Factor equations like \( x^2 + 5x + 6 = 0 \) into two binomials.
We want to write the quadratic in the form \( (x + m)(x + n) = 0 \) such that \( m \cdot n = c \) and \( m + n = b \).
\( x^2 + 5x + 6 = 0 \)
We find two numbers that multiply to 6 and add to 5 → (2 and 3)
So: \( (x + 2)(x + 3) = 0 \)
\( x^2 - 7x + 12 = 0 \)
We find two numbers that multiply to 12 and add to -7 → (-3 and -4)
So: \( (x - 3)(x - 4) = 0 \)
Solve equations like \( x^2 + 3x - 10 = 0 \) using factored form.
Once the quadratic is factored, use the Zero Product Property:
\( (x + a)(x + b) = 0 \Rightarrow x + a = 0 \text{ or } x + b = 0 \)
\( x^2 + 3x - 10 = 0 \)
Factor: \( (x + 5)(x - 2) = 0 \)
Solutions: \( x = -5 \), \( x = 2 \)
\( x^2 - 4x - 12 = 0 \)
Factor: \( (x - 6)(x + 2) = 0 \)
Solutions: \( x = 6 \), \( x = -2 \)
Solve quadratics like \( x^2 = 49 \) using square roots.
Use it when the quadratic is in the form \( x^2 = a \) or \( (x + c)^2 = a \).
Take the square root of both sides and include ± in the solution.
\( x^2 = 49 \)
\( x = \pm\sqrt{49} = \pm7 \)
\( x^2 = 20 \)
\( x = \pm\sqrt{20} = \pm2\sqrt{5} \)
\( (x - 3)^2 = 36 \)
\( x - 3 = \pm6 \) → \( x = 9 \) or \( x = -3 \)
\( (x + 4)^2 = 12 \)
\( x + 4 = \pm\sqrt{12} = \pm2\sqrt{3} \) → \( x = -4 \pm 2\sqrt{3} \)
Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve any quadratic.
For any equation \( ax^2 + bx + c = 0 \), use:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\( x^2 - 6x + 5 = 0 \)
\( a = 1, b = -6, c = 5 \)
\( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(5)}}{2(1)} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = 4, 2 \)
\( x^2 + 4x + 1 = 0 \)
\( x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2} = -2 \pm \sqrt{3} \)
\( 2x^2 + 3x - 2 = 0 \)
\( x = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4} \Rightarrow x = \frac{1}{2}, -2 \)
\( x^2 + 2x + 2 = 0 \)
\( x = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i \)