Quadratic Equations

Q: How can I study Quadratic Equations effectively?

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 21 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Q: Is this Quadratic Equations study guide free?

Yes — all study notes, flashcards, and practice problems for Quadratic Equations on Study Mondo are free to access. No account is needed.

Q: What course covers Quadratic Equations?

Quadratic Equations is part of the SAT Prep course on Study Mondo, specifically in the Advanced Math section. You can explore the full course for more related topics and practice resources.

Q: Are there practice problems for Quadratic Equations?

Yes, this page includes 21 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Solve quadratics by factoring, completing the square, and quadratic formula

Written and reviewed by the Study Mondo Education TeamLast updated February 16, 2026

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Quadratic Equations (SAT)

Standard Form

$ax^2 + bx + c = 0$

Method 1: Factoring

Example: $x^2 + 5x + 6 = 0$

Factor: $(x + 2)(x + 3) = 0$

Solutions: $x = -2$ or $x = -3$

Method 2: Quadratic Formula

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

Example: $x^2 - 3x - 10 = 0$

Here: $a = 1, b = -3, c = -10$

$x = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm 7}{2}$

Solutions: $x = 5$ or $x = -2$

The Discriminant

$\Delta = b^2 - 4ac$

If $\Delta > 0$ : Two real solutions
If $\Delta = 0$ : One real solution
If $\Delta < 0$ : No real solutions

Vertex Form

$y = a(x - h)^2 + k$

Vertex is at $(h, k)$

SAT Tips

Factor when possible (fastest method)
Quadratic formula works every time
Watch for: $x^2 = 16$ means $x = \pm 4$ (two solutions!)

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve: $x^2 - 9 = 0$

💡 Show Solution

Solution:

Add 9 to both sides: $x^2 = 9$

Take square root (remember ±): $x = \pm 3$

Answer: $x = 3$ or $x = -3$

SAT Tip: Don't forget the negative solution!

2Problem 2easy

❓ Question:

Solve: $x^2 - 9 = 0$

💡 Show Solution

Solution:

Add 9 to both sides: $x^2 = 9$

Take square root (remember ±): $x = \pm 3$

Answer: $x = 3$ or $x = -3$

SAT Tip: Don't forget the negative solution!

3Problem 3medium

❓ Question:

Solve by factoring: $x^2 + 7x + 12 = 0$

💡 Show Solution

Solution:

Factor (find two numbers that multiply to 12 and add to 7): $(x + 3)(x + 4) = 0$

Set each factor to zero: $x + 3 = 0 \quad \Rightarrow \quad x = -3$ $x + 4 = 0 \quad \Rightarrow \quad x = -4$

Answer: $x = -3$ or $x = -4$

4Problem 4medium

❓ Question:

Solve by factoring: $x^2 + 7x + 12 = 0$

💡 Show Solution

Solution:

Factor (find two numbers that multiply to 12 and add to 7): $(x + 3)(x + 4) = 0$

Set each factor to zero: $x + 3 = 0 \quad \Rightarrow \quad x = -3$ $x + 4 = 0 \quad \Rightarrow \quad x = -4$

Answer: $x = -3$ or $x = -4$

5Problem 5hard

❓ Question:

Use the quadratic formula to solve: $2x^2 - 5x - 3 = 0$

💡 Show Solution

Solution:

Identify: $a = 2, b = -5, c = -3$

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$

$x = \frac{5 \pm \sqrt{25 + 24}}{4}$

$x = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$

$x = \frac{12}{4} = 3 \quad \text{or} \quad x = \frac{-2}{4} = -\frac{1}{2}$

Answer: $x = 3$ or $x = -\frac{1}{2}$

6Problem 6hard

❓ Question:

Use the quadratic formula to solve: $2x^2 - 5x - 3 = 0$

💡 Show Solution

Solution:

Identify: $a = 2, b = -5, c = -3$

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$

$x = \frac{5 \pm \sqrt{25 + 24}}{4}$

$x = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$

$x = \frac{12}{4} = 3 \quad \text{or} \quad x = \frac{-2}{4} = -\frac{1}{2}$

Answer: $x = 3$ or $x = -\frac{1}{2}$

7Problem 7easy

❓ Question:

Solve: $x^2 - 9 = 0$

💡 Show Solution

Method: Difference of Squares

$x^2 - 9 = 0$ $x^2 = 9$ $x = \pm 3$

Or by factoring: $(x-3)(x+3) = 0$ , so $x = 3$ or $x = -3$ .

