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# 5.13: Deriving and Using the Quadratic Formula

Difficulty Level: At Grade Created by: CK-12
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The profit on your school fundraiser is represented by the quadratic expression \begin{align*}-3p^2 + 200p - 3000\end{align*}, where p is your price point. What is your break-even point (i.e., the price point at which you will begin to make a profit)? Hint: Set the equation equal to zero.

### Guidance

The last way to solve a quadratic equation is the Quadratic Formula. This formula is derived from completing the square for the equation \begin{align*}ax^2+bx+c=0\end{align*} (see #13 from the Problem Set in the previous concept). We will derive the formula here.

#### Investigation: Deriving the Quadratic Formula

Walk through each step of completing the square of \begin{align*}ax^2+bx+c=0\end{align*}.

1. Move the constant to the right side of the equation. \begin{align*}ax^2+bx=-c\end{align*}

2. “Take out” \begin{align*}a\end{align*} from everything on the left side of the equation. \begin{align*}a\left(x^2+\frac{b}{a}x\right)=-c\end{align*}

3. Complete the square using \begin{align*}\frac{b}{a}\end{align*}. \begin{align*}\left(\frac{b}{2}\right)^2=\left(\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}\end{align*}

4. Add this number to both sides. Don’t forget on the right side, you need to multiply it by \begin{align*}a\end{align*} (to account for the \begin{align*}a\end{align*} outside the parenthesis). \begin{align*}a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\right)=-c+\frac{b^2}{4a}\end{align*}

5. Factor the quadratic equation inside the parenthesis and give the right hand side a common denominator. \begin{align*}a\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a}\end{align*}

6. Divide both sides by \begin{align*}a\end{align*}. \begin{align*}\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\end{align*}

7. Take the square root of both sides. \begin{align*}x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\end{align*}

8. Subtract \begin{align*}\frac{b}{2a}\end{align*} from both sides to get \begin{align*}x\end{align*} by itself. \begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}

This formula will enable you to solve any quadratic equation as long as you know \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*} (from \begin{align*}ax^2+bx+c=0\end{align*}).

#### Example A

Solve \begin{align*}9x^2-30x+26=0\end{align*} using the Quadratic Formula.

Solution: First, make sure one side of the equation is zero. Then, find \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*}. \begin{align*}a = 9, b = -30, c = 26\end{align*}. Now, plug in the values into the formula and solve for \begin{align*}x\end{align*}.

\begin{align*}x &=\frac{-(-30)\pm \sqrt{(-30)^2-4(9)(26)}}{2(9)}\\ &=\frac{30\pm \sqrt{900-936}}{18}\\ &=\frac{30\pm \sqrt{-36}}{18}\\ &=\frac{30\pm 6i}{18}\\ &=\frac{5}{3} \pm \frac{1}{3}i\end{align*}

#### Example B

Solve \begin{align*}2x^2+5x-15=-x^2+7x+2\end{align*} using the Quadratic Formula.

Solution: Let’s get everything onto the left side of the equation.

\begin{align*}2x^2+5x-15 &=-x^2+7x+2\\ 3x^2-2x-13 &=0\end{align*}

Now, use \begin{align*}a = 3, b = -2,\end{align*} and \begin{align*}c = -13\end{align*} and plug them into the Quadratic Formula.

\begin{align*}x &=\frac{-(-2) \pm \sqrt{(-2)^2-4(3)(-13)}}{2(3)}\\ &=\frac{2 \pm \sqrt{4+156}}{6}\\ &=\frac{2 \pm \sqrt{160}}{6}\\ &=\frac{2 \pm 4\sqrt{10}}{3}\end{align*}

#### Example C

Solve \begin{align*}x^2+20x+51=0\end{align*} by factoring, completing the square, and the Quadratic Formula.

Solution: While it might not look like it, 51 is not a prime number. Its factors are 17 and 3, which add up to 20.

\begin{align*}x^2+20x+51 &=0\\ (x+17)(x+13) &=0\\ x &=-17, -3\end{align*}

Now, solve by completing the square.

\begin{align*}x^2+20x+51 &=0\\ x^2+20x &=-51\\ x^2+20x+100 &=-51+100\\ (x+10)^2 &=49\\ x+10 &=\pm 7\\ x &=-10 \pm 7 \rightarrow -17, -3\end{align*}

Lastly, let’s use the Quadratic Formula. \begin{align*}a = 1, b = 20, c = 51\end{align*}.

\begin{align*}x &=\frac{-20 \pm \sqrt{20^2-4 (1)(51)}}{2(1)}\\ &=\frac{-20 \pm \sqrt{400-204}}{2}\\ &=\frac{-20 \pm \sqrt{196}}{2}\\ &=\frac{-20 \pm 14}{2}\\ &=-17, -3\end{align*}

Notice that no matter how you solve this, or any, quadratic equation, the answer will always be the same.

