Solving Quadratic Equations

Introduction

Quadratic equations are one of the most heavily tested skills in the Advanced Math domain, which accounts for roughly 35% of SAT Math questions. You'll encounter quadratics in pure-algebra form and in real-world contexts, and the SAT will test whether you can choose the fastest method rather than always defaulting to the same one. This note covers all four solution methods and the elegant shortcut formulas for the sum and product of the solutions.

Core Concept

A quadratic equation in one variable has the standard form:

ax^2 + bx + c = 0, \quad a \neq 0

Every such equation has exactly two solutions in the complex numbers (counting multiplicity). The SAT tests four methods for finding those solutions:

Method	Use when…
Factoring	The expression factors over the integers (small, clean coefficients)
Taking square roots	The equation has the form $(x - h)^2 = k$ or $ax^2 = k$ (no $x$ -term)
Completing the square	$a = 1$ and $b$ is even; also useful to derive vertex form
Quadratic formula	Always works; best when the equation doesn't factor cleanly

Choosing fast matters. On a timed test, recognize the structure first: if the right side is a perfect square or there's no $x$ -term, take square roots immediately. If the trinomial factors in seconds, factor it. Reach for the formula only when needed.

Key Formulas & Rules

Quadratic formula (must be memorized — NOT on the reference sheet):

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

Completing the square (for $x^2 + bx + c = 0$ ):

\left(x + \frac{b}{2}\right)^2 = \left(\frac{b}{2}\right)^2 - c

Vieta's formulas — if $r_1$ and $r_2$ are the two solutions of $ax^2 + bx + c = 0$ :

r_1 + r_2 = -\frac{b}{a} \qquad r_1 \cdot r_2 = \frac{c}{a}

These let you find the sum or product of the solutions without solving for each one — a major time-saver when the SAT asks for "the sum of all values of $x$ " or "the product of the solutions."

Note: The discriminant $b^2 - 4ac$ determines how many real solutions exist. That's covered in the sibling note The Discriminant and Number of Solutions — here we focus on actually finding the solutions.

Worked Examples

Example 1

Which values of $x$ satisfy $x^2 - 2x - 15 = 0$ ?

A) $x = 5$ and $x = 3$

B) $x = -5$ and $x = 3$

C) $x = 5$ and $x = -3$

D) $x = -5$ and $x = -3$

Solution:

Recognize the structure: The coefficients are small integers — try factoring first.

Find two numbers whose product is $-15$ and whose sum is $-2$ : that's $-5$ and $+3$ .

x^2 - 2x - 15 = (x - 5)(x + 3) = 0

Set each factor to zero:

x - 5 = 0 \implies x = 5 \qquad x + 3 = 0 \implies x = -3

Verify: $5^2 - 2(5) - 15 = 25 - 10 - 15 = 0$ ✓ and $(-3)^2 - 2(-3) - 15 = 9 + 6 - 15 = 0$ ✓

Why each distractor fails:

A) $x = 5$ and $x = 3$ : treated the constant term as $+15$ instead of $-15$ , choosing two positive integers with product $15$ and difference $2$ — giving $(x-5)(x-3) = x^2 - 8x + 15 \neq x^2 - 2x - 15$ .
B) $x = -5$ and $x = 3$ : factors flipped — $(x+5)(x-3) = x^2 + 2x - 15 \neq x^2 - 2x - 15$ .
D) Both signs wrong — $(x+5)(x+3) = x^2 + 8x + 15 \neq x^2 - 2x - 15$ .

Answer: C

Example 2

What are the solutions of $x^2 + 4x - 1 = 0$ ?

A) $x = -2 \pm \sqrt{5}$

B) $x = 2 \pm \sqrt{5}$

C) $x = -2 \pm \sqrt{3}$

D) $x = -2 \pm 2\sqrt{5}$

Solution:

Recognize the structure: $a = 1$ and $b = 4$ is even — completing the square is efficient.

x^2 + 4x = 1

Add $\left(\dfrac{4}{2}\right)^2 = 4$ to both sides:

x^2 + 4x + 4 = 1 + 4 = 5

(x + 2)^2 = 5

x + 2 = \pm\sqrt{5}

x = -2 \pm \sqrt{5}

Verify (sum and product check): Sum $= (-2 + \sqrt{5}) + (-2 - \sqrt{5}) = -4 = -b/a = -4/1$ ✓. Product $= (-2+\sqrt{5})(-2-\sqrt{5}) = 4 - 5 = -1 = c/a = -1/1$ ✓.