Answer: $x = 3$ or $x = -3$

8Problem 8easy

❓ Question:

Solve: $x^2 - 9 = 0$

💡 Show Solution

Method: Difference of Squares

$x^2 - 9 = 0$ $x^2 = 9$ $x = \pm 3$

Or by factoring: $(x-3)(x+3) = 0$ , so $x = 3$ or $x = -3$ .

Answer: $x = 3$ or $x = -3$

9Problem 9easy

❓ Question:

Solve: $x^2 - 9 = 0$

💡 Show Solution

Method: Difference of Squares

$x^2 - 9 = 0$ $x^2 = 9$ $x = \pm 3$

Or by factoring: $(x-3)(x+3) = 0$ , so $x = 3$ or $x = -3$ .

Answer: $x = 3$ or $x = -3$

10Problem 10medium

❓ Question:

Solve by factoring: $x^2 + 5x - 14 = 0$

💡 Show Solution

Step 1: Find two numbers that multiply to $-14$ and add to $+5$ . $7 \times (-2) = -14$ and $7 + (-2) = 5$ ✓

Step 2: Factor: $(x + 7)(x - 2) = 0$

Step 3: Apply zero product property: $x + 7 = 0 \implies x = -7$ $x - 2 = 0 \implies x = 2$

Check: $(-7)^2 + 5(-7) - 14 = 49 - 35 - 14 = 0$ ✓ $(2)^2 + 5(2) - 14 = 4 + 10 - 14 = 0$ ✓

Answer: $x = -7$ or $x = 2$

11Problem 11medium

❓ Question:

Solve by factoring: $x^2 + 5x - 14 = 0$

💡 Show Solution

Step 1: Find two numbers that multiply to $-14$ and add to $+5$ . $7 \times (-2) = -14$ and $7 + (-2) = 5$ ✓

Step 2: Factor: $(x + 7)(x - 2) = 0$

Step 3: Apply zero product property: $x + 7 = 0 \implies x = -7$ $x - 2 = 0 \implies x = 2$

Check: $(-7)^2 + 5(-7) - 14 = 49 - 35 - 14 = 0$ ✓ $(2)^2 + 5(2) - 14 = 4 + 10 - 14 = 0$ ✓

Answer: $x = -7$ or $x = 2$

12Problem 12medium

❓ Question:

Solve by factoring: $x^2 + 5x - 14 = 0$

💡 Show Solution

Step 1: Find two numbers that multiply to $-14$ and add to $+5$ . $7 \times (-2) = -14$ and $7 + (-2) = 5$ ✓

Step 2: Factor: $(x + 7)(x - 2) = 0$

Step 3: Apply zero product property: $x + 7 = 0 \implies x = -7$ $x - 2 = 0 \implies x = 2$

Check: $(-7)^2 + 5(-7) - 14 = 49 - 35 - 14 = 0$ ✓ $(2)^2 + 5(2) - 14 = 4 + 10 - 14 = 0$ ✓

Answer: $x = -7$ or $x = 2$

13Problem 13medium

❓ Question:

The graph of $y = (x-3)^2 - 4$ is a parabola. What are its vertex, axis of symmetry, and $x$ -intercepts?

💡 Show Solution

This is in vertex form: $y = (x-h)^2 + k$ where vertex = $(h, k)$

Vertex: $(3, -4)$

Axis of symmetry: $x = 3$ (vertical line through the vertex)

$x$ -intercepts: Set $y = 0$ : $(x-3)^2 - 4 = 0$ $(x-3)^2 = 4$ $x - 3 = \pm 2$ $x = 5 \text{ or } x = 1$

Answer: Vertex $(3, -4)$ , axis $x = 3$ , $x$ -intercepts at $(1, 0)$ and $(5, 0)$ .

Since the coefficient of $(x-3)^2$ is positive ( $+1$ ), the parabola opens upward.

