The Quadratic Formula and the Discriminant: Additional Practice (Set 1)

No real solutions means $\Delta < 0$, with $a = 1$, $b = -2$, $c = k$.

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

At $k = 1$ the discriminant is $0$ (one root), and for $k < 1$ it is positive (two roots).

With $a = 3$, $b = -5$, $c = 1$, the discriminant is $25 - 12 = 13$.

$x = \frac{5 \pm \sqrt{13}}{6}$

Since $13$ is prime the radical does not simplify, and the denominator is $2a = 6$.

The roots are a mirror pair about the axis, so the axis is their average.

$x = \frac{\text{sum of roots}}{2} = \frac{10}{2} = 5$

The axis of symmetry is the vertical line $x = 5$, halfway between the two roots.

With $a = 4$, $b = 0$, $c = 1$, the discriminant is $0 - 16 = -16 < 0$.

$4x^2 + 1 = 0 \ \Rightarrow \ x^2 = -\tfrac{1}{4}$

No real number squares to a negative, so there are no real solutions.

Two distinct real solutions needs $\Delta > 0$, with $a = 1$, $b = k$, $c = 16$.

$\Delta = k^2 - 64 > 0 \ \Rightarrow \ k^2 > 64 \ \Rightarrow \ k < -8 \ \text{ or } \ k > 8$

Between $-8$ and $8$ the discriminant is negative (no real roots), and at $k = \pm 8$ it is zero (one root).

The number of crossings is the number of real roots, decided by the discriminant.

$\Delta = (-4)^2 - 4(1)(7) = 16 - 28 = -12 < 0$

A negative discriminant means no real roots, so the upward parabola stays entirely above the axis and never meets it.

First $\Delta = (-6)^2 - 4(1)(11) = 36 - 44 = -8$. The vertex height with $a = 1$ is the minimum, since the parabola opens upward.

$y = -\frac{\Delta}{4a} = -\frac{-8}{4} = 2$

The minimum value is $2$; because $\Delta < 0$ the vertex sits above the axis and the graph never reaches $0$.

With $a = 1$, $b = -14$, $c = 49$, the discriminant is $196 - 196 = 0$, so there is one repeated root.

$x = \frac{14 \pm \sqrt{0}}{2} = \frac{14}{2} = 7$

The left side is the perfect square $(x - 7)^2$, so $7$ is a double root.

Two distinct real solutions needs a positive discriminant; check each.

$x^2 - 3x + 1: \ \Delta = 9 - 4 = 5 > 0$

The others give $\Delta = -3$, $\Delta = 0$, and $\Delta = -8$, so only $x^2 - 3x + 1 = 0$ has two distinct real roots.

Tangent to the axis means one repeated root, so $\Delta = 0$ with $a = 1$, $b = 4$.

$\Delta = 16 - 4c = 0 \ \Rightarrow \ c = 4$

Then $x^2 + 4x + 4 = (x + 2)^2$, tangent to the axis at $x = -2$.

A zero discriminant gives exactly one repeated root, sitting at the axis.

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

With rational $a$ and $b$, this value $-\tfrac{b}{2a}$ is rational, so the single repeated root is rational.

The formula is not asserted; it is proved by completing the square on the general equation.

$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \ \Rightarrow \ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Because that derivation holds for every $a \ne 0$, $b$, $c$ at once, the result is a proven theorem you own outright, not a recipe to accept on faith.