Section 4.3
The Quadratic Formula
Consider the general quadratic equation in "standard form" —$ax^2+bx+c=0$ where $a,\;b,$ and $c$ are constants with $a$ called the leading coefficient and $c$ is the constant term. The Quadratice Formula is derived, or produced, by completing the square of a general quadratic equation in standard form.
To complete the square, we need the leading coefficient to be 1, so divide both sides by $a$. That gives $$x^2+\frac{b}{a}x+\frac{c}{a}=0.$$ Subtract $\dfrac{c}{a}$ from both sides and complete the square.
\begin{align}
x^2+\frac{b}{a}x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&=-\frac{c}{a}\\
x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2&=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2\\
\left(x+\frac{b}{2a}\right)^2&=-\frac{c}{a}+\frac{b^2}{4a^2}\\
\left(x+\frac{b}{2a}\right)^2&=\frac{b^2-4ac}{4a^2}\\
x+\frac{b}{2a}&=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\\
x+\frac{b}{2a}&=\pm \frac{\sqrt{b^2-4ac}}{2a}\\
x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
\end{align}
This formula can be used to solve any quadratic equation, once the equation has been written in standard form.
Examples:
a. To solve the equation $x^2-5x-4=0$, we note that $a=1$, $b=-5$, and $c=-4$. Then,
\begin{align}
x&=\frac{-(-5)\pm \sqrt{(-5)^2-4\cdot 1\cdot(-4)}}{2\cdot 1}\\
&=\frac{5\pm \sqrt{25+16}}{2}\\
&=\frac{5\pm \sqrt{41}}{2}
\end{align}
b. In the equation $2x^2+7x+2=0$, we have $a=2$, $b=7$, and $c=2$. Then,
\begin{align}
x&=\frac{-(7)\pm \sqrt{7^2-4\cdot 2\cdot 2}}{2\cdot 2}\\
&=\frac{-7\pm \sqrt{49-16}}{4}\\
&=\frac{-7\pm \sqrt{33}}{4}
\end{align}
c. In the quadratic equation $x^2+8x-2=0$, we have $a=1$, $b=8$ and $c=-2$. This quadratic can most easily be solved by completing the square because $a=1$ and $b$ is even. However, we can use the Quadratic Formula. We get
\begin{align}
x&=\frac{-8\pm \sqrt{64-4\cdot (-2)}}{2}\\
&=\frac{-8\pm \sqrt{72}}{2}\\
&=\frac{-8\pm \sqrt{36\cdot 2}}{2}\\
&=\frac{-8 \pm 6\sqrt{2}}{2}\\
&=\frac{2(-4\pm 3\sqrt{2}}{2}\\
&=-4\pm 3\sqrt{2}
\end{align}
d. To solve the equation $x^2+3x+7=0$, we note that $a=1$, $b=3$, and $c=7$. Then,
\begin{align}
x&=\frac{-3\pm\sqrt{9-4\cdot 1 \cdot 7}}{2\cdot 1}\\
&=\frac{-3\pm \sqrt{9-28}}{2}
\end{align}
The number under the square root sign is clearly negative. Therefore, this equation has no real solutions.
Completing the square for quadratic formula: Completing the Square 4
Exercises
Solve the following quadratic equations, if possible.\begin{array}{l l l}
1. \;\;{{x}^{2}}+5x-3=0\;\;& 2. \;\;{{x}^{2}}-2x-7=0\;\;& 3. \;\;2{{x}^{2}}+x-1=0\;\; \\
4. \;\;{{x}^{2}}-3x-7=0\;\;& 5. \;\;2{{x}^{2}}+5x-1=0\;\;& 6. \;\;4{{x}^{2}}-4x+1=0\;\; \\
7. \;\;{{x}^{2}}-3x+7=0\;\;& 8. \;\;2{{x}^{2}}-6x-3=0\;\;& 9. \;\;4{{x}^{2}}-10x+3=0\;\; \\
10. \;\;3{{x}^{2}}-5x=0\;\;& 11. \;\;2{{x}^{2}}-3=0\;\;& 12. \;\;4{{x}^{2}}-x+3=0\;\;
\end{array}
\begin{array}{l l}
13. \;\;{{(x-2)}^{2}}-3(x+1)=0\;\;& 14. \;\;(x-1)(2x-3)={{(x+1)}^{2}}\;\; \\
15. \;\;{{(x+5)}^{2}}-10(x+2)=0\;\;& 16. \;\;{{(x+2)}^{2}}=3{{x}^{2}}+7x+2\;\; \\
17. \;\;{{x}^{2}}=6x+2\;\;& 18. \;\;(2x+1)(x-3)=(x+5)(x-3)\;\; \\
19. \;\;\left( {{x}^{2}}-1 \right)-{{(x-1)}^{2}}=(x-1)(x+5)\;\;& 20. \;\;{{x}^{2}}+8x=3\;\; \\
21. \;\;{{x}^{2}}-5x+6=2{{x}^{2}}+x-1\;\; & 22. \;\;{{(2x-1)}^{2}}-(x+1)(x-2)=6\;\; \\
23. \;\;2.4{{x}^{2}}-7.2x=-3\;\;& 24. \;\;(3x-1)(x+4)=10x+2\;\; \\
25. \;\;3x(x-1)=6-15{{x}^{2}}\;\;& 26. \;\;6{{x}^{2}}-5x+4=0\;\; \\
27. \;\;{{x}^{2}}-x=5x-2\;\;& 28. \;\;(3x-5)(2x+4)=1\;\; \\
29. \;\;4x(x+3)=7\;\;& 30. \;\;(3x-1)(x+4)=10x+2\;\; \\
31. \;\;\frac{1}{x}=\frac{x}{x+1}\;\;& 32. \;\;\frac{x+3}{x+2}=\frac{x+2}{1}\;\; \\
33. \;\;\frac{x+3}{x+2}=\frac{x-2}{2x+7}\;\; & 34. \;\;\frac{1}{x+5}=\frac{x-5}{2x+7}\;\; \\
35. \;\;\frac{1}{x+5}+\frac{1}{x-5}=1\;\;& 36. \;\;\frac{x}{x+5}+\frac{x}{x-5}=\frac{1}{{{x}^{2}}-25}\;\;
\end{array}
37. It takes 239 feet of fencing to enclose the barnyard in the figure to the right. What is $x$?
38. The length of a rectangle is 2 more than the width. The area and the perimeter are numerically the same (the units, of course, are different). What is the width of the rectangle?