4.3 The Quadratic Formula
Introduction
The highly trained atheletes of the Red Bull Cliff Diving World Series execute incredible acrobatics from heights of up to \(88\) feet above the water. Such dives require focus and skill, along with timing.
How do the divers know how much time they have to perform the twists and flips that make up their dives? From those heights you don't want to guess, you want to be sure you know how long it takes to hit the water.
In this section will cover a formula that can answer that question. What's more, this magic formula can solve every possible quadratic equation.
The Quadratic Formula
We know that while many quadratic equations can be solved by factor, some quadratics are "prime" and cannot be factord. Additionally, there might be times when a quadratic could be factored but we simply can't spot the factors.
Fortunately, there is a formula that will solve all quadratic equations, always. The foundations of this formula stretch back thousands of years beginning, we think, with ancient Babylonian and Egyptian mathematicians and continuing on through developments by Greek, Chinese, Indian, Persian, and European scholars.
The quadratic formula, in the form we know it today, is given below.
If \(a x^2+ b x+ c=0\) where \(a \ne 0\), then the two solutions are:
\[ x = \frac{- b \pm \sqrt{b^2-4 a c}}{2 a} \]The quadratic formula is something you should memorize. There are lots of ways to do that, and you should pick a method that suits your learning style, but many find that singing the formula works.
Using the Quadratic Formula
Simply knowing the quadratic formula is not enough, however, we need to be able to insert values and evaluate the formula without making errors. Here are a few tips for avoiding common mistakes.
- Arrange the Equation: It must be in standard form and equal to zero: \(ax^2 + bx + c = 0\).
- Mind the Signs: The values \(a\), \(b\), and \(c\) do not include the variable \(x\) but they do keep their signs. Put parenthesis around each one and remember that \(b^2\) is always positive.
- Evaluate Systematically: Work on the square root first. Then simplify the rest of the numerator. Don't forget the \(\pm\) sign and write two equations if needed. Lastly, divide everything in the numerator by \(2a\).
Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses, especially when inserting a negative number. Also make sure that both terms in the numerator are divided by the denominator.
Since there are lots of places were simple mistakes can be made, we'll walk through a number of examples together.
Solve \(x^2+2x-15=0\) for \(x\) using the quadratic formula.
Solution
The equation is in the standard form, equal to zero, and the coefficients are \(a = 1\), \(b = 2\), and \(c = -15\). We can solve this with the quadratic formula.
\begin{align*} &&x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} & \quad \small \color{#5fa2ce} {\text{The quadratic formula}} \\ &&x = \frac{-2 \pm \sqrt{(2)^2-4(1)(-15)}}{2(1)} & \quad \small \color{#5fa2ce} {\text{Substitute } a = 1, b = 2, c = -15} \\ &&x = \frac{-2 \pm \sqrt{64}}{2} & \quad \small \color{#5fa2ce} {\text{Simplify inside the square root}} \\ &&x = \frac{-2 \pm 8}{2} & \quad \small \color{#5fa2ce} {\text{Evaluate the square root}} \\ &&x = \frac{-2 + 8}{2} \quad \text{ or } \quad x = \frac{-2 - 8}{2} & \quad \small \color{#5fa2ce} {\text{Write as two equations}} \\ &&x = \frac{6}{2} \quad \text{ or } \quad x = \frac{-10}{2} & \quad \small \color{#5fa2ce} {\text{Evaluate}} \\ &&x = 3 \quad \text{ or } \quad x = -5 \quad \quad & \quad \small \color{#5fa2ce} {\text{Simplify}} \end{align*}The equation used in Example 1 can also be solved by factoring and applying the zero-product property.
\begin{align} x^2+2x-15 = 0 &&& \small \color{#5fa2ce} {\text{Original equation}} \\ (x - 3)(x + 5) = 0 &&& \small \color{#5fa2ce} {\text{Factor}} \\ x - 3 = 0 \quad \text{ or } \quad x + 5 = 0 &&& \small \color{#5fa2ce} {\text{Use zero factor principle}} \\ x = 3 \quad \text{ or } \quad x = -5 &&& \small \color{#5fa2ce} {\text{Solve}} \end{align}Since we obtain the same solutions using either method, you can choose which ever technique you prefer. However, if a quadratic equation can be factored, that will always be faster than using the quadratic formula.
In the next example, however, the quadratic equation that cannot be factored, making the quadratic formula our best option for solving.
Solve \(3x^2=9x-4\) for \(x\) using the quadratic formula.
Solution
This equation is not in the standard form, so that will be the first step.
\begin{align} 3x^2 = 9x - 4 & \quad \small \color{#5fa2ce} {\text{ Original equation}} \\ 3x^2 - 9x = -4 & \quad \small \color{#5fa2ce} {\text{Subtract } 9x} \\ 3x^2 - 9x + 4 = 0 & \quad \small \color{#5fa2ce} {\text{Add } 4} \end{align}Now that the equation is in the standard form it is easier to identify the coefficients: \(a = 3\), \(b = -9\), \(c = 4\) and we are ready to start using the quadratic formula.
\begin{align} &&x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} & \quad \small \color{#5fa2ce} {\text{The quadratic formula}} \\ &&x = \frac{-(-9) \pm \sqrt{(-9)^2-4(3)(4)}}{2(3)} & \quad \small \color{#5fa2ce} {\text{Substitute } a = 3, b = -9, c = 4} \\ &&x = \frac{9 \pm \sqrt{33}}{6} & \quad \small \color{#5fa2ce} {\text{Simplify inside the square root.}} \\ &&x = \frac{9 + \sqrt{33}}{6} \quad \text{ or } \quad x = \frac{9 - \sqrt{33}}{6} & \quad \small \color{#5fa2ce} {\text{Write as two equations.}} \\ &&x = 2.457 \quad \text{ or } \quad x = 0.543 & \quad \small \color{#5fa2ce} {\text{Decimal approximation}} \end{align}Notice in the second to last step, the fraction bar is extended below both terms in the numerator: both are divided by \(6\). When using a calculator to simplify an expression such as this, we must use two separate steps or put parentheses around the entire numerator to get the correct value.
