 ## Solving Quadratic Equations and Inequalities The solutions of a quadratic equation are the $x$-intercepts of the graph of the related function.
A quadratic equation is a polynomial equation, where $a$, $b$, and $c$ are real numbers and $a\neq0$, that can be written in the form:
$ax^2+bx+c=0$
A solution of an equation is a value of the variable that makes the equation true, which means both sides are equal. A quadratic equation can have zero, one, or two real solutions. A solution to a quadratic equation is also called a root. The real solutions of a quadratic equation represent zeros of the related quadratic function:
$f(x)=ax^2+bx+c$
A zero of a function is any input value that makes the output of the function equal zero. The zeros also describe the $x$-intercepts of the graph of the function. An x-intercept is the value of $x$, where the graph touches or crosses the $x$-axis.

To solve a quadratic equation by graphing, graph the related function $f(x)=ax^2+bx+c$. The real solutions are represented by the $x$-intercepts of the graph. A quadratic function may have zero, one, or two $x$-intercepts. For example:

• For the equation $x^2+x-6=0$, the graph of its related function, $f(x)=x^2+x-6$, has two $x$-intercepts, $x=-3$ and $x=2$. So, the equation has two real solutions: $x=-3$ and $x=2$.
• For the equation $x^2-8x+16=0$, the graph of its related function, $g(x)=x^2-8x+16$, has one $x$-intercept at $x=4$. So, the equation has one real solution at $x=4$.
• For the equation $-x^2-12x-38=0$, the graph of the related function, $h(x)=-x^2-12x-38$, has no $x$-intercepts. So, the equation has no real solutions.
On a graphing calculator, TRACE can be used to estimate the $x$-intercepts. When graphing by hand, it may be difficult to identify exact solutions. Even when the graph appears to pass through a point, verify that the $x$-intercepts read from the graph are solutions of the equation.
Solutions or Roots Zeros $x$-Intercepts
An equation has solutions, which are also called roots. A solution is a value of the variable that makes the equation true. A function has zeros. A zero is an input value that makes the value of the function zero. A graph has $x$-intercepts. An $x$-intercept is an $x$-coordinate of a point where the graph touches or crosses the $x$-axis.

Step-By-Step Example
Graphing to Solve a Quadratic Equation
Solve the quadratic equation by graphing:
$x^2+6x=-5$
Step 1
Write the quadratic equation in standard form, where $a$ represents the coefficient of $x^2$, $b$ represents the coefficient of $x$, and $c$ represents the last term.
$ax^2+bx+c=0$
Add 5 to both sides of the equation:
\begin{aligned}x^2+6x&={-5}\\{x^2}+6x+5&=0\end{aligned}
Step 2
Write the related function of the quadratic equation:
\begin{aligned}f(x)&={ax^2+bx+c}\\{y}&={x^2}+6x+5\end{aligned}
Identify the values of $a$, $b$, and $c$.
\begin{aligned}a&={1}\\{b}&={6}\\{c}&=5\end{aligned}
Step 3

Use the values of $a$, $b$, and $c$ to determine the vertex and axis of symmetry.

To determine the vertex, use the formulas for the $x$- and $y$-coordinates of the graph of a quadratic equation.
\begin{aligned}{\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)}&={\left(-\frac{6}{{2}\cdot{1}},5-\frac{6^2}{{4}\cdot{1}}\right)}\\&={\left(-\frac{6}{2},5-\frac{36}{4}\right)}\\&={(-3,5-9)}\\&=(-3,-4)\end{aligned}
To determine the location of the axis of symmetry, identify the $x$-coordinate of the vertex. The graph of the given equation has an axis of symmetry at $x=-3$.
Step 4
Identify at least one other point on the graph by determining the $y$-intercept. Substitute zero for $x$:
\begin{aligned}y&=x^2+{6x+5}\\&=0^2+6(0)+{5}\\&=5\end{aligned}
The graph passes through the point $(0,5)$.

Another point that is easy to locate is 1 unit right of the vertex. From the vertex, move 1 unit right and $a$ units up (if $a$ is positive) or down (if $a$ is negative). Since $a=1$, move 1 unit right and 1 unit up from the vertex. The graph passes through $(-2,-3)$.

Use symmetry to identify points reflected across the axis of symmetry. The axis of symmetry is $x=-3$, so the mirror images of the points $(0,5)$ and $(-2,-3)$ with respect to axis of symmetry are also points on the graph. Therefore, the graph passes through $(-6,5)$ and $(-4,-3)$.

Step 5
Plot the vertex, and draw the axis of symmetry. Then plot the additional points and sketch the graph, which is a parabola.
Solution

The $x$-intercepts represent the solutions of the equation. Use the graph to identify the $x$-intercepts.

The graph appears to pass through the points $(-5,0)$ and $(-1,0)$, so the $x$-intercepts appear to be –5 and –1.
Verify the solutions by substituting the values of the $x$-intercepts in the original quadratic equation. Using the point $(-5, 0)$:
\begin{aligned}x^2+6x&={-5}\\(-5)^2+6(-5)&=-5\;\;\;\;\; \rm{?}\\25-30&=-5\;\;\;\;\;\rm{?}\\-5&=-5\;\;\;\;\;\checkmark\end{aligned}
Using the point $(-1, 0)$:
\begin{aligned}x^2+6x&={-5}\\(-1)^2+6(-1)&=-5 \;\;\;\;\;\rm{?}\\1-6&=-5\;\;\;\;\;\rm{?}\\-5&=-5\;\;\;\;\;\checkmark\end{aligned}
The solutions are $x=-5$ and $x=-1$.