Lesson 4

What values of the variables make each equation true?

\(6 + 2a = 0\)

\(7b=0\)

\(7(c-5)=0\)

\(g \boldcdot h=0\)

For each equation, find its solution or solutions. Be prepared to explain your reasoning.

1. \(x-3=0\)
2. \(x+11=0\)
3. \(2x+11=0\)
4. \(x(2x+11)=0\)
5. \((x-3)(x+11)=0\)
6. \((x-3)(2x+11)=0\)
7. \(x(x+3)(3x-4)=0\)

We have seen quadratic functions modeling the height of a projectile as a function of time.

Here are two ways to define the same function that approximates the height of a projectile in meters, \(t\) seconds after launch:

\(\displaystyle h(t)=\text-5t^2+27t+18 \qquad \qquad h(t)=(\text-5t-3)(t-6)\)

1. Which way of defining the function allows us to use the zero product property to find out when the height of the object is 0 meters?
2. Without graphing, determine at what time the height of the object is 0 meters. Show your reasoning.

The zero product property says that if the product of two numbers is 0, then one of the numbers must be 0. In other words, if \(a\boldcdot b=0,\) then either \(a=0\) or \(b=0\). This property is handy when an equation we want to solve states that the product of two factors is 0.

Suppose we want to solve \(m(m+9)=0\). This equation says that the product of \(m\) and \((m+9)\) is 0. For this to be true, either \(m=0\) or \(m+9=0\), so both 0 and -9 are solutions.

Here is another equation: \((u-2.345)(14u+2)=0\). The equation says the product of \((u-2.345)\) and \((14u+2)\) is 0, so we can use the zero product property to help us find the values of \(u\). For the equation to be true, one of the factors must be 0.

• For \(u-2.345=0\) to be true, \(u\) would have to be 2.345.
• For \(14u+2=0\) or \(14u = \text-2\) to be true, \(u\) would have to be \(\text-\frac{2}{14}\) or \(\text-\frac17\).

The solutions are 2.345 and \(\text-\frac17\).

In general, when a quadratic expression in factored form is on one side of an equation and 0 is on the other side, we can use the zero product property to find its solutions.