Desmos check

Desmos check: Graph $y = x^2 + 4x - 1$ and read the two $x$ -intercepts. They appear at approximately $x \approx 0.236$ and $x \approx -4.236$ , matching $-2 \pm 2.236$ .

Why each distractor fails:

B) $x = 2 \pm \sqrt{5}$ : wrote $(x - 2)^2$ instead of $(x + 2)^2$ — flipped the sign of the half-coefficient.
C) $x = -2 \pm \sqrt{3}$ : added only $\dfrac{b}{2} = 2$ (not $\left(\dfrac{b}{2}\right)^2 = 4$ ) to the right side, giving $(x+2)^2 = 3$ .
D) $x = -2 \pm 2\sqrt{5}$ : applied the quadratic formula and reached $\dfrac{-4 \pm 2\sqrt{5}}{2}$ , then divided only the $-4$ by $2$ while leaving the $2\sqrt{5}$ undivided (partial division of the numerator — see Common Mistake #4), yielding $-2 \pm 2\sqrt{5}$ instead of the correct $-2 \pm \sqrt{5}$ .

Answer: A

Common Mistakes & Traps

Sign error in the quadratic formula. The numerator is $-b \pm \sqrt{\ldots}$ , not $b \pm \sqrt{\ldots}$ . When $b$ is negative, $-b$ is positive — work this out explicitly each time.
Forgetting $\pm$ when taking square roots. $(x-3)^2 = 20$ gives $x - 3 = \pm\sqrt{20}$ , yielding two solutions. Students who write only $+\sqrt{20}$ lose one answer.
Not adding to both sides when completing the square. When you add $\left(\frac{b}{2}\right)^2$ to the left, you must add the same value to the right.
Dividing by $2a$ only partially. In the quadratic formula, $2a$ divides the entire numerator $-b \pm \sqrt{b^2 - 4ac}$ , not just the $-b$ term.
Confusing Vieta's sum and product. The sum is $-b/a$ (note the negative sign) and the product is $c/a$ (no negative sign unless $c$ is negative). Many students mix up which formula gets the minus sign.
Stopping at one solution. Quadratics have two solutions. If you factor and find one root, don't forget the second.
Applying the quadratic formula without $a = 1$ . If $a \neq 1$ , divide the whole equation by $a$ before completing the square, or keep careful track of $a$ in the formula. Dividing by only $a$ instead of $2a$ in the formula is a very common SAT distractor.

Practice Questions

Question 1 (Student-produced response)

For the equation $3x^2 - 7x + 4 = 0$ , what is the product of its two solutions? Enter your answer as a fraction or decimal.

Show answer

Answer: $\dfrac{4}{3}$ (or 1.333…)

Solution: By Vieta's product formula, the product of the solutions equals $\dfrac{c}{a} = \dfrac{4}{3}$ .

To verify by factoring: $3x^2 - 7x + 4 = (3x - 4)(x - 1) = 0$ , giving $x = \tfrac{4}{3}$ and $x = 1$ . Product $= \tfrac{4}{3} \times 1 = \tfrac{4}{3}$ ✓.

Acceptable entries: 4/3 or 1.33 (the test accepts any correctly rounded or truncated decimal, but 4/3 is exact).

Question 2 (Multiple choice)

What is the sum of the solutions of $5x^2 + 3x - 2 = 0$ ?

A) $-\dfrac{3}{5}$

B) $\dfrac{3}{5}$

C) $\dfrac{2}{5}$

D) $-\dfrac{2}{5}$

Show answer

Answer: A

Solution: By Vieta's sum formula: $r_1 + r_2 = -\dfrac{b}{a} = -\dfrac{3}{5}$ .

To verify: factor $5x^2 + 3x - 2 = (5x - 2)(x + 1) = 0$ , giving $x = \tfrac{2}{5}$ and $x = -1$ . Sum $= \tfrac{2}{5} + (-1) = \tfrac{2}{5} - \tfrac{5}{5} = -\tfrac{3}{5}$ ✓.

Why other options fail:

B) $\tfrac{3}{5}$ : used $+b/a$ instead of $-b/a$ — dropped the negative sign.
C) $\tfrac{2}{5}$ : found only one solution ( $x = \tfrac{2}{5}$ ) and stopped.
D) $-\tfrac{2}{5}$ : used the product formula $c/a = -\tfrac{2}{5}$ and mistook it for the sum.

Question 3 (Multiple choice)

What are the solutions of $(x - 3)^2 = 20$ ?

