# Quadratics on the SAT Canonical: https://duckyhelper.com/learn/sat-math/quadratics/ Updated: 2026-10-01 A quadratic has an \(x^2\) term, like \(ax^2 + bx + c\), and its graph is a parabola. Solve \(ax^2 + bx + c = 0\) by factoring, by square roots, or with the quadratic formula. The SAT also asks for the vertex (the highest or lowest point), the intercepts, and how many real solutions there are, which the discriminant \(b^2 - 4ac\) tells you without solving. ## The key idea The same parabola can be written three ways. Each form shows different features at a glance: **Three forms of a quadratic** | Form | Looks like | Shows you | | --- | --- | --- | | Standard | \(y = ax^2 + bx + c\) | y-intercept \(c\); vertex at \(x = -\frac{b}{2a}\) | | Factored | \(y = a(x - r)(x - s)\) | x-intercepts r and s | | Vertex | \(y = a(x - h)^2 + k\) | Vertex \((h, k)\) | If \(a > 0\) the parabola opens up and the vertex is a minimum. If \(a < 0\) it opens down and the vertex is a maximum. When factoring does not work, the quadratic formula always does: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ The part under the root, \(b^2 - 4ac\), is the discriminant. Positive means two real solutions, zero means exactly one, negative means none. ## Worked examples **Example 1: factoring** Problem: Solve \(x^2 - 5x - 14 = 0\). 1. Find two numbers that multiply to \(-14\) and add to \(-5\): they are \(-7\) and 2. $$ (x - 7)(x + 2) = 0 $$ 2. A product is 0 only if a factor is 0. $$ x - 7 = 0 $$ 3. Or the other factor: $$ x + 2 = 0 $$ Answer: \(x = 7\) or \(x = -2\) **Example 2: the quadratic formula** Problem: Solve \(2x^2 + 3x - 4 = 0\). 1. Here \(a = 2\), \(b = 3\), \(c = -4\). Find the discriminant first. $$ b^2 - 4ac = 9 - 4(2)(-4) = 41 $$ 2. 41 is positive, so there are two real solutions. Put everything into the formula. $$ x = \frac{-3 \pm \sqrt{41}}{4} $$ Answer: \(x = \frac{-3 + \sqrt{41}}{4}\) or \(x = \frac{-3 - \sqrt{41}}{4}\) **Example 3: the vertex** Problem: What is the minimum value of \(y = 2x^2 - 12x + 7\)? 1. The vertex is at \(x = -\frac{b}{2a}\). $$ x = \frac{12}{2 \cdot 2} = 3 $$ 2. Put \(x = 3\) back in to get the y value. $$ 2(3)^2 - 12(3) + 7 = -11 $$ 3. Since \(a = 2 > 0\), the parabola opens up, so the vertex \((3, -11)\) is the lowest point. Answer: The minimum value is \(-11\), at \(x = 3\). **Example 4 (SAT-hard): a constant and the discriminant** Problem: For what value of c does \(3x^2 - 12x + c = 0\) have exactly one real solution? 1. Exactly one real solution means the discriminant is 0. $$ (-12)^2 - 4(3)(c) = 0 $$ 2. Simplify. $$ 144 - 12c = 0 $$ 3. Solve for c. $$ c = 12 $$ 4. Check: \(3x^2 - 12x + 12 = 3(x - 2)^2\), which is 0 only at \(x = 2\). Answer: \(c = 12\) ## Common mistakes - **Dividing both sides by x.** From \(x^2 = 5x\), dividing by x loses the solution \(x = 0\). Fix: move everything to one side and factor: \(x(x - 5) = 0\). - **Reading the vertex sign wrong.** \(y = (x + 4)^2 - 9\) has vertex \((-4, -9)\), not \((4, -9)\). Fix: write it as \((x - (-4))^2\). - **Forgetting the \(\pm\).** \(x^2 = 49\) has two solutions, 7 and \(-7\). Fix: write \(\pm\) as soon as you take a square root. - **Sign slips in the discriminant.** With \(c = -4\), \(-4ac\) becomes \(+32\). Fix: put negative values in parentheses. - **Setting factors equal to the wrong number.** In \((x - 7)(x + 2) = 0\), the solutions are 7 and \(-2\), the opposite signs of the numbers in the factors. ## Quick methods > **Tip: Sum and product of solutions** > > For \(ax^2 + bx + c = 0\), the solutions add to \(-\frac{b}{a}\) and multiply to \(\frac{c}{a}\). If a question asks only for the sum or product, you do not need to solve. This is exact and works even when the solutions are messy. > **Note: Desmos shows vertex and zeros** > > Graph \(y = 2x^2 - 12x + 7\) in Desmos and tap the curve. Gray dots appear at the vertex and the x-intercepts, with their coordinates. It is the fastest way to find a vertex when the numbers are not friendly. See [Desmos on the SAT](https://duckyhelper.com/learn/sat-math/desmos-calculator/). ## Practice **5 SAT-style questions** 1. What are the solutions to \(x^2 + 2x - 24 = 0\)? A. \(-6\) and \(4\) B. \(6\) and \(-4\) C. \(-8\) and \(3\) D. \(-12\) and \(2\) Answer: \(-6\) and \(4\). \((x + 6)(x - 4) = 0\), so \(x = -6\) or \(x = 4\). \(6\) and \(-4\) use the signs inside the factors instead of their opposites. The other pairs multiply to \(-24\) but do not add to 2. 