14Problem 14medium

❓ Question:

The graph of $y = (x-3)^2 - 4$ is a parabola. What are its vertex, axis of symmetry, and $x$ -intercepts?

💡 Show Solution

This is in vertex form: $y = (x-h)^2 + k$ where vertex = $(h, k)$

Vertex: $(3, -4)$

Axis of symmetry: $x = 3$ (vertical line through the vertex)

$x$ -intercepts: Set $y = 0$ : $(x-3)^2 - 4 = 0$ $(x-3)^2 = 4$ $x - 3 = \pm 2$ $x = 5 \text{ or } x = 1$

Answer: Vertex $(3, -4)$ , axis $x = 3$ , $x$ -intercepts at $(1, 0)$ and $(5, 0)$ .

Since the coefficient of $(x-3)^2$ is positive ( $+1$ ), the parabola opens upward.

15Problem 15medium

❓ Question:

The graph of $y = (x-3)^2 - 4$ is a parabola. What are its vertex, axis of symmetry, and $x$ -intercepts?

💡 Show Solution

This is in vertex form: $y = (x-h)^2 + k$ where vertex = $(h, k)$

Vertex: $(3, -4)$

Axis of symmetry: $x = 3$ (vertical line through the vertex)

$x$ -intercepts: Set $y = 0$ : $(x-3)^2 - 4 = 0$ $(x-3)^2 = 4$ $x - 3 = \pm 2$ $x = 5 \text{ or } x = 1$

Answer: Vertex $(3, -4)$ , axis $x = 3$ , $x$ -intercepts at $(1, 0)$ and $(5, 0)$ .

Since the coefficient of $(x-3)^2$ is positive ( $+1$ ), the parabola opens upward.

16Problem 16hard

❓ Question:

For the equation $2x^2 - 5x + k = 0$ to have exactly one real solution, what must be the value of $k$ ?

💡 Show Solution

Key concept: A quadratic has exactly one real solution when the discriminant equals zero.

Discriminant formula: $b^2 - 4ac$

Here: $a = 2$ , $b = -5$ , $c = k$

$b^2 - 4ac = 0$ $(-5)^2 - 4(2)(k) = 0$ $25 - 8k = 0$ $k = \frac{25}{8}$

Check: $2x^2 - 5x + \frac{25}{8} = 0$ $\Delta = 25 - 4(2)(\frac{25}{8}) = 25 - 25 = 0$ ✓ (one solution)

Answer: $k = \frac{25}{8}$

Discriminant summary:

$\Delta > 0$ : two real solutions
$\Delta = 0$ : one real solution (double root)
$\Delta < 0$ : no real solutions

17Problem 17hard

❓ Question:

For the equation $2x^2 - 5x + k = 0$ to have exactly one real solution, what must be the value of $k$ ?

💡 Show Solution

Key concept: A quadratic has exactly one real solution when the discriminant equals zero.

Discriminant formula: $b^2 - 4ac$

Here: $a = 2$ , $b = -5$ , $c = k$

$b^2 - 4ac = 0$ $(-5)^2 - 4(2)(k) = 0$ $25 - 8k = 0$ $k = \frac{25}{8}$

Check: $2x^2 - 5x + \frac{25}{8} = 0$ $\Delta = 25 - 4(2)(\frac{25}{8}) = 25 - 25 = 0$ ✓ (one solution)

Answer: $k = \frac{25}{8}$

Discriminant summary:

$\Delta > 0$ : two real solutions
$\Delta = 0$ : one real solution (double root)
$\Delta < 0$ : no real solutions

18Problem 18hard

❓ Question:

For the equation $2x^2 - 5x + k = 0$ to have exactly one real solution, what must be the value of $k$ ?

💡 Show Solution

Key concept: A quadratic has exactly one real solution when the discriminant equals zero.

