## Solving Quadratic Equations and Inequalities

Completing the square for the standard form of a quadratic equation results in the quadratic formula.
The quadratic formula is used to solve a quadratic equation of the form $ax^2+bx+c=0$, where $a\neq0$, by using its coefficients:
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
The quadratic formula can be used to solve any quadratic equation. It is found by completing the square for the standard form of a quadratic equation.

### How to Derive the Quadratic Formula

Standard Form Steps
$ax^2+bx+c=0$, $a\neq0$ The equation is in standard form.
\begin{aligned}\frac{a}{a}x^2+\frac{b}{a}x+\frac{c}{a}&=\frac{0}{a}\\x^2+\frac{b}{a}x+\frac{c}{a}&=0\end{aligned}
Divide both sides by $a$.
$x^2+\frac{b}{a}x=-\frac{c}{a}$
Subtract $\frac{c}{a}$ from both sides.
$x^2+\frac{b}{a}x+\left ( \frac{b}{2a} \right )^{2}=-\frac{c}{a}+\left ( \frac{b}{2a} \right )^{2}$
Add $\left ( \frac{b}{2a} \right )^{2}$ to the left side to complete the square. Add $\left ( \frac{b}{2a} \right )^{2}$ to the right side to keep the equation balanced.
$\left (x+\frac{b}{2a}\right )^2=\frac{b^2}{4a^2}-\frac{c}{a}$
Factor and simplify.
\begin{aligned}\left (x+\frac{b}{2a}\right )^2&=\frac{b^2}{4a^2}-\frac{4ac}{4a^2}\\ \left (x+\frac{b}{2a}\right )^2&=\frac{b^2-4ac}{4a^2}\end{aligned}
Use a common denominator on the right side, and simplify.
\begin{aligned}x+\frac{b}{2a}&=\pm\sqrt{\frac{b^2-4ac}{4a^2}}\\x+\frac{b}{2a}&=\pm\frac{\sqrt{b^2-4ac}}{2a}\end{aligned}
Take the square root of both sides, and simplify.
$x=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$
Subtract $\frac{b}{2a}$ from both sides.
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Simplify.

The quadratic formula can be applied to solve any quadratic equation. The discriminant of the related quadratic equation can be used to determine the number and type of roots of the quadratic function.
The discriminant is the part of the quadratic formula that is under the radical sign:
$b^2-4ac$
Recall that a quadratic equation may have zero, one, or two real solutions. The value of the discriminant determines how many real solutions a quadratic equation has:
• If $b^2-4ac$ is positive, then the equation has two real solutions.
• If $b^2-4ac$ is zero, then the equation has one real solution.
• If $b^2-4ac$ is negative, then the equation has zero real solutions.
Step-By-Step Example
Using the Quadratic Formula with Two Real Solutions
$x^2-4x-6=0$
Step 1
The equation is in the standard form, with $a=1$, $b=-4$, and $c=-6$:
\begin{aligned}ax^2+bx+c&=0\\x^2-4x-6&=0\end{aligned}
The discriminant from the quadratic formula is the expression under the radical symbol:
$x=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}$
Use the discriminant from the quadratic formula to determine the number of solutions:
$\begin{gathered}b^2-4ac\\(-4)^2-4(1)(-6)\\16+24\\40\end{gathered}$
The discriminant is positive. So, the equation has two real solutions.
Step 2
Using the quadratic formula, substitute the value of the discriminant under the radical. Then substitute the values for $a$ and $b$.
\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-(-4)\pm\sqrt{40}}{2(1)}\end{aligned}
Step 3
Simplify the expression.
\begin{aligned}x&=\frac{4\pm\sqrt{40}}{2}\\x&=\frac{4\pm2\sqrt{10}}{2}\\x&=2\pm\sqrt{10}\end{aligned}
Solution
The two solutions of the equation are:
$x=2+\sqrt{10}\;\;\;\;\;\text{or}\;\;\;\;\;x=2-\sqrt{10}$
Step-By-Step Example
Using the Quadratic Formula with One Real Solution
$2x^2+8x+8=0$
Step 1
The equation is in the standard form, with $a=2$, $b=8$, and $c=8$.
\begin{aligned}ax^2+bx+c&=0\\2x^2+8x+8&=0\end{aligned}
The discriminant from the quadratic formula is the expression under the radical symbol:
$x=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}$
Use the discriminant from the quadratic formula to determine the number of solutions:
$\begin{gathered}b^2-4ac\\(8)^2-4(2)(8)\\64-64\\0\end{gathered}$
The discriminant is zero. So, the equation has 1 real solution.
Step 2
Using the quadratic formula, substitute the value of the discriminant under the radical. Then substitute the values for $a$ and $b$.
\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-8\pm\sqrt{0}}{2(2)}\end{aligned}
Solution
Simplify the expression.
\begin{aligned}x&=\frac{-8\pm0}{4}\\x&=\frac{-8}{4}\\x&=-2\end{aligned}
The solution is $x=-2$.
Step-By-Step Example
Using the Quadratic Formula with No Real Solutions
$x^2-6x+11=0$
The equation is in the standard form, with $a=1$, $b=-6$, and $c=11$.
\begin{aligned}ax^2+bx+c&=0\\x^2-6x+11&=0\end{aligned}
$x=\frac{-b\pm\sqrt{{\color{#c42126}{b^2-4ac}}}}{2a}$
$\begin{gathered}b^2-4ac\\(-6)^2-4(1)(11)\end{gathered}$
$\begin{gathered}(-6)^2-4(1)(11)\\(-6)(-6)-(4)(11)\\36-44\\-8\end{gathered}$