - 4t + 3 = 0 \ \Rightarrow \ (t - 1)(t - 3) = 0$

So $t = 1$ or $t = 3$, both positive: the ball passes height $15$ once on the way up and once on the way down.

The count is decided by the discriminant $b^2 - 4ac$, and here $a = 1$.

$\Delta = b^2 - 4(1)(c) = b^2 - 4c$

Its sign gives two, one, or no real roots, so $b^2 - 4c$ is the deciding quantity.