Discriminant formula: $b^2 - 4ac$

Here: $a = 2$ , $b = -5$ , $c = k$

$b^2 - 4ac = 0$ $(-5)^2 - 4(2)(k) = 0$ $25 - 8k = 0$ $k = \frac{25}{8}$

Check: $2x^2 - 5x + \frac{25}{8} = 0$ $\Delta = 25 - 4(2)(\frac{25}{8}) = 25 - 25 = 0$ ✓ (one solution)

Answer: $k = \frac{25}{8}$

Discriminant summary:

$\Delta > 0$ : two real solutions
$\Delta = 0$ : one real solution (double root)
$\Delta < 0$ : no real solutions

19Problem 19expert

❓ Question:

If the sum of the solutions of $3x^2 - 12x + c = 0$ is twice the product of the solutions, find $c$ .

💡 Show Solution

Step 1: Use Vieta's formulas for $ax^2 + bx + c = 0$ :

Sum of solutions: $x_1 + x_2 = -\frac{b}{a} = -\frac{-12}{3} = 4$
Product of solutions: $x_1 \cdot x_2 = \frac{c}{a} = \frac{c}{3}$

Step 2: Apply the condition: sum = twice the product $4 = 2 \cdot \frac{c}{3}$ $4 = \frac{2c}{3}$ $12 = 2c$ $c = 6$

Check: $3x^2 - 12x + 6 = 0 \implies x^2 - 4x + 2 = 0$ $x = \frac{4 \pm \sqrt{16-8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$ Sum = $4$ ✓, Product = $(2+\sqrt{2})(2-\sqrt{2}) = 4-2 = 2$ ✓ Is $4 = 2(2)$ ? Yes ✓

Answer: $c = 6$

20Problem 20expert

❓ Question:

If the sum of the solutions of $3x^2 - 12x + c = 0$ is twice the product of the solutions, find $c$ .

💡 Show Solution

Step 1: Use Vieta's formulas for $ax^2 + bx + c = 0$ :

Sum of solutions: $x_1 + x_2 = -\frac{b}{a} = -\frac{-12}{3} = 4$
Product of solutions: $x_1 \cdot x_2 = \frac{c}{a} = \frac{c}{3}$

Step 2: Apply the condition: sum = twice the product $4 = 2 \cdot \frac{c}{3}$ $4 = \frac{2c}{3}$ $12 = 2c$ $c = 6$

Answer: $c = 6$

21Problem 21expert

❓ Question:

If the sum of the solutions of $3x^2 - 12x + c = 0$ is twice the product of the solutions, find $c$ .

💡 Show Solution

Step 1: Use Vieta's formulas for $ax^2 + bx + c = 0$ :

Sum of solutions: $x_1 + x_2 = -\frac{b}{a} = -\frac{-12}{3} = 4$
Product of solutions: $x_1 \cdot x_2 = \frac{c}{a} = \frac{c}{3}$

Step 2: Apply the condition: sum = twice the product $4 = 2 \cdot \frac{c}{3}$ $4 = \frac{2c}{3}$ $12 = 2c$ $c = 6$

Answer: $c = 6$

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❓ Frequently Asked Questions

What is Quadratic Equations?▾

Solve quadratics by factoring, completing the square, and quadratic formula

How can I study Quadratic Equations effectively?▾

Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 21 problems provided, checking solutions as you go. Regular review and active practice are key to retention.

Is this Quadratic Equations study guide free?▾

Yes — all study notes, flashcards, and practice problems for Quadratic Equations on Study Mondo are free to access. No account is needed.

What course covers Quadratic Equations?▾

Quadratic Equations is part of the SAT Prep course on Study Mondo, specifically in the Advanced Math section. You can explore the full course for more related topics and practice resources.

Are there practice problems for Quadratic Equations?▾

Yes, this page includes 21 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.

Quadratic Equations

Try the Interactive Version!

Quadratic Equations (SAT)

Standard Form

Method 1: Factoring

Method 2: Quadratic Formula

The Discriminant

Vertex Form

SAT Tips

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2easy

❓ Question:

3Problem 3medium

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4Problem 4medium

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5Problem 5hard

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6Problem 6hard

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7Problem 7easy

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8Problem 8easy

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9Problem 9easy

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10Problem 10medium

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11Problem 11medium

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12Problem 12medium

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13Problem 13medium

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14Problem 14medium

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15Problem 15medium

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16Problem 16hard

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17Problem 17hard

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18Problem 18hard

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19Problem 19expert

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20Problem 20expert

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21Problem 21expert

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❓ Frequently Asked Questions