5.13: Deriving and Using the Quadratic Formula
The profit on your school fundraiser is represented by the quadratic expression
Watch This
Khan Academy: Quadratic Formula 1
Guidance
The last way to solve a quadratic equation is the Quadratic Formula. This formula is derived from completing the square for the equation
Investigation: Deriving the Quadratic Formula
Walk through each step of completing the square of
1. Move the constant to the right side of the equation.
2. “Take out”
3. Complete the square using
4. Add this number to both sides. Don’t forget on the right side, you need to multiply it by
5. Factor the quadratic equation inside the parenthesis and give the right hand side a common denominator.
6. Divide both sides by
7. Take the square root of both sides.
8. Subtract
This formula will enable you to solve any quadratic equation as long as you know
Example A
Solve
Solution: First, make sure one side of the equation is zero. Then, find
Example B
Solve
Solution: Let’s get everything onto the left side of the equation.
Now, use
Example C
Solve
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.
Now, solve by completing the square.
Lastly, let’s use the Quadratic Formula.
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
Now, use
Therefore, there are two break-even points:
Guided Practice
1. Solve
2. Solve
Answers
1.
2. Factoring:
\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*}
Quadratic Formula
\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
- Quadratic Formula
- 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.
- \begin{align*}x^2+8x+9=0\end{align*}
- \begin{align*}4x^2-13x-12=0\end{align*}
- \begin{align*}-2x^2+x+5=0\end{align*}
- \begin{align*}7x^2-11x+12=0\end{align*}
- \begin{align*}3x^2+4x+5=0\end{align*}
- \begin{align*}x^2-14x+49=0\end{align*}
Choose any method to solve the equations below.
- \begin{align*}x^2+5x-150=0\end{align*}
- \begin{align*}8x^2-2x-3=0\end{align*}
- \begin{align*}-5x^2+18x-24=0\end{align*}
- \begin{align*}10x^2+x-2=0\end{align*}
- \begin{align*}x^2-16x+4=0\end{align*}
- \begin{align*}9x^2-196=0\end{align*}
Solve the following equations using all three methods.
- \begin{align*}4x^2+20x+25=0\end{align*}
- \begin{align*}x^2-18x-63=0\end{align*}
- 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?
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Show More |
Binomial
A binomial is an expression with two terms. The prefix 'bi' means 'two'.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 term.Quadratic Formula
The quadratic formula states that for any quadratic equation in the form , .Roots
The roots of a function are the values of x that make y equal to zero.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
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.Image Attributions
Here you'll derive the Quadratic Formula and use it to solve any quadratic equation.