Quadratic Equations

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

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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 2medium

❓ 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 $x = -4$

3Problem 3hard

❓ 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 $x = -\frac{1}{2}$

4Problem 4easy

❓ 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 $x = -3$

5Problem 5easy

❓ 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 $x = -3$

6Problem 6medium

❓ 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 $x = 2$

7Problem 7medium

❓ 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 $x = 2$

8Problem 8medium

❓ 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.

9Problem 9medium

❓ 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.

10Problem 10hard

❓ 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

11Problem 11hard

❓ 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

12Problem 12expert

❓ 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: $c = 6$

13Problem 13expert

❓ 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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