ORCCA (Draft)Open Resources for Community College Algebra

We have learned how to factor trinomials like \(x^2+5x+6\) into \((x+2)(x+3)\text{.}\) This skill is needed to solve an equation like \(x^2+5x+6=0\text{,}\) which is called a quadratic equation, because its leading term has a degree of \(2\text{.}\)

Consider this scenario: A physics class launches a tennis ball from a rooftop that is \(96\) feet above the ground. They fire it directly upward at a speed of \(16\) feet per second and measure the time it takes for the ball to hit the ground below. We can model the height of the tennis ball, \(h\text{,}\) in feet, with the quadratic equation, \(h=-16t^2+16t+96\text{,}\) where \(t\) represents the time in seconds after the launch. Using the model we can predict when the ball will hit the ground.

The ground has a height of \(0\text{,}\) or \(h=0\text{.}\) We will substitute \(0\) for \(h\) in the equation and we have

We need to solve this quadratic equation for \(t\) to find when the ball will hit the ground.

Let's review how to solve linear equations. For example,

The key strategy for solving a linear equation is to separate the variable terms from the constant terms on either side of the equal sign. Let's try the same strategy to solve \(-16t^2+16t+96=0\text{.}\)

We are stuck here because there is no way to isolate the \(t\) or get rid of the square in \(t^2\text{.}\) This is why we learned how to factor. For example, if we factor \(x^2+5x+6\) into \((x+2)(x+3)\text{,}\) the factors are linear.

Once we factor the polynomial in our example, we will be able to solve it using the linear factors. We will look for a common factor first, and see that we can factor out \(-16\text{.}\) Then we can finish factoring the trinomial:

In order to solve the equation, we need to understand a property:

To understand this property, let's look at a few products:

If none of the factors are zero, it is impossible to get a product of zero. The only way to get a product of zero is if one of the factors is zero. This property is unique to the number zero and can be used no matter how many numbers are multiplied together.

Now we can see the value of factoring. We have three factors in our equation \(-16(t+2)(t-3)=0\text{.}\) The first factor is the number \(-16\text{.}\) The second and third factors, \(t+2\) and \(t-3\) are expressions that represent numbers. Since the product of the three factors is equal to zero, one of the factors must be zero.

Since \(-16\) cannot be \(0\text{,}\) either \(t+2\) or \(t-3\) must be \(0\text{.}\) This gives us two equations to solve:

We have found two solutions, \(t=-2\) and \(t=3\text{.}\) A quadratic equation will have two linear factors, not including any constants, so it can have up to two solutions.

Let's check each of our two solutions \(t=-2\) and \(t=3\text{:}\)

We have verified that our solutions are correct. While there are two solutions to the equation, the solution \(t=-2\) is not relevant to the physics model because it is a negative time. The solution \(t=3\) does make sense. According to the model, the tennis ball will hit the ground \(3\) seconds after it is launched.

Here are a few more examples of solving quadratic equations. For problems without a context we will write the solutions in solution set notation.

If the two factors of a polynomial happen to be the same, the equation will only have one solution. Let's look at an example of that.

Here is an example where we need to factor out a common factor first.

Here is an example where we will factor using the AC method.

So far the equations have been written in standard form, which is

If an equation is not given in standard form then we must rearrange it in order to use the Zero Product Property.

We will conclude this section with another application problem.