Intro Problem Revisit The break-even point is the point at which the equation equals zero. So use the Quadratic Formula to solve \begin{align*}-3p^2 + 200p - 3000\end{align*} for p.

\begin{align*}-3p^2 + 200p - 3000 = 0\end{align*}

Now, use \begin{align*}a = -3, b = 200,\end{align*} and \begin{align*}c = -3000\end{align*} and plug them into the Quadratic Formula.

\begin{align*}p &=\frac{-(200) \pm \sqrt{(200)^2-4(-3)(-3000)}}{2(-3)}\\ &=\frac{-200\pm \sqrt{40000-36000}}{-6}\\ &=\frac{-200\pm \sqrt{4000}}{-6}\\ &=\frac{-200\pm 20\sqrt{10}}{-6}\\ &=\frac{100}{3} \pm \frac{10\sqrt{10}}{3}\end{align*}

Therefore, there are two break-even points: \begin{align*}\frac{100}{3} \pm \frac{10\sqrt{10}}{3}\end{align*}.

### Guided Practice

1. Solve \begin{align*}-6x^2+15x-22=0\end{align*} using the Quadratic Formula.

2. Solve \begin{align*}2x^2-x-15=0\end{align*} using all three methods.

1. \begin{align*}a = -6, b = 15,\end{align*} and \begin{align*}c = -22\end{align*}

\begin{align*}x &=\frac{-15 \pm \sqrt{15^2-4(-6)(-22)}}{2(-6)}\\ &=\frac{-15 \pm \sqrt{225-528}}{-12}\\ &=\frac{-15 \pm i \sqrt{303}}{-12}\\ &=\frac{5}{4} \pm \frac{\sqrt{303}}{12}i\end{align*}

2. Factoring: \begin{align*}ac = -30\end{align*}. The factors of -30 that add up to -1 are -6 and 5. Expand the \begin{align*}x-\end{align*}term.

\begin{align*}2x^2-6x+5x-15 &=0\\ 2x(x-3)+5(x-3) &=0\\ (x-3)(2x+5) &=0\\ x &=3, -\frac{5}{2}\end{align*}

Complete the square

\begin{align*}2x^2-x-15 &=0\\ 2x^2-x &=15\\ 2\left(x^2-\frac{1}{2}x\right) &=15\\ 2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right) &=15+\frac{1}{8}\\ 2\left(x-\frac{1}{4}\right)^2 &=\frac{121}{8}\\ \left(x-\frac{1}{4}\right)^2 &=\frac{121}{16}\\ x-\frac{1}{4} &= \pm \frac{11}{4}\\ x &=\frac{1}{4} \pm \frac{11}{4} \rightarrow 3, -\frac{5}{2}\end{align*}

\begin{align*}x &=\frac{1 \pm \sqrt{1^2-4(2)(-15)}}{2(2)}\\ &=\frac{1 \pm \sqrt{1+120}}{4}\\ &=\frac{1 \pm \sqrt{121}}{4}\\ &=\frac{1 \pm 11}{4}\\ &=\frac{12}{4}, -\frac{10}{4} \rightarrow3, -\frac{5}{2}\end{align*}

### Vocabulary

For any quadratic equation in the form \begin{align*}ax^2+bx+c=0\end{align*}, \begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}.

### Practice

Solve the following equations using the Quadratic Formula.

1. \begin{align*}x^2+8x+9=0\end{align*}
2. \begin{align*}4x^2-13x-12=0\end{align*}
3. \begin{align*}-2x^2+x+5=0\end{align*}
4. \begin{align*}7x^2-11x+12=0\end{align*}
5. \begin{align*}3x^2+4x+5=0\end{align*}
6. \begin{align*}x^2-14x+49=0\end{align*}

Choose any method to solve the equations below.

1. \begin{align*}x^2+5x-150=0\end{align*}
2. \begin{align*}8x^2-2x-3=0\end{align*}
3. \begin{align*}-5x^2+18x-24=0\end{align*}
4. \begin{align*}10x^2+x-2=0\end{align*}
5. \begin{align*}x^2-16x+4=0\end{align*}
6. \begin{align*}9x^2-196=0\end{align*}

Solve the following equations using all three methods.

1. \begin{align*}4x^2+20x+25=0\end{align*}
2. \begin{align*}x^2-18x-63=0\end{align*}
3. Writing Explain when you would use the different methods to solve different types of equations. Would the type of answer (real or imaginary) help you decide which method to use? Which method do you think is the easiest?

### Vocabulary Language: English

Binomial

Binomial

A binomial is an expression with two terms. The prefix 'bi' means 'two'.
Completing the Square

Completing the Square

Completing the square is a common method for rewriting quadratics. It refers to making a perfect square trinomial by adding the square of 1/2 of the coefficient of the $x$ term.

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Roots

Roots

The roots of a function are the values of x that make y equal to zero.
Square Root

Square Root

The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9.
Vertex

Vertex

The vertex of a parabola is the highest or lowest point on the graph of a parabola. The vertex is the maximum point of a parabola that opens downward and the minimum point of a parabola that opens upward.

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Date Created:
Mar 12, 2013
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