A) $x = 3 \pm 2\sqrt{5}$

B) $x = 3 \pm \sqrt{20}$

C) $x = -3 \pm 2\sqrt{5}$

D) $x = 3 \pm 4\sqrt{5}$

Show answer

Answer: A

Solution: The equation is already in perfect-square form — take square roots immediately.

x - 3 = \pm\sqrt{20} = \pm 2\sqrt{5}

x = 3 \pm 2\sqrt{5}

Verify: $(3 + 2\sqrt{5} - 3)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20$ ✓.

Why other options fail:

B) $x = 3 \pm \sqrt{20}$ : correct process but failed to simplify $\sqrt{20} = 2\sqrt{5}$ .
C) $x = -3 \pm 2\sqrt{5}$ : sign error — moved 3 to the right with the wrong sign ( $x = -3$ instead of $x = 3$ ).
D) $x = 3 \pm 4\sqrt{5}$ : computed $\sqrt{20} = 4\sqrt{5}$ (confused $\sqrt{4 \cdot 5} = \sqrt{4} \cdot \sqrt{5} = 4\sqrt{5}$ — took $\sqrt{4} = 4$ instead of $2$ ).

Question 4 (Multiple choice)

Which of the following gives the solutions of $x^2 - 5x + 3 = 0$ ?

A) $x = \dfrac{5 \pm \sqrt{13}}{2}$

B) $x = \dfrac{-5 \pm \sqrt{13}}{2}$

C) $x = \dfrac{5 \pm \sqrt{37}}{2}$

D) $x = 5 \pm \sqrt{13}$

Show answer

Answer: A

Solution: The equation doesn't factor over the integers (no integer pair multiplies to $3$ and sums to $-5$ ), so use the quadratic formula with $a = 1$ , $b = -5$ , $c = 3$ :

x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(3)}}{2(1)} = \frac{5 \pm \sqrt{25 - 12}}{2} = \frac{5 \pm \sqrt{13}}{2}

Verify (Vieta's): Sum $= \tfrac{5+\sqrt{13}}{2} + \tfrac{5-\sqrt{13}}{2} = \tfrac{10}{2} = 5 = -b/a = 5$ ✓. Product $= \tfrac{(5)^2 - 13}{4} = \tfrac{12}{4} = 3 = c/a$ ✓.

Desmos check

Desmos check: Graph $y = x^2 - 5x + 3$ and read the $x$ -intercepts. They fall near $x \approx 0.697$ and $x \approx 4.303$ , which match $(5 \pm \sqrt{13})/2 \approx (5 \pm 3.606)/2$ .

Why other options fail:

B) $\dfrac{-5 \pm \sqrt{13}}{2}$ : used $b = -5$ instead of $-b = 5$ in the numerator (dropped the negative-of-negative).
C) $\dfrac{5 \pm \sqrt{37}}{2}$ : computed $b^2 + 4ac = 25 + 12 = 37$ (used $+4ac$ instead of $-4ac$ , i.e., ignored that the formula subtracts).
D) $5 \pm \sqrt{13}$ : divided only by $a = 1$ instead of $2a = 2$ — dropped the factor of 2 in the denominator.

Question 5 (Student-produced response)

If $4x^2 - 36 = 0$ and $x > 0$ , what is the value of $x$ ?

Show answer

Answer: 3

Solution: Isolate $x^2$ directly — no $x$ -term means taking square roots is the fastest method.

4x^2 = 36 \implies x^2 = 9 \implies x = \pm 3

Since $x > 0$ , the answer is $x = 3$ .

Verify: $4(3)^2 - 36 = 36 - 36 = 0$ ✓.

Connections

Prerequisite — Factoring Polynomials: Factoring quadratics rests on recognizing factor pairs and the AC method. Review that skill if factoring trinomials with $a \neq 1$ feels slow.
Sibling — The Discriminant and Number of Solutions: Once you know how to solve, that note covers how many real solutions exist and what the discriminant tells you about the graph.
Sibling — Systems of Linear and Nonlinear Equations: Substituting a linear expression into a quadratic produces a quadratic equation — every method here applies there.
Sibling — Solving Formulas for a Variable: Sometimes the SAT asks you to isolate a squared variable inside a formula, which reduces to taking square roots or the quadratic formula.
Sibling — Absolute Value, Radical, and Rational Equations: Squaring both sides to eliminate a radical creates a quadratic; you'll need to check for extraneous solutions.
Geometry connection: The SAT sometimes embeds quadratics in area or distance problems — setting up the equation is half the work; these methods finish it.

Introduction

Core Concept

Key Formulas & Rules

Worked Examples

Example 1

Example 2

Common Mistakes & Traps

Practice Questions

Connections

Figures

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