Example Card: Solving a Quadratic Equation by Factoring
Solve quadratic equations by factoring in Grade 9 algebra: rearrange to standard form ax²+bx+c=0, factor completely, then apply the Zero Product Property to each factor to find solutions.
Key Concepts
Let's tackle a quadratic equation that isn't in standard form yet. This example shows one of the most important key ideas from this lesson: Solving Quadratic Equations by Factoring .
Example Problem Find the roots of the equation $3x^2 14x = 5$.
Step by Step 1. First, we need to set the equation equal to 0 by writing it in the standard form $ax^2 + bx + c = 0$. $$3x^2 14x 5 = 0$$ 2. Next, we factor the quadratic expression on the left side. $$(3x+1)(x 5) = 0$$ 3. Now we apply the Zero Product Property. If the product of these two factors is zero, then at least one of the factors must be zero. We set each factor equal to zero. $$ \begin{array}{c|c} 3x + 1 = 0 & x 5 = 0 \\ \end{array} $$ 4. Finally, we solve each of these linear equations for $x$. $$ \begin{array}{c|c} 3x = 1 & x = 5 \\ x = \frac{1}{3} & \\ \end{array} $$ 5. Check your answers by substituting them back into the original equation. $$ \begin{array}{c|c} 3x^2 14x = 5 & 3x^2 14x = 5 \\ 3( \frac{1}{3})^2 14( \frac{1}{3}) \stackrel{?}{=} 5 & 3(5)^2 14(5) \stackrel{?}{=} 5 \\ 3(\frac{1}{9}) + \frac{14}{3} \stackrel{?}{=} 5 & 3(25) 70 \stackrel{?}{=} 5 \\ \frac{1}{3} + \frac{14}{3} \stackrel{?}{=} 5 & 75 70 \stackrel{?}{=} 5 \\ \frac{15}{3} = 5 \quad \checkmark & 5 = 5 \quad \checkmark \\ \end{array} $$ The roots are $ \frac{1}{3}$ and $5$.
Common Questions
What is the first step when solving a quadratic equation by factoring?
Always rearrange the equation to standard form: ax² + bx + c = 0. Move all terms to one side so the other side equals zero. This is essential — the Zero Product Property only works when one side equals exactly zero.
How do you solve x² + 3x = 18 by factoring?
Rearrange: x² + 3x - 18 = 0. Factor: (x + 6)(x - 3) = 0. Apply Zero Product Property: x + 6 = 0 → x = -6, or x - 3 = 0 → x = 3. Verify both: (-6)² + 3(-6) = 36 - 18 = 18 ✓ and 9 + 9 = 18 ✓.
Why do quadratic equations typically have two solutions?
Quadratic equations are degree 2, and the Fundamental Theorem of Algebra states a degree-n polynomial has exactly n roots (counting multiplicity). Factoring into two linear factors gives two values — one from each factor set to zero.