Using the Principle of Zero Product to Solve Quadratic Equations
We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself.
Zero Product Property
If \(a \cdot b=0\), then either \(a=0\) or \(b=0\) or both.
We will now use the Zero Product Property, to solve a quadratic equation.
Example 7.69
How to Use the Zero Product Property to Solve a Quadratic Equation
Solve: \((x+1)(x-4)=0\).
Solution
|
Step 1. Set each factor equal to zero. |
The product equals zero, so at least one factor must equal zero. |
\( \begin{array}{c} (x+1)(x-4)=0 \\ x+1=0 \quad \text { or } \quad x-4=0 \end{array} \) |
|
Step 2. Solve the linear equations. |
Solve each equation. | \(x=-1 \quad\) or \(\quad x=4\) |
|
Step 3. Check. |
Substitute each solution separately into the original equation. |
\(\begin{aligned} x &=-1 \\(x+1)(x-4) &=0 \\(-1+1)(-1-4) & \stackrel{?}{=} 0 \\(0)(-5) & \stackrel{?}{=} 0 \\ 0 &=0 \text{✓} \end{aligned}\) \(\begin{aligned} x &=4 \\(x+1)(x-4) &=0 \\(4+1)(4-4) & \stackrel{?}{=} 0 \\(5)(0) & \stackrel{?}{=} 0 \\ 0 &=0 \text{✓} \end{aligned}\) |
Try It 7.137
Solve: \((x−3)(x+5)=0\).
Try It 7.138
Solve: \((y−6)(y+9)=0\).
We usually will do a little more work than we did in this last Example to solve the linear equations that result from using the Zero Product Property.
Example 7.70
Solve: \((5 n-2)(6 n-1)=0\).
Solution
| \((5 n-2)(6 n-1)=0\) | ||
| Use the Zero Product Property to set each factor to 0. |
\(5n−2=0\) | \(6n−1=0\) |
| Solve the equations. | \(n=\frac{2}{5}\) | \(n=\frac{1}{6}\) |
| Check your answers. | ||
| \(\begin{array}{rlrl}n & =\frac{2}{5} & & n=\frac{1}{6} \\ (5 n-2)(6 n-1) & =0 & (5 n-2)(6 n-1) & =0 \\ \left(5 \cdot \frac{2}{5}-2\right)\left(6 \cdot \frac{2}{5}-1\right) & \stackrel{?}{=} 0 & \left(5 \cdot \frac{1}{6}-2\right)\left(6 \cdot \frac{1}{6}-1\right) & \stackrel{?}{=} 0 \\ (2-2)\left(\frac{12}{5}-\frac{5}{5}\right) & \stackrel{?}{=} 0 & \left(\frac{5}{6}-\frac{12}{6}\right)(1-1) & \underline{?} 0 \\ (0)\left(\frac{7}{5}\right) & \stackrel{?}{=} 0 & \left(-\frac{7}{6}\right)(0) \stackrel{?}{=} 0 \\ 0 & =0 \text{✓} & 0=0 \text{✓} \end{array}\) | ||
Try It 7.139
Solve: \((3m−2)(2m+1)=0\).
Try It 7.140
Solve: \((4p+3)(4p−3)=0\).
Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions.
Example 7.71
Solve: \(3p(10p+7)=0\).
Solution
| \(3 p(10 p+7)=0\) | ||
| Use the Zero Product Property to set each factor to 0. |
\(3p=0\) | \(10p+7=0\) |
| Solve the equations. | \(p=0\) | \(10p=−7\) |
| \(p=-\frac{7}{10}\) | ||
| Check your answers. | ||
| \(\begin{array}{rrr}p & =0 & p=\frac{7}{10} \\ 3 p(10 p+7) & =0 & 3 p(10 p+7) & =0 \\ 3 \cdot 0(10 \cdot 0+7) & \stackrel{?}{=} 0 & \qquad 3\left(\frac{7}{10}\right) 10\left(\frac{7}{10}\right)+7 &\stackrel{?}{=} 0 \\ 0(0+7) & \stackrel{?}{=} 0 & \left(-\frac{21}{10}\right)(-7+7) & \stackrel{?}{=} 0 \\ 0(7) & \stackrel{?}{=} 0 & \left(\frac{21}{10}\right)(0) & \stackrel{?}{=} 0 \\ 0 & =0 \text{✓} & 0 &=0 \text{✓}\end{array}\) | ||
Try It 7.141
Solve: \(2u(5u−1)=0\).
Try It 7.142
Solve: \(w(2w+3)=0\).
It may appear that there is only one factor in the next example. Remember, however, that \((y-8)^{2}\) means \((y-8)(y-8)\).
Example 7.72
Solve: \((y-8)^{2}=0\).
Solution
| \((y-8)^{2}=0\) | ||
| Rewrite the left side as a product. | \((y−8)(y−8)=0\) | |
| Use the Zero Product Property and set each factor to 0. |
\(y−8=0\) | \(y−8=0\) |
| Solve the equations. | \(y=8\) | \(y=8\) |
| When a solution repeats, we call it a double root. |
||
| Check your answer. | ||
| \(\begin{aligned} y &=8 \\(y-8)^{2} &=0 \\(8-8)^{2} & \stackrel{?}{=} 0 \\(0)^{2} & \stackrel{?}{=} 0 \\ 0 &=0 . \end{aligned}\) | ||
Try It 7.143
Solve: \((x+1)^{2}=0\).
Try It 7.144
Solve: \((v-2)^{2}=0\).
Source: OpenStax, https://openstax.org/books/elementary-algebra/pages/7-6-quadratic-equations
This work is licensed under a Creative Commons Attribution 4.0 License.