ORCCA The Quadratic Formula

Section8.4The Quadratic Formula

ObjectivesPCC Course Content and Outcome Guide

Item C.2.1:4.c
Item C.2.1:4.d
Item C.2.1:4.g
Item C.2.1:4.f

We have learned how to solve quadratic equations with the square root property and by factoring. In this section, we will learn a third method, using the Quadratic Formula.

Figure8.4.1Alternative Video Lesson

Subsection8.4.1Solving Quadratic Equations by Quadratic Formula

The standard form of a quadratic equation is

\begin{equation*} ax^2+bx+c=0 \end{equation*}

Let's look at two examples as a review of solving quadratic equations.

When \(b=0\text{,}\) the equation looks like \(ax^2+c=0\text{.}\) The easiest way to solve this type of equations is to use the square root property.

Example8.4.2

Solve for \(x\) in \(x^2-4=0\text{.}\)

Solution

\begin{align*} x^2-4\amp=0\\ x^2\amp=4\\ \sqrt{x^2}\amp=\pm\sqrt{4}\\ x\amp=\pm2 \end{align*}

If \(b\ne0\text{,}\) and if it's easy to factor the left side of \(ax^2+bx+c=0\text{,}\) we can solve the equation by factoring.

Example8.4.3

Solve for \(x\) in \(x^2-4x-12=0\text{.}\)

Solution

\begin{align*} x^2-4x-12\amp=0\\ (x-6)(x+2)\amp=0\\ x-6\amp=0\text{ or }x+2=0\\ x\amp=6\text{ or }x=-2 \end{align*}

If \(b\ne0\text{,}\) and if it's not easy or not possible to factor the left side of \(ax^2+bx+c=0\text{,}\) we use the quadratic formula:

\begin{equation*} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{equation*}

We will learn how to derive this formula in later math courses. For now, we will simply learn how to use it.

We did the following example when we learned how to solve quadratic equations by factoring. In that example, numbers were made nice, so we could factor. In real life, numbers are usually decimals.

Example8.4.4

At the height of \(90.2\) feet above sea level, an object is launched at a vertical velocity of \(14.4\) feet per second. The object's height can be modeled by the equation \(h=-16t^2+14.4t+90.2\text{,}\) where \(h\) represents the object's height in feet, and \(t\) represents time passed since the launch in seconds. When will the object fall to the ground (at sea level)? Round numbers to two decimal places.

When the object falls to the sea level, its height is \(0\text{.}\) Substituting \(h\) in the equation with \(0\text{,}\) we can solve the quadratic equation:

\begin{equation*} 0=-16t^2+14.4t+90.2 \end{equation*}

We cannot solve this equation with the square root property or factoring. We have to resort to the Quadratic Formula. Indentify that \(a=-16\text{,}\) \(b=14.4\) and \(c=90.2\text{,}\) and substitute them into the formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(14.4)\pm\sqrt{(14.4)^2-4(-16)(90.2)}}{2(-16)}\\ x\amp=\frac{-14.4\pm\sqrt{201.64-(-5772.8)}}{-32}\\ x\amp=\frac{-14.4\pm\sqrt{201.64+5772.8}}{-32}\\ x\amp=\frac{-14.4\pm\sqrt{5974.44}}{-32}\\ x\amp\approx\frac{-14.4\pm77.2945}{-32}\\ x\amp\approx\frac{-14.4+77.2945}{-32}\text{ or }x\approx\frac{-14.4-77.2945}{-32}\\ x\amp\approx\frac{63.0945}{-32}\text{ or }x\approx\frac{-91.4945}{-32}\\ x\amp\approx-1.97\text{ or }x\approx2.86 \end{align*}

We don't take the negative solution as it doesn't make sense in this context. Solution: The object will fall to the ground approximately \(2.86\) seconds since its launch.

The Quadratic Formula can be used to solve all quadratic equations. In the next two examples, we will use the Quadratic Formula to solve the same two equations we solved earlier, although it's easier to use the square root property or factoring.

