Quadratic equations are the backbone of Advanced Math on the digital SAT, and there are exactly three ways to solve them. This guide walks through factoring, the quadratic formula, and completing the square on the same equation so you can see precisely how each method works, then shows a second example where factoring simply isn't an option.
Why Quadratics Deserve Extra Attention
Quadratic equations sit at the center of two Advanced Math subtopics — nonlinear functions and nonlinear equations — and they resurface inside word problems, systems of equations, and graph-reading questions. Since the reference sheet on the digital SAT doesn't include the quadratic formula or completing-the-square steps, you need all three methods memorized and ready to deploy depending on what the numbers look like.
Method 1: Factoring
Factoring is fastest when the roots are integers or simple fractions. The strategy: get the equation equal to zero, then find two numbers that multiply to \(ac\) and add to \(b\).
Worked Example: Solve \(x^2 - 2x - 8 = 0\) by Factoring
- Confirm the equation is set to zero: \(x^2 - 2x - 8 = 0\).
- Find two numbers that multiply to \(-8\) and add to \(-2\): those numbers are \(-4\) and \(2\).
- Write the factored form: \((x-4)(x+2) = 0\).
- Set each factor to zero: \(x - 4 = 0\) or \(x + 2 = 0\).
- Solve each: \(x = 4\) or \(x = -2\).
Method 2: The Quadratic Formula
The quadratic formula always works, regardless of whether the roots are ugly or the trinomial doesn't factor nicely. For \(ax^2+bx+c=0\):
\(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\)
Worked Example: Solve \(x^2 - 2x - 8 = 0\) by Formula
- Identify coefficients: \(a=1\), \(b=-2\), \(c=-8\).
- Substitute into the formula: \(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}\).
- Simplify inside the square root: \(x = \frac{2 \pm \sqrt{4+32}}{2} = \frac{2 \pm \sqrt{36}}{2}\).
- Take the square root: \(x = \frac{2 \pm 6}{2}\).
- Solve both cases: \(x = \frac{8}{2} = 4\) or \(x = \frac{-4}{2} = -2\).
Same equation, same answers as factoring — which is exactly the point of comparing methods side by side.
Method 3: Completing the Square
Completing the square is the slowest by hand but it's essential for converting standard form into vertex form, which is often what a graph-reading question actually needs.
Worked Example: Solve \(x^2 - 2x - 8 = 0\) by Completing the Square
- Move the constant to the other side: \(x^2 - 2x = 8\).
- Take half of the \(x\)-coefficient and square it: \(\left(\frac{-2}{2}\right)^2 = 1\).
- Add that value to both sides: \(x^2 - 2x + 1 = 8 + 1\).
- Write the left side as a perfect square: \((x-1)^2 = 9\).
- Take the square root of both sides: \(x - 1 = \pm 3\).
- Solve: \(x = 1+3 = 4\) or \(x = 1-3 = -2\).
Worked Example: A Quadratic That Won't Factor Nicely
Solve \(2x^2 + 3x - 4 = 0\).
- Try factoring first: you need two numbers multiplying to \(ac = 2 \times -4 = -8\) that add to \(b=3\). No integer pair works, so factoring is ruled out quickly.
- Switch to the quadratic formula with \(a=2\), \(b=3\), \(c=-4\): \(x = \frac{-3 \pm \sqrt{9-4(2)(-4)}}{2(2)} = \frac{-3 \pm \sqrt{9+32}}{4}\).
- Simplify inside the root: \(x = \frac{-3 \pm \sqrt{41}}{4}\).
- Since \(41\) isn't a perfect square, leave the answer in exact radical form: \(x = \frac{-3+\sqrt{41}}{4}\) or \(x = \frac{-3-\sqrt{41}}{4}\).
- As decimals, these are approximately \(x \approx 0.85\) and \(x \approx -2.35\).
This is the pattern to recognize on test day: spend no more than a few seconds checking for clean integer factors, and the moment they don't appear, move to the formula rather than continuing to guess. A grid-in version of this question would ask for the decimal approximation, rounded, since the grid can't accept a radical expression directly.
Comparing the Three Methods
| Method | Best when... | Speed |
|---|---|---|
| Factoring | Roots are integers and \(a=1\) or small | Fastest |
| Quadratic formula | Numbers are messy or don't factor cleanly | Reliable, moderate speed |
| Completing the square | You need vertex form or the vertex itself | Slower, but reveals the vertex |
Tip: try factoring first — spend no more than 10-15 seconds checking for integer factors. If nothing obvious appears, switch straight to the quadratic formula rather than wasting time forcing a factorization that doesn't exist.
Verifying Roots With Desmos
Whichever method you use, the embedded Desmos calculator gives you a fast final check: graph \(y = ax^2+bx+c\) and look at where the curve crosses the \(x\)-axis. The crossing points should match your algebraic solutions, including irrational ones displayed as decimals. This is particularly useful on a question like the one above, where it's easy to make an arithmetic slip under the square root — a quick graph confirms whether your two roots are in the right neighborhood before you commit to an answer.
Practicing These Methods Under Timed Conditions
Reading about three methods is one thing; recognizing which one to use in eight seconds during a timed section is another. Our topic-wise SAT Math practice sets include Nonlinear Equations problems built specifically around these three solving methods, with instant scoring and a full explanation for every question so you can see which approach the explanation used versus the one you picked. For a broader view of how quadratics fit into the wider domain, see our guide on nonlinear equations and systems of equations.
Frequently asked questions
Which method should I default to if I'm not sure?
Try factoring briefly first since it's fastest when it works. If the coefficients look awkward or you can't find integer factors quickly, move directly to the quadratic formula rather than guessing at factors.
Why does completing the square matter if the formula always works?
Completing the square converts standard form into vertex form, \(f(x) = a(x-h)^2+k\), which is what you need when a question asks for the maximum, minimum, or vertex coordinates directly.
What if the discriminant is negative?
A negative discriminant, \(b^2-4ac < 0\), means there are no real solutions — the parabola never crosses the \(x\)-axis. On the SAT this usually shows up as a question about the number of solutions rather than asking you to find complex roots.
How should I handle a grid-in answer that comes out irrational?
Convert it to a decimal and enter as many digits as the grid allows, or check whether the question specifies a rounding instruction. Never try to enter a radical symbol directly into a grid-in box.