Solving Quadratics By Factoring Worksheet - Chapter 14-1 Page 2
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Section 14–1
Solving Quadratics by Factoring
The equation is now in general form, with a
5, b
12, and c
21.
◆◆◆
Number of Roots
We’ll see later in this chapter that the maximum number of solutions that certain types of equa-
tions may have is equal to the degree of the equation. Thus a quadratic equation, being of degree
2, has two solutions or roots. The two roots are sometimes equal, or they may be imaginary or
complex numbers. However, in applications, we’ll see that one of the two roots must sometimes
be discarded.
◆◆◆
Example 3:
The quadratic equation
x
2
4
has two roots, x
2 and x
2.
◆◆◆
Solving Pure Quadratics
To solve a pure quadratic, we simply isolate the x
2
term and then take the square root of both
sides, as in the following example.
2
◆◆◆
Example 4:
Solve 3x
75
0.
Solution:
Adding 75 to both sides and dividing by 3, we obtain
2
3x
75
x
2
25
Taking the square root yields
x
25
5
Check:
We check our solution the same way as for other equations, by substituting back into
the original equation. Now, however, we must check two solutions.
2
Substitute
5: 3(5)
75
0
75
75
0 (checks)
2
Substitute
5: 3( 5)
75
0
75
75
0 (checks)
◆◆◆
When taking the square root, be sure to keep both the plus and the minus
values. Both will satisfy the equation.
At this point students usually grumble: “First we’re told that
4
2 only, not
2
(Sec. 1–5). Now we’re told to keep both plus and minus. What’s going on here?” Here’s the
difference: When we solve a quadratic, we know that it must have two roots, so we keep both
values. When we evaluate a square root, such as
4 , which is not the solution of a quadratic,
we agree to take only the positive value to avoid ambiguity.
The roots might sometimes be irrational, as in the following example.
2
◆◆◆
Example 5:
Solve 4x
15
0.
Solution:
Following the same steps as before, we get
2
4x
15
15
x
2
4
15
15
x
◆◆◆
4
2
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