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!)

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!

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

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

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

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

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

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 $x = 2$

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 $x = 2$

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.

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.

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

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

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

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

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: $c = 6$

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: $c = 6$