To solve the equation [tex]\(\frac{2m}{2m + 3} - \frac{2m}{2m - 3} = 1\)[/tex] and determine the number of extraneous solutions, let's go through a step-by-step process.
### Step 1: Combine the Fractions
First, let's combine the fractions on the left-hand side of the equation:
[tex]\[
\frac{2m}{2m + 3} - \frac{2m}{2m - 3}
\][/tex]
To combine these fractions, we'll need a common denominator, which is [tex]\((2m + 3)(2m - 3)\)[/tex].
[tex]\[
\frac{2m(2m - 3) - 2m(2m + 3)}{(2m + 3)(2m - 3)}
\][/tex]
### Step 2: Simplify the Numerator
Now, let's expand and simplify the numerator:
[tex]\[
2m(2m - 3) - 2m(2m + 3) = 4m^2 - 6m - (4m^2 + 6m)
\][/tex]
Combine like terms in the numerator:
[tex]\[
4m^2 - 6m - 4m^2 - 6m = -12m
\][/tex]
So the combined fraction is:
[tex]\[
\frac{-12m}{(2m + 3)(2m - 3)}
\][/tex]
### Step 3: Solve the Equation
The equation now reads:
[tex]\[
\frac{-12m}{(2m + 3)(2m - 3)} = 1
\][/tex]
Multiply both sides by [tex]\((2m + 3)(2m - 3)\)[/tex] to clear the fraction:
[tex]\[
-12m = (2m + 3)(2m - 3)
\][/tex]
Recognize that [tex]\((2m + 3)(2m - 3)\)[/tex] is a difference of squares:
[tex]\[
(2m + 3)(2m - 3) = 4m^2 - 9
\][/tex]
So the equation becomes:
[tex]\[
-12m = 4m^2 - 9
\][/tex]
### Step 4: Rearrange and Solve the Quadratic Equation
Rearrange the equation to standard quadratic form:
[tex]\[
4m^2 - 12m - 9 = 0
\][/tex]
Using the quadratic formula [tex]\(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 4\)[/tex], [tex]\(b = -12\)[/tex], and [tex]\(c = -9\)[/tex]:
[tex]\[
m = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot (-9)}}{2 \cdot 4}
\][/tex]
Calculate the discriminant:
[tex]\[
\Delta = 144 + 144 = 288
\][/tex]
So:
[tex]\[
m = \frac{12 \pm \sqrt{288}}{8}
\][/tex]
Simplify [tex]\(\sqrt{288}\)[/tex]:
[tex]\[
\sqrt{288} = \sqrt{144 \cdot 2} = 12\sqrt{2}
\][/tex]
Thus:
[tex]\[
m = \frac{12 \pm 12\sqrt{2}}{8} = \frac{12(1 \pm \sqrt{2})}{8} = \frac{3(1 \pm \sqrt{2})}{2}
\][/tex]
The solutions are:
[tex]\[
m = \frac{3(1 + \sqrt{2})}{2} \quad \text{and} \quad m = \frac{3(1 - \sqrt{2})}{2}
\][/tex]
### Step 5: Check for Extraneous Solutions
We must ensure that neither solution makes the original denominators zero. We originally stated:
[tex]\[
2m + 3 = 0 \quad \text{or} \quad 2m - 3 = 0
\][/tex]
1. For [tex]\(m = \frac{3(1 + \sqrt{2})}{2}\)[/tex]:
[tex]\[
2m = 3(1+\sqrt{2}) \quad \Rightarrow \quad 2m + 3 = 3(1+\sqrt{2}) + 3 \neq 0 \quad \text{and} \quad 2m - 3 = 3(1+\sqrt{2}) - 3 \neq 0
\][/tex]
2. For [tex]\(m = \frac{3(1 - \sqrt{2})}{2}\)[/tex]:
[tex]\[
2