SAT Math · Advanced Math

Quadratics on the SAT

A quadratic has an x2x^2 term, like ax2+bx+cax^2 + bx + c, and its graph is a parabola. Solve ax2+bx+c=0ax^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 b2−4acb^2 - 4ac tells you without solving.

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The key idea

The same parabola can be written three ways. Each form shows different features at a glance:

Three forms of a quadratic
FormLooks likeShows you
Standardy=ax2+bx+cy = ax^2 + bx + cy-intercept cc; vertex at x=−b2ax = -\frac{b}{2a}
Factoredy=a(x−r)(x−s)y = a(x - r)(x - s)x-intercepts r and s
Vertexy=a(x−h)2+ky = a(x - h)^2 + kVertex (h,k)(h, k)

If a>0a > 0 the parabola opens up and the vertex is a minimum. If a<0a < 0 it opens down and the vertex is a maximum.

When factoring does not work, the quadratic formula always does:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The part under the root, b2−4acb^2 - 4ac, is the discriminant. Positive means two real solutions, zero means exactly one, negative means none.

Worked examples

Example 1: factoring

Problem Solve x2−5x−14=0x^2 - 5x - 14 = 0.

  1. Find two numbers that multiply to −14-14 and add to −5-5: they are −7-7 and 2.
    (x−7)(x+2)=0(x - 7)(x + 2) = 0
  2. A product is 0 only if a factor is 0.
    x−7=0x - 7 = 0
  3. Or the other factor:
    x+2=0x + 2 = 0

Answer x=7x = 7 or x=−2x = -2

Example 2: the quadratic formula

Problem Solve 2x2+3x−4=02x^2 + 3x - 4 = 0.

  1. Here a=2a = 2, b=3b = 3, c=−4c = -4. Find the discriminant first.
    b2−4ac=9−4(2)(−4)=41b^2 - 4ac = 9 - 4(2)(-4) = 41
  2. 41 is positive, so there are two real solutions. Put everything into the formula.
    x=−3±414x = \frac{-3 \pm \sqrt{41}}{4}

Answer x=−3+414x = \frac{-3 + \sqrt{41}}{4} or x=−3−414x = \frac{-3 - \sqrt{41}}{4}

Example 3: the vertex

Problem What is the minimum value of y=2x2−12x+7y = 2x^2 - 12x + 7?

  1. The vertex is at x=−b2ax = -\frac{b}{2a}.
    x=122⋅2=3x = \frac{12}{2 \cdot 2} = 3
  2. Put x=3x = 3 back in to get the y value.
    2(3)2−12(3)+7=−112(3)^2 - 12(3) + 7 = -11
  3. Since a=2>0a = 2 > 0, the parabola opens up, so the vertex (3,−11)(3, -11) is the lowest point.

Answer The minimum value is −11-11, at x=3x = 3.

Example 4 (SAT-hard): a constant and the discriminant

Problem For what value of c does 3x2−12x+c=03x^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(-12)^2 - 4(3)(c) = 0
  2. Simplify.
    144−12c=0144 - 12c = 0
  3. Solve for c.
    c=12c = 12
  4. Check: 3x2−12x+12=3(x−2)23x^2 - 12x + 12 = 3(x - 2)^2, which is 0 only at x=2x = 2.

Answer c=12c = 12

Common mistakes

  • Dividing both sides by x. From x2=5xx^2 = 5x, dividing by x loses the solution x=0x = 0. Fix: move everything to one side and factor: x(x−5)=0x(x - 5) = 0.
  • Reading the vertex sign wrong. y=(x+4)2−9y = (x + 4)^2 - 9 has vertex (−4,−9)(-4, -9), not (4,−9)(4, -9). Fix: write it as (x−(−4))2(x - (-4))^2.
  • Forgetting the ±\pm. x2=49x^2 = 49 has two solutions, 7 and −7-7. Fix: write ±\pm as soon as you take a square root.
  • Sign slips in the discriminant. With c=−4c = -4, −4ac-4ac becomes +32+32. Fix: put negative values in parentheses.
  • Setting factors equal to the wrong number. In (x−7)(x+2)=0(x - 7)(x + 2) = 0, the solutions are 7 and −2-2, the opposite signs of the numbers in the factors.

Quick methods

Practice

5 SAT-style questions

  1. What are the solutions to x2+2x−24=0x^2 + 2x - 24 = 0?

    1. −6-6 and 44
    2. 66 and −4-4
    3. −8-8 and 33
    4. −12-12 and 22
    Show answer

    Answer: −6-6 and 44

    (x+6)(x−4)=0(x + 6)(x - 4) = 0, so x=−6x = -6 or x=4x = 4. 66 and −4-4 use the signs inside the factors instead of their opposites. The other pairs multiply to −24-24 but do not add to 2.

  2. What is the vertex of the graph of y=−(x−4)2+9y = -(x - 4)^2 + 9?

    1. (4,9)(4, 9)
    2. (−4,9)(-4, 9)
    3. (9,4)(9, 4)
    4. (4,−9)(4, -9)
    Show answer

    Answer: (4,9)(4, 9)

    In y=a(x−h)2+ky = a(x - h)^2 + k, the vertex is (h,k)=(4,9)(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 2x2+5x+4=02x^2 + 5x + 4 = 0 have?

    1. Zero
    2. Exactly one
    3. Exactly two
    4. Infinitely many
    Show answer

    Answer: Zero

    The discriminant is 52−4(2)(4)=25−32=−75^2 - 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 3x2−18x+5=03x^2 - 18x + 5 = 0?

    1. −6-6
    2. 53\frac{5}{3}
    3. 66
    4. 1818
    Show answer

    Answer: 66

    The sum is −ba=−−183=6-\frac{b}{a} = -\frac{-18}{3} = 6. 53\frac{5}{3} is the product, ca\frac{c}{a}. −6-6 forgets the minus sign in −ba-\frac{b}{a}.

  5. Student-produced response: a ball's height in feet t seconds after it is thrown is h(t)=−16t2+64t+5h(t) = -16t^2 + 64t + 5. What is the greatest height the ball reaches, in feet?

    Show answer

    Answer: 69

    The vertex is at t=−642(−16)=2t = -\frac{64}{2(-16)} = 2. Then h(2)=−64+128+5=69h(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. b2−4ac>0b^2 - 4ac > 0 means two, =0= 0 means one (the vertex touches the x-axis), and <0< 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 −b2a-\frac{b}{2a}; plug it in to get y. In vertex form a(x−h)2+ka(x - h)^2 + k, the vertex is (h,k)(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

  1. College Board: SAT Math, Advanced Math skills, accessed October 1, 2026
  2. College Board: SAT Math overview (questions per domain), accessed October 1, 2026

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.”

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