Solve \(4x^2 - 20x + 25 = 0\) for \(x\) using the quadratic formula.
Solution
The equation is in the standard form, equal to zero, and the coefficients are \(a = 4\), \(b = -20\), and \(c = 25\). We can solve this with the quadratic formula.
\begin{align*} &&x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} & \quad \small \color{#5fa2ce} {\text{The quadratic formula}} \\ &&x = \frac{-(-20) \pm \sqrt{(-20)^2-4(4)(25)}}{2(4)} & \quad \small \color{#5fa2ce} {\text{Substitute } a = 4, b = -20, c = 25} \\ &&x = \frac{20 \pm \sqrt{0}}{8} & \quad \small \color{#5fa2ce} {\text{Simplify inside the square root}} \\ &&x = \frac{20}{8} & \quad \small \color{#5fa2ce} {\text{Simplify}} \\ &&x = \frac{5}{2} & \quad \small \color{#5fa2ce} {\text{Reduce fraction}} \end{align*}In Examples 1 and 2 we had two solutions, but this time, in Example 3, there was only one solution: \(x = \frac{5}{2}\). In the next example, there will not be any real number solutions.
Solve \(-5x^2 - 5 = -8x\) for \(x\) using the quadratic formula.
Solution
If we rearrange the equation to be in the standard form: \( -5x^2 + 8x - 5 = 0\) then we can use the quadratic formula with \(a = -5\), \(b = 8\), and \(c = -5\).
\begin{align*} &&x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} & \quad \small \color{#5fa2ce} {\text{The quadratic formula}} \\ &&x = \frac{-(8) \pm \sqrt{(8)^2-4(-5)(-5)}}{2(-5)} & \quad \small \color{#5fa2ce} {\text{Substitute } a = -5, b = 8, c = -5} \\ &&x = \frac{-8 \pm \sqrt{-36}}{-10} & \quad \small \color{#5fa2ce} {\text{Simplify inside the square root}} \end{align*}We stop at this point because \(\sqrt{-36}\) is not a real number.
Since \(\sqrt{-36}\) is not a real number, the solutions will involve imaginary and complex numbers, which are not part of this course. The solutions exist, we simply don't have the tools to work with them.
There are two takeaways from this example. The first is that anytime your quadratic formula involves the square root of a negative number then, in this course at least, we can skip to the end and say there are no real solutions.
The second is a reminder to be careful with the signs as we work though the quadratic formula. If we had mistakenly used \(5\) for \(a\) or for \(c\) then the formula would have produced two incorrect real number solutions.
Now that we've see a few variations of what can happen, let's return to the situation that started this section.
A Practial Application
Let's suppose that for a particular dive the quadratic equation \(y=-16x^2+7x+88\) gives the height (in feet) of the diver above the surface of the water \(x\) seconds after leaving the platform. Equations like this come from the laws of motion you might study in a physics course.
Since the surface of the water corresponds to a height of \(y=0\), the quadratic formula can help us determine how much time the diver has before they hit the water. All we need to do is solve \(0=-16x^2+7x+88\).
Solve \(0=-16x^2+7x+88\) for \(x\) using the quadratic formula.
Solution
The equation is in the standard form, equal to zero, and the coefficients are \(a = -16\), \(b = 7\), and \(c = 88\).
\begin{align*} &&x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} & \quad \small \color{#5fa2ce} {\text{The quadratic formula}} \\ &&x = \frac{-(7) \pm \sqrt{(7)^2-4(-16)(88)}}{2(-16)} & \quad \small \color{#5fa2ce} {\text{Substitute } a = -16, b = 7, c = 88} \\ &&x = \frac{-7 \pm \sqrt{5681}}{-32} & \quad \small \color{#5fa2ce} {\text{Simplify inside the square root}} \\ &&x = \frac{-7 \pm 75.372}{-32} & \quad \small \color{#5fa2ce} {\text{Apprximate the square root}} \\ &&x = \frac{-7 + 75.372}{-32} \quad \text{ or } \quad x = \frac{-7 - 75.372}{-32} & \quad \small \color{#5fa2ce} {\text{Write as two equations}} \\ &&x = \frac{68.372}{-32} \quad \text{ or } \quad x = \frac{-82.372}{-32} & \quad \small \color{#5fa2ce} {\text{Add}} \\ &&x = -2.137 \quad \text{ or } \quad x = 2.574 \quad \quad & \quad \small \color{#5fa2ce} {\text{Divide}} \end{align*}In practial applications, it's not unusual for only one of the solutions to make sense in the given context. Since \(x\) represents the time in seconds after the diver left the platform, the only value of the two that makes sense is \(x=2.574\). This would indicate that the diver will hit the water in about \(2.5\) seconds. They will need to plan, practice and train with that time limit in mind.
Conclusion
In the world of mathematics it is somewhat rare to find a universal key that unlocks solutions, and that's exactly what the quadratic formula does. While factoring and other methods might be simpler and faster, they do not have the universal application of the quadratic formula. It's widely used in both theoretical and practial settings, and there's no doubt you'll run into it again.
As we move into the next chapter, where graphing comes into play, we'll uncover another lens through which to view these equations and interpret their solutions.