2. What is the vertex of the graph of \(y = -(x - 4)^2 + 9\)? A. \((4, 9)\) B. \((-4, 9)\) C. \((9, 4)\) D. \((4, -9)\) Answer: \((4, 9)\). In \(y = a(x - h)^2 + k\), the vertex is \((h, k) = (4, 9)\). The minus in front of the square only makes the parabola open down, so 9 is the maximum. 3. How many real solutions does \(2x^2 + 5x + 4 = 0\) have? A. Zero B. Exactly one C. Exactly two D. Infinitely many Answer: Zero. The discriminant is \(5^2 - 4(2)(4) = 25 - 32 = -7\). It is negative, so the parabola never touches the x-axis and there are no real solutions. 4. What is the sum of the solutions of \(3x^2 - 18x + 5 = 0\)? A. \(-6\) B. \(\frac{5}{3}\) C. \(6\) D. \(18\) Answer: \(6\). The sum is \(-\frac{b}{a} = -\frac{-18}{3} = 6\). \(\frac{5}{3}\) is the product, \(\frac{c}{a}\). \(-6\) forgets the minus sign in \(-\frac{b}{a}\). 5. Student-produced response: a ball's height in feet t seconds after it is thrown is \(h(t) = -16t^2 + 64t + 5\). What is the greatest height the ball reaches, in feet? Answer: 69. The vertex is at \(t = -\frac{64}{2(-16)} = 2\). Then \(h(2) = -64 + 128 + 5 = 69\). Entering 2 gives the time of the peak, not its height. ## Frequently asked questions ### Should I factor or use the quadratic formula? Try factoring for about ten seconds when a is 1 and the numbers are small. If nothing works, use the quadratic formula, which always works. On the SAT, graphing in Desmos is a third option when the answer choices are decimals. ### What does the discriminant tell you? It tells you how many real solutions the equation has, without solving it. \(b^2 - 4ac > 0\) means two, \(= 0\) means one (the vertex touches the x-axis), and \(< 0\) means none. SAT questions often use it to find an unknown constant. ### How do I find the vertex of a parabola? In standard form, the x value is \(-\frac{b}{2a}\); plug it in to get y. In vertex form \(a(x - h)^2 + k\), the vertex is \((h, k)\). In factored form, it is halfway between the x-intercepts. ### How many quadratic questions are on the SAT? College Board does not publish a count per topic. Quadratics belong to the Advanced Math domain, which has 13 to 15 of the 44 math questions, and quadratic ideas also appear in systems and word problems. ## Sources - [College Board: SAT Math, Advanced Math skills](https://satsuite.collegeboard.org/sat/whats-on-the-test/math/types/advanced), accessed 2026-10-01 - [College Board: SAT Math overview (questions per domain)](https://satsuite.collegeboard.org/sat/whats-on-the-test/math/overview), accessed 2026-10-01 ## Related - [Equivalent expressions on the SAT](https://duckyhelper.com/learn/sat-math/equivalent-expressions/) - [Nonlinear equations and systems on the SAT](https://duckyhelper.com/learn/sat-math/nonlinear-equations-and-systems/) - [Completing the square](https://duckyhelper.com/learn/algebra-2/completing-the-square/) - [How to solve quadratic equations](https://duckyhelper.com/learn/algebra-1/solving-quadratics/) - [SAT Math study guides](https://duckyhelper.com/learn/sat-math/) ## Try asking Ducky - "I got x = 7 and x = 2. Why is it -2?" - "When do I use the discriminant instead of solving?" - "Show me how to find the vertex in Desmos." - "Give me three more questions like the one with c." ## Get DuckyHelper Free to start. The web app works in any browser, Chromebooks included; the Mac app can also draw on your real screen. [Try it free in your browser](https://app.duckyhelper.com/?utm_source=duckyhelper.com&utm_medium=learn) or [Get the Mac app](https://duckyhelper.com/download/)