Example8.4.5

Solve for \(x\) in \(x^2-4=0\text{.}\)

Solution

Identify that \(a=1\text{,}\) \(b=0\) and \(c=-4\text{.}\) We will substitute them into the Quadratic Formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(0)\pm\sqrt{(0)^2-4(1)(-4)}}{2(1)}\\ x\amp=\frac{\pm\sqrt{16}}{2}\\ x\amp=\frac{\pm4}{2}\\ x\amp=\frac{4}{2}\text{ or }x=\frac{-4}{2}\\ x\amp=2\text{ or }x=-2 \end{align*}

Example8.4.6

Solve for \(x\) in \(x^2-4x-12=0\text{.}\)

Solution

Identify that \(a=1\text{,}\) \(b=-4\) and \(c=-12\text{.}\) We will substitute them into the Quadratic Formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-12)}}{2(1)}\\ x\amp=\frac{4\pm\sqrt{16+48}}{2}\\ x\amp=\frac{4\pm\sqrt{64}}{2}\\ x\amp=\frac{4\pm8}{2}\\ x\amp=\frac{4+8}{2}\text{ or }x=\frac{4-8}{2}\\ x\amp=6\text{ or }x=-2 \end{align*}

Before we use the Quadratic Formula, we must change a quadratic equation into the standard form \(ax^2+bx+c=0\) to identify the values of \(a\text{,}\) \(b\) and \(c\text{.}\) For example, to solve \(4x+12=x^2\text{,}\) we must first convert it to the standard form:

\begin{align*} 4x+12\amp=x^2\\ 4x+12\subtractright{4x}\subtractright{12}\amp=x^2\subtractright{4x}\subtractright{12}\\ 0\amp=x^2-4x-12 \end{align*}

Now we can identify \(a=1\text{,}\) \(b=-4\) and \(c=-12\text{.}\)

The next equation has only one solution. Let's see how the Quadratic Formula handles this situation.

Example8.4.7

Solve for \(x\) in \(4x^2+9=12x\text{.}\)

Solution

First, we convert the equation into standard form:

\begin{align*} 4x^2+9\amp=12x\\ 4x^2+9\subtractright{12x}\amp=12x\subtractright{12x}\\ 4x^2-12x+9\amp=0 \end{align*}

Now we can identify that \(a=4\text{,}\) \(b=-12\) and \(c=9\text{.}\) We will substitute them into the Quadratic Formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-12)\pm\sqrt{(-12)^2-4(4)(9)}}{2(4)}\\ x\amp=\frac{12\pm\sqrt{144-144}}{8}\\ x\amp=\frac{12\pm\sqrt{0}}{8}\\ x\amp=\frac{12\pm0}{8}\\ x\amp=\frac{12}{8}\\ x\amp=\frac{3}{2} \end{align*}

Sometimes the solutions are complex numbers. Let's look at the next two examples.

Example8.4.8

Solve for \(y\) in \(y^2-4y+8=0\text{.}\)

Solution

Identify that \(a=1\text{,}\) \(b=-4\) and \(c=8\text{.}\) We will substitute them into the Quadratic Formula:

\begin{align*} y\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ y\amp=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(8)}}{2(1)}\\ y\amp=\frac{4\pm\sqrt{16-32}}{2}\\ y\amp=\frac{4\pm\sqrt{-16}}{2} \end{align*}

The square root of a negative number is a complex number. Without any context, we assume \(y\) is a real number, so this equation has no real solutions.

Example8.4.9

Solve for \(y\) in \(y^2-4y+8=0\text{,}\) where \(y\) is a complex number.

