How do you find the roots of a quadratic on the ACT?

First try factoring: find two numbers that multiply to the constant and add to the middle coefficient, then set each factor to zero. If it does not factor cleanly, use the quadratic formula x = (-b ± √(b² - 4ac)) / (2a), which works for every quadratic.

What does the discriminant tell you?

The discriminant b² - 4ac tells you how many real solutions a quadratic has without solving it. Positive means two distinct real roots, zero means one repeated real root, and negative means no real roots (the solutions are complex).

How do you find the vertex of a parabola?

The x-coordinate of the vertex is x = -b/(2a). Substitute that value back into the equation to get the y-coordinate. If the equation is already in vertex form a(x - h)² + k, the vertex is simply (h, k).

How do you find the maximum height in a projectile problem?

Projectile height is a downward-opening parabola, so the maximum is the vertex. Find the time t = -b/(2a) when the peak occurs, then substitute that time back into the height equation to get the maximum height.

CramClub • Quadratics and the Parabola

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

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

$y = a(x - h)^2 + k$

The Quadratic and Its Graph

A quadratic is an expression of the form $ax^2 + bx + c$ with $a \ne 0$ . Its graph is a parabola — a symmetric U-shaped curve that opens upward when $a > 0$ and downward when $a < 0$ . Quadratics appear all over ACT Math: solving for roots, finding a maximum height, reading a vertex off a graph, and counting real solutions. Knowing three forms and one formula covers nearly every question.

Solving by Factoring

If a quadratic factors, factoring is the fastest way to its roots. The idea rests on the zero-product property: if a product equals zero, at least one factor is zero.

Solve $x^2 - 5x + 6 = 0$ . Find two numbers that multiply to $6$ and add to $-5$ : those are $-2$ and $-3$ .

$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

Set each factor to zero: $x = 2$ or $x = 3$ . Those two values are the roots (also called zeros or $x$ -intercepts) of the parabola.

The Quadratic Formula Always Works

When a quadratic does not factor nicely, use the quadratic formula. For $ax^2 + bx + c = 0$ :

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Solve $2x^2 + 3x - 4 = 0$ . Here $a = 2$ , $b = 3$ , $c = -4$ :

x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)} \\ x &= \frac{-3 \pm \sqrt{9 + 32}}{4} \\ x &= \frac{-3 \pm \sqrt{41}}{4} \end{aligned}$$ Both roots are irrational, which is perfectly acceptable on the ACT. ## The Discriminant: Counting Real Solutions The expression under the radical, $b^2 - 4ac$, is the **discriminant**, and its sign tells you how many real solutions exist without solving: | Discriminant $b^2 - 4ac$ | Real solutions | |---|---| | Positive | **Two** distinct real roots | | Zero | **One** (repeated) real root | | Negative | **No** real roots (two complex) | ## The Vertex and Axis of Symmetry Every parabola has a **vertex** — its highest or lowest point — and a vertical **axis of symmetry** through it. For $y = ax^2 + bx + c$, the axis of symmetry (and the $x$-coordinate of the vertex) is: $$x = -\frac{b}{2a}$$ Plug that $x$ back in to get the $y$-coordinate. For $y = x^2 - 6x + 5$, the vertex is at $x = -\frac{-6}{2(1)} = 3$, and $y = 3^2 - 6(3) + 5 = -4$, so the vertex is $(3, -4)$. Because $a = 1 > 0$, the parabola opens upward and $(3, -4)$ is the **minimum**. ## Vertex Form Writing a quadratic in **vertex form** exposes the vertex directly: $$y = a(x - h)^2 + k, \quad \text{vertex } (h, k)$$ If a question hands you $y = 2(x - 4)^2 - 7$, you can read the vertex as $(4, -7)$ with no computation. Watch the sign: the $x$-coordinate is $+4$ because the form subtracts $h$. ## Application: Projectile Height A classic ACT setup models height with a downward parabola such as $h(t) = -16t^2 + 48t + 5$. The **maximum height** occurs at the vertex, at time $t = -\frac{b}{2a} = -\frac{48}{2(-16)} = 1.5$ seconds. Substituting $t = 1.5$ gives the peak height. The object returns to the ground when $h(t) = 0$, which you solve with the quadratic formula. ## A Quick Strategy Guide - **Factors easily?** Factor and use the zero-product property. - **Doesn't factor?** Use the quadratic formula — it never fails. - **Only need the count of solutions?** Compute the discriminant $b^2 - 4ac$. - **Need the max, min, or turning point?** Find the vertex with $x = -\frac{b}{2a}$. - **Given vertex form?** Read $(h, k)$ straight off. ## Common Mistakes to Avoid - Forgetting the $\pm$ in the quadratic formula, which loses one of the two roots. - Sign errors with $-4ac$ when $c$ is negative — a double negative becomes addition. - Reading the vertex of $a(x - h)^2 + k$ as $(-h, k)$ instead of $(h, k)$. - Dividing only the $-b$ term by $2a$ instead of the whole numerator.

Solve $x^2 + 7x + 12 = 0$ .

Solve $x^2 - 49 = 0$ .

Solve $2x^2 + 3x - 4 = 0$ using the quadratic formula.

Find the vertex of the parabola $y = x^2 - 6x + 5$ .

For what value of $k$ does $x^2 + kx + 9 = 0$ have exactly one real solution?

A ball is thrown so its height in feet after $t$ seconds is $h(t) = -16t^2 + 64t + 6$ . What is the maximum height the ball reaches?

Dropping the $\pm$ in the quadratic formula and reporting only one root.

The $\pm$ produces both roots. Always compute the $+$ and the $-$ versions unless the discriminant is zero.

Sign errors in the discriminant when $c$ is negative, e.g. treating $-4(2)(-4)$ as $-32$ .

Multiply signs carefully: $-4ac$ with a negative $c$ becomes positive. Here $-4(2)(-4) = +32$ .

Misreading vertex form: taking $y = 2(x - 4)^2 - 7$ to have vertex $(-4, -7)$ .

Vertex form is $a(x - h)^2 + k$ with vertex $(h, k)$ . Because it subtracts $h$ , $(x - 4)$ gives $h = +4$ : the vertex is $(4, -7)$ .

Dividing only $-b$ by $2a$ and adding the radical afterward, so the $\sqrt{\;}$ term is not divided.

The entire numerator $-b \pm \sqrt{b^2 - 4ac}$ sits over $2a$ . Keep it as one fraction.

What are the solutions to $x^2 - 9x + 20 = 0$ ?

What is the $x$ -coordinate of the vertex of the parabola $y = x^2 - 8x + 3$ ?

The parabola $y = 3(x - 2)^2 + 5$ is written in vertex form. What is its vertex?

How many distinct real solutions does $x^2 + 4x + 7 = 0$ have?

For what value of $c$ does $x^2 - 10x + c = 0$ have exactly one real solution?

A projectile has height $h(t) = -16t^2 + 80t + 4$ feet after $t$ seconds. What is its maximum height, in feet?