Solving Quadratic Equations and Inequalities

Graphing Quadratic Equations

The solutions of a quadratic equation are the xx-intercepts of the graph of the related function.
A quadratic equation is a polynomial equation, where aa, bb, and cc are real numbers and a0a\neq0, that can be written in the form:
ax2+bx+c=0ax^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)=ax2+bx+cf(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 xx-intercepts of the graph of the function. An x-intercept is the value of xx, where the graph touches or crosses the xx-axis.

To solve a quadratic equation by graphing, graph the related function f(x)=ax2+bx+cf(x)=ax^2+bx+c. The real solutions are represented by the xx-intercepts of the graph. A quadratic function may have zero, one, or two xx-intercepts. For example:

  • For the equation x2+x6=0x^2+x-6=0, the graph of its related function, f(x)=x2+x6f(x)=x^2+x-6, has two xx-intercepts, x=3x=-3 and x=2x=2. So, the equation has two real solutions: x=3x=-3 and x=2x=2.
  • For the equation x28x+16=0x^2-8x+16=0, the graph of its related function, g(x)=x28x+16g(x)=x^2-8x+16, has one xx-intercept at x=4x=4. So, the equation has one real solution at x=4x=4.
  • For the equation x212x38=0-x^2-12x-38=0, the graph of the related function, h(x)=x212x38h(x)=-x^2-12x-38, has no xx-intercepts. So, the equation has no real solutions.
The xx-intercepts of a graph represent real solutions of the related equation.
On a graphing calculator, TRACE can be used to estimate the xx-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 xx-intercepts read from the graph are solutions of the equation.
Solutions or Roots Zeros xx-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 xx-intercepts. An xx-intercept is an xx-coordinate of a point where the graph touches or crosses the xx-axis.

Step-By-Step Example
Graphing to Solve a Quadratic Equation
Solve the quadratic equation by graphing:
x2+6x=5x^2+6x=-5
Step 1
Write the quadratic equation in standard form, where aa represents the coefficient of x2x^2, bb represents the coefficient of xx, and cc represents the last term.
ax2+bx+c=0ax^2+bx+c=0
Add 5 to both sides of the equation:
x2+6x=5x2+6x+5=0\begin{aligned}x^2+6x&={-5}\\{x^2}+6x+5&=0\end{aligned}
Step 2
Write the related function of the quadratic equation:
f(x)=ax2+bx+cy=x2+6x+5\begin{aligned}f(x)&={ax^2+bx+c}\\{y}&={x^2}+6x+5\end{aligned}
Identify the values of aa, bb, and cc.
a=1b=6c=5\begin{aligned}a&={1}\\{b}&={6}\\{c}&=5\end{aligned}
Step 3

Use the values of aa, bb, and cc to determine the vertex and axis of symmetry.

To determine the vertex, use the formulas for the xx- and yy-coordinates of the graph of a quadratic equation.
(b2a,cb24a)=(621,56241)=(62,5364)=(3,59)=(3,4)\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 xx-coordinate of the vertex. The graph of the given equation has an axis of symmetry at x=3x=-3.
Step 4
Identify at least one other point on the graph by determining the yy-intercept. Substitute zero for xx:
y=x2+6x+5=02+6(0)+5=5\begin{aligned}y&=x^2+{6x+5}\\&=0^2+6(0)+{5}\\&=5\end{aligned}
The graph passes through the point (0,5)(0,5).

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

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

The graph appears to pass through the points (5,0)(-5,0) and (1,0)(-1,0), so the xx-intercepts appear to be –5 and –1.
Verify the solutions by substituting the values of the xx-intercepts in the original quadratic equation. Using the point (5,0)(-5, 0):
x2+6x=5(5)2+6(5)=5?2530=5?5=5\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)(-1, 0):
x2+6x=5(1)2+6(1)=5?16=5?5=5\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=5x=-5 and x=1x=-1.