Solve quadratic equations by factoring, the quadratic formula and square roots, and use the discriminant to count real solutions (Algebra).

What this topic is asking

A quadratic equation has the form $ax^{2} + bx + c = 0$. The ACT tests three solving methods, factoring, the quadratic formula, and taking square roots, plus reading the discriminant to count real solutions. Choosing the fastest method for each equation is the skill that saves time.

Factoring

Factoring is fastest when the quadratic has integer roots.

The quadratic formula

When no integer factors work, the formula solves any quadratic.

The formula is $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$. The two error-prone spots are the sign of $-b$ and the sign of $-4ac$. For $2x^{2} + 3x - 2 = 0$ with $a = 2$, $b = 3$, $c = -2$: $x = \frac{-3 \pm \sqrt{9 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$, giving $x = \frac{1}{2}$ or $x = -2$.

Solving by square roots

When there is no linear ($bx$) term, take square roots directly. For $x^{2} = 49$, $x = \pm 7$ (both signs). For $(x - 3)^{2} = 16$, $x - 3 = \pm 4$, so $x = 7$ or $x = -1$. Forgetting the negative root is a frequent slip; a squared quantity has two square roots.

The discriminant

The discriminant $b^{2} - 4ac$ (the part under the radical) counts the real solutions without full solving:

• $b^{2} - 4ac > 0$: two distinct real solutions.
• $b^{2} - 4ac = 0$: one real solution (a double root).
• $b^{2} - 4ac < 0$: no real solutions (two complex solutions).

When a question only asks how many real solutions, compute the discriminant alone; it is much faster than solving.

Completing the square

A fourth method, completing the square, both solves a quadratic and reveals its vertex. To solve $x^{2} + 6x + 5 = 0$, move the constant ($x^{2} + 6x = -5$), add the square of half the linear coefficient ($\left(\frac{6}{2}\right)^{2} = 9$) to both sides ($x^{2} + 6x + 9 = 4$), write the left as a perfect square ($(x + 3)^{2} = 4$), and take square roots ($x + 3 = \pm 2$, so $x = -1$ or $x = -5$). The ACT rarely demands this method for solving, since factoring or the formula is faster, but it is the bridge to vertex form $a(x - h)^{2} + k$, which the Functions area uses to read a parabola's turning point.

Quadratics in word problems

Quadratics model situations with a squared relationship, such as area or projectile height. A typical ACT item gives a height function like $h(t) = -16t^{2} + 32t$ and asks when the object lands ($h = 0$). Factor: $-16t(t - 2) = 0$, so $t = 0$ (launch) or $t = 2$ (landing). Translating the words into $ax^{2} + bx + c = 0$, then choosing the physically meaningful root, is the whole task; discard a negative time or length as not sensible in context.

Choosing a method efficiently

A quick decision saves time: if the quadratic obviously factors (small integer $a$, recognisable factor pair), factor. If it has no $bx$ term, take square roots. Otherwise, use the formula. Always set the equation to zero first, because the methods assume the right side is zero. A final check is to substitute a solution back, or to confirm the two roots multiply to $\frac{c}{a}$ and add to $-\frac{b}{a}$, a fast consistency test.

Try this

Q1. Solve $x^{2} - 9 = 0$. [1 point]

• Cue. $x^{2} = 9$, so $x = \pm 3$ (both signs).

Q2. Solve $x^{2} - 7x + 12 = 0$ by factoring. [1 point]

• Cue. $-3$ and $-4$ multiply to 12 and add to $-7$: $(x - 3)(x - 4) = 0$, so $x = 3$ or $x = 4$.