The quadratic equation \begin{align*}x^2+x-5=0\end{align*} describes the shape of the cut-away of a ditch. You can graph the function to get a general idea of the shape. However you would want more precise data in order to learn the length of bridge you will need to cross it. The roots of the function would represent the borders of the ditch. How can you accurately identify the roots of this quadratic function that does not easily factor? The Quadratic Formula is a useful tool for identifying accurate roots of quadratic functions that aren't easy to factor.

### Quadratic Formula

There are a number of methods to solve a quadratic equation:

- Graphing to find the zeros
- Using square roots
- Completing the square

Here you will learn a fourth way to solve a quadratic equation: using the quadratic formula.

#### History of the Quadratic Formula

As early as 1200 BC, people were interested in solving quadratic equations. The Babylonians solved simultaneous equations involving quadratics. In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit formula to solve a quadratic equation. The quadratic formula was written as it is today by the Arabic mathematician Al-Khwarizmi. It is his name upon which the word “Algebra” is based.

The solution to any quadratic equation in standard form, \begin{align*}0=ax^2+bx+c\end{align*}, is:

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

This formula is called the **quadratic formula.**

#### Let's solve the following equations using the quadratic formula:

- \begin{align*}x^2+10x+9=0\end{align*}

Applying the quadratic formula and \begin{align*}a=1, b=10\end{align*}, and \begin{align*}c=9\end{align*}, we get:

\begin{align*}x &= \frac{-10 \pm \sqrt{(10)^2-4(1)(9)}}{2(1)}\\ x &= \frac{-10 \pm \sqrt{100-36}}{2}\\ x &= \frac{-10 \pm \sqrt{64}}{2}\\ x &= \frac{-10 \pm 8}{2}\\ x &= \frac{-10 + 8}{2} \ or \ x=\frac{-10-8}{2}\\ x &= -1 \ or \ x=-9\end{align*}

- \begin{align*}-4x^2+x+1=0\end{align*}

\begin{align*}\text{Quadratic formula} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \text{Substitute} \ a=-4, b=1, c=1 && x& =\frac{-1 \pm \sqrt{(1)^2-4(-4)(1)}}{2(-4)}\\ \text{Simplify} && x & =\frac{-1 \pm \sqrt{1+16}}{-8}=\frac{-1 \pm \sqrt{17}}{-8}\\ \text{Separate the two options} && x&=\frac{-1+\sqrt{17}}{-8} \ \text{and} \ x=\frac{-1-\sqrt{17}}{-8}\\ \text{Solve} && x & \approx -.39 \ \text{and} \ x \approx .64\end{align*}

- \begin{align*}8t^2+10t+3=0\end{align*}

\begin{align*}\text{Quadratic formula} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \text{Plug in the values} \ a=8, b=10, c=3 && x& =\frac{-10 \pm \sqrt{(10)^2-4(8)(3)}}{2(8)}\\ \text{Simplify} && x & =\frac{-10 \pm \sqrt{100+96}}{16}=\frac{-10 \pm \sqrt{196}}{16}\\ \text{Separate the two options} && x&=\frac{-10+\sqrt{196}}{16} \ \text{and} \ x=\frac{-10-\sqrt{196}}{16}\\ && x&=\frac{-10+14}{16} \ \text{and} \ x=\frac{-10-14}{16}\\ \text{Solve} && x & =\frac{1}{4} \ \text{and} \ x =-\frac{3}{2}\end{align*}

### Examples

#### Example 1

Earlier, you were told that the quadratic equation \begin{align*}x^2+x-5=0\end{align*} describes the shape of the cut-away of a ditch. To find the borders of the ditch, you need to know the roots of the function. How can you accurately identify the roots of this quadratic function?

The quadratic formula is a useful tool for identifying accurate roots of quadratic equations that aren't easy to factor.

\begin{align*}\text{Quadratic formula} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\
\text{Plug in the values} \ a=1, b=1, c=-5 && x& =\frac{-1 \pm \sqrt{(1)^2-4(1)(-5)}}{2(1)}\\
\text{Simplify} && x & =\frac{-1 \pm \sqrt{1+20}}{2}=\frac{-1 \pm \sqrt{21}}{2}\\
\text{Separate the two options} && x&=\frac{-1+\sqrt{21}}{2} \ \text{and} \ x=\frac{-1-\sqrt{21}}{2}\\
\text{Solve} && x & =-\frac{1}{2}+\frac{\sqrt{21}}{2} \text{and} \ -\frac{1}{2}-\frac{\sqrt{21}}{2}\end{align*}

Notice that 21 is not a perfect square so we will leave the square root in the results to keep the answer more exact. If necessary, the approximate solutions are -2.79 and 1.79. The borders of the ditch are at (-2.79, 0) and (1.79, 0).

#### Example 2

Solve \begin{align*}3k^2+11k=4\end{align*} using the quadratic formula.

First, we must make it so one side is equal to zero:

\begin{align*}3k^2+11k=4 \Rightarrow 3k^2+11k-4=0\end{align*}

Now it is in the correct form for using the quadratic formula.

\begin{align*}\text{Quadratic formula} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \text{Plug in the values} \ a=3, b=11, c=-4 && x& =\frac{-11 \pm \sqrt{(11)^2-4(3)(-4)}}{2(3)}\\ \text{Simplify} && x & =\frac{-11 \pm \sqrt{121+48}}{6}=\frac{-11 \pm \sqrt{169}}{6}\\ \text{Separate the two options} && x&=\frac{-11+\sqrt{169}}{6} \ \text{and} \ x=\frac{-11-\sqrt{169}}{6}\\ && x&=\frac{-11+13}{6} \ \text{and} \ x=\frac{-11-13}{6}\\ \text{Solve} && x & =\frac{1}{3} \ \text{and} \ x =-4\end{align*}

### Review

- What is the quadratic formula? When is the most appropriate situation to use this formula?
- When was the first known solution of a quadratic equation recorded?

Solve the following quadratic equations using the quadratic formula.

- \begin{align*}x^2+4x-21=0\end{align*}
- \begin{align*}x^2-6x=12\end{align*}
- \begin{align*}3x^2-\frac{1}{2}x=\frac{3}{8}\end{align*}
- \begin{align*}2x^2+x-3=0\end{align*}
- \begin{align*}-x^2-7x+12=0\end{align*}
- \begin{align*}-3x^2+5x=0\end{align*}
- \begin{align*}4x^2=0\end{align*}
- \begin{align*}x^2+2x+6=0\end{align*}

### Review (Answers)

To see the Review answers, open this PDF file and look for section 10.7.