Lesson 16
The Quadratic Formula
 Let’s learn a formula for finding solutions to quadratic equations.
16.1: Evaluate It
Each expression represents two numbers. Evaluate the expressions and find the two numbers.
 \(1 \pm \sqrt{49}\)
 \(\displaystyle \frac{8 \pm 2}{5}\)
 \(\pm \sqrt{(\text5)^24 \boldcdot 4 \boldcdot 1}\)
 \(\displaystyle \frac{\text18 \pm \sqrt{36}}{2 \boldcdot 3}\)
16.2: Pesky Equations
Choose one equation to solve, either by rewriting it in factored form or by completing the square. Be prepared to explain your choice of method.
 \(x^22x1.25=0\)
 \(5x^2+9x44=0\)
 \(x^2+1.25x=0.375\)
 \(4x^228x+29=0\)
16.3: Meet the Quadratic Formula
Here is a formula called the quadratic formula.
\(\displaystyle x=\dfrac{\text b \pm \sqrt{b^24ac}}{2a}\)
The formula can be used to find the solutions to any quadratic equation in the form of \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are numbers and \(a\) is not 0.
This example shows how it is used to solve \(x^2  8x + 15=0\), in which \(a=1\), \(b=\text8\), and \(c=15\).
\(\displaystyle \begin {align} x &=\dfrac{\text b \pm \sqrt{b^24ac}}{2a} &\qquad &\text{original equation}\\ x &=\dfrac{\text (\text8) \pm \sqrt{(\text8)^24(1)(15)}}{2(1)} &\qquad &\text{substitute the values of }a, b, \text{and }c &\\ x &=\dfrac{8 \pm \sqrt{6460}}{2} &\qquad &\text {evaluate each part of the expression} \\ x &=\dfrac{8 \pm \sqrt{4}}{2} \\x &=\dfrac{8 \pm 2}{2}\\ x &=\frac{10}{2} \qquad \text {or} \qquad x =\frac{6}{2} \\x &=\text{ }5 \qquad \text{ } \text {or} \qquad x =\text{ }3 \end{align}\)
Here are some quadratic equations and their solutions. Use the quadratic formula to show that the solutions are correct.
 \(x^2 + 4x  5 = 0\). The solutions are \(x=\text5\) and \(x=1\).
 \(x^2 + 7x + 12 = 0\). The solutions are \(x=\text3\) and \(x=\text4\).
 \(x^2 + 10x + 18 = 0\). The solutions are \(x={\text5} \pm \frac{\sqrt{28}}{2}\).
 \(x^2  8x + 11 = 0\). The solutions are \(x=4 \pm\frac{ \sqrt{20}}{2}\).
 \(9x^26x+1=0\). The solution is \(x=\frac13\).
 \(6x^2+9x15=0\). The solutions are \(x=\text\frac52\) and \(x=1\).
 Use the quadratic formula to solve \(ax^2+c=0\). Let’s call the resulting equation P.

Solve the equation \(3x^227=0\) in two ways, showing your reasoning for each:
 Without using any formulas.
 Using equation P.

Check that you got the same solutions using each method.
 Use the quadratic formula to solve \(ax^2+bx=0\). Let’s call the resulting equation Q.

Solve the equation \(2x^2+5x=0\) in two ways, showing your reasoning for each:
 Without using any formulas.
 Using equation P.
 Check that you got the same solutions using each method.
Summary
We have learned a couple of methods for solving quadratic equations algebraically:
 by rewriting the equation as \(\text{factored form}=0\) and using the zero product property
 by completing the square
Some equations can be solved quickly with one of these methods, but many cannot. Here is an example: \(5x^23x1=0\). The expression on the left cannot be rewritten in factored form with rational coefficients. Because the coefficient of the squared term is not a perfect square, and the coefficient of the linear term is an odd number, completing the square would be inconvenient and would result in a perfect square with fractions.
The quadratic formula can be used to find the solutions to any quadratic equation, including those that are tricky to solve with other methods.
For an equation of the form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are numbers and \(a \neq 0\), the solutions are given by:
\(\displaystyle x=\dfrac{\text b \pm \sqrt{b^24ac}}{2a}\)
For the equation \(5x^23x1=0\), we see that \(a=5\), \(b=\text3\), and \(c=\text1\). Let’s solve it!
\(\displaystyle \begin {align} x &=\dfrac{\text b \pm \sqrt{b^24ac}}{2a} &\qquad &\text{the quadratic formula}\\ x &=\dfrac{\text(\text3) \pm \sqrt{(\text3)^24(5)(\text1)}}{2(5)} &\qquad &\text{substitute the values of }a, b, \text{and }c &\\ x &=\dfrac{3 \pm \sqrt{9+20}}{10} &\qquad &\text {evaluate each part of the expression} \\ x &=\dfrac{3 \pm \sqrt{29}}{10}\end{align}\)
A calculator gives approximate solutions of 0.84 and 0.24 for \(\frac{3 + \sqrt{29}}{10}\) and \(\frac{3  \sqrt{29}}{10}\).
We can also use the formula for simpler equations like \(x^29x+8=0\), but it may not be the most efficient way. If the quadratic expression can be easily rewritten in factored form or made into a perfect square, those methods may be preferable. For example, rewriting \(x^29x+8=0\) as \((x1)(x8)=0\) immediately tells us that the solutions are 1 and 8.
Glossary Entries

quadratic formula
The formula \(x = {\textb \pm \sqrt{b^24ac} \over 2a}\) that gives the solutions of the quadratic equation \(ax^2 + bx + c = 0\), where \(a\) is not 0.