As seen on the quadratic equations introduction page, quadratic equations have the following general form.
ax^{\tt{2}} + bx + c = 0The factor by grouping page showed how to effectively factor quadratics when possible, this page will show how this approach can be used to solve quadratics by factoring.
Solutions of a Quadratic Equation:
Let’s consider a basic example of a quadratic equation. x^{\tt{2}} + x \space {\text{–}} \space 2 = 0
This quadratic equation factors nicely. (x + 2)(x \space {\text{–}} \space 1) = 0
If we set both binomials equal to zero and solve, we obtain two x values.
x + 2 = 0 => x = {\text{-}}2
x \space {\text{–}} \space 1 = 0 => x = 1
Now if we look at the graph of y =x^{\tt{2}} + x \space {\text{–}} \space 2,
we can see what these x values represent.
The x values found are the solutions of the quadratic equation. Often also called ‘roots’ or ‘zeroes.
The values with which the graph touches the x-axis.
Usually it’s good practice to input any root values found into the equation to check that they do definitely hold. When solving quadratics by factoring.
({\text{-}}2)^2 + ({\text{-}}2) \space {\text{–}} \space 2 \space = \space 4 \space {\text{–}} \space 2 \space {\text{–}} \space 2 \space = \space 0 ( CORRECT )
(1)^2 + 1 \space {\text{–}} \space 2 \space = 1 + 1 \space {\text{–}} \space 2 \space = \space 0 ( CORRECT )
Solving Quadratics by Factoring Summary:
– With the quadratic equation equal to zero, factor the quadratic expression on the left hand side.
– Set the factor terms found equal to zero, and solve them.
– Double check the solutions obtained by inputting into the original quadratic equation.
Solving Quadratics by Factoring Examples
(1.1)
Solve x^{\tt{2}} + 7x \space {\text{–}} \space 8 = 0.
Solution
x^{\tt{2}} + 7x \space {\text{–}} \space 8 = 0
factors to (x \space {\text{–}} \space 1)(x + 8) = 0.
(x \space {\text{–}} \space 1) = 0 => x = 1
(x + 8) = 0 => x = {\text{-}}8
Now to check the found solutions.
x = 1 : 1^{\tt{2}} + 7(1) \space {\text{–}} \space 8 = 1 + 7 \space {\text{–}} \space 8 = 0
x = {\text{-}}8 : ({\text{-}}8)^{\tt{2}} + 7({\text{-}}8) \space {\text{–}} \space 8 = 64 \space {\text{–}} \space 56 \space {\text{–}} \space 8 = 0
(1.2)
Solve 4x^{\tt{2}} + x \space {\text{–}} \space 5 = 0.
Solution
4x^{\tt{2}} + x \space {\text{–}} \space 5 = 0
factors to (4x + 5)(x \space {\text{–}} \space 1) = 0.
(4x + 5) = 0 => x = {\text{-}}{\tt{\frac{5}{4}}}
(x \space {\text{–}} \space 1) = 0 => x = 1
Now to check the found solutions.
x = {\text{-}}{\tt{\frac{5}{4}}} : 4({\text{-}}{\tt{\frac{5}{4}}})^{\tt{2}} + ({\text{-}}{\tt{\frac{5}{4}}}) \space {\text{–}} \space 5 = 4({\tt{\frac{25}{16}}}) \space {\text{–}} \space {\tt{\frac{5}{4}}} \space {\text{–}} \space 5 = 0 = {\tt{\frac{100}{16}}} \space {\text{–}} \space {\tt{\frac{20}{16}}} \space {\text{–}} \space {\tt{\frac{80}{16}}} = 0
x = 1 : 4(1)^{\tt{2}} + 1 \space 5 = 4 + 1 \space {\text{–}} \space 5 = 0
(1.3)
Solve 4x^{\tt{2}} + 12x + 9 = 0.
Solution
4x^{\tt{2}} + 12x + 9 = 0
factors to (2x + 3)(2x + 3) = 0.
(2x + 3) = 0 => x = {\text{-}}{\tt{\frac{3}{2}}}
Two identical roots means there is just one solution overall, which can be called a ‘double root’.
x = {\text{-}}{\tt{\frac{3}{2}}} : 4({\text{-}}{\tt{\frac{3}{2}}})^{\tt{2}} + 12({\text{-}}{\tt{\frac{3}{2}}}) + 9 = 4({\tt{\frac{9}{4}}}) \space {\text{–}} \space {\tt{\frac{36}{2}}} + 9 = {\tt{\frac{36}{4}}} \space {\text{–}} \space {\tt{\frac{72}{4}}} + {\tt{\frac{36}{4}}} = 0