Solution

Identify that \(a=1\text{,}\) \(b=-4\) and \(c=8\text{.}\) We will substitute them into the Quadratic Formula:

\begin{align*} y\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ y\amp=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(8)}}{2(1)}\\ y\amp=\frac{4\pm\sqrt{16-32}}{2}\\ y\amp=\frac{4\pm\sqrt{-16}}{2}\\ y\amp=\frac{4\pm\sqrt{-1\cdot16}}{2}\\ y\amp=\frac{4\pm\sqrt{-1}\cdot\sqrt{16}}{2}\\ y\amp=\frac{4\pm i\cdot4}{2}\\ y\amp=\frac{4\pm4i}{2}\\ y\amp=\frac{4}{2}\pm\frac{4i}{2}\\ y\amp=2\pm 2i \end{align*}

Notice that we broke one fraction into two:

\begin{equation*} \frac{4\pm4i}{2}=\frac{4}{2}\pm\frac{4i}{2} \end{equation*}

We can do this because

\begin{equation*} \frac{1}{5}+\frac{2}{5}=\frac{1+2}{5} \end{equation*}

If we reverse this equation, we have:

\begin{equation*} \frac{1+2}{5}=\frac{1}{5}+\frac{2}{5} \end{equation*}

This is why we can break one fraction into two.

Subsection8.4.2Summary of Solving Linear and Quadratic Equations

In this section, we will review the difference in solving linear and quadratic equations, and see how to choose the best method to solve a quadratic equation.

Let's compare the proper methods used to solve the following linear and quadratic equations:

Example8.4.10

Solve for \(x\) in \(x=-2x+3\text{.}\)

\begin{align*} x\addright{2x}\amp=-2x+3\addright{2x}\\ 3x\amp=3\\ \divideunder{3x}{3}\amp=\divideunder{3}{3}\\ x\amp=1 \end{align*}

Example8.4.11

Solve for \(x\) in \(x^2=-2x^2+3\text{.}\)

\begin{align*} x^2\addright{2x^2}\amp=-2x+3\addright{2x^2}\\ 3x^2\amp=3\\ \divideunder{3x^2}{3}\amp=\divideunder{3}{3}\\ x^2\amp=1\\ x\amp=\pm\sqrt{1}\\ x\amp=\pm1 \end{align*}

Example8.4.12

Solve for \(x\) in \(x^2=-2x+3\text{.}\)

\begin{align*} x^2\addright{2x-3}\amp=-2x+3\addright{2x-3}\\ x^2+2x-3\amp=0\\ (x+3)(x-1)\amp=0\\ x+3=0\amp\text{ or }x-1=0\\ x=-3\amp\text{ or }x=1 \end{align*}

Example8.4.13

Solve for \(x\) in \(x^2=-2x+2\text{.}\)

\begin{align*} x^2\addright{2x-2}\amp=-2x+2\addright{2x-2}\\ x^2+2x-2\amp=0\\ x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(2)\pm\sqrt{(2)^2-4(1)(-2)}}{2(1)}\\ x\amp=\frac{-2\pm\sqrt{4+8}}{2}\\ x\amp=\frac{-2\pm\sqrt{12}}{2}\\ x\amp=\frac{-2\pm\sqrt{2^2\cdot3}}{2}\\ x\amp=\frac{-2\pm2\sqrt{3}}{2}\\ x\amp=\frac{-2}{2}\pm\frac{2\sqrt{3}}{2}\\ x\amp=-1\pm \sqrt{3} \end{align*}

We have the following observations:

When we solve a linear equation (without \(x^2\) terms) like in Example 8.4.10, we move all \(x\) terms to one side of the equals sign, and move all number terms to the other side. Then, we use division to solve for \(x\text{.}\)
When we solve a quadratic equation (with \(x^2\) terms) like in Example 8.4.11, we move all \(x^2\) terms to one side of the equals sign, and move all number terms to the other side. Then, we use the square root property to solve for \(x\text{.}\)
When we solve a quadratic equation like in Example 8.4.12, we move all terms to one side of the equals sign (making the other side \(0\)), and then use factoring to solve the equation if the polynomial can be easily factored.
When we solve a quadratic equation like in Example 8.4.13, where we cannot use the square root property or factoring, we can use the Quadratic Formula to solve the equation. Again, we can always choose to use the formula to solve any quadratic equation, although the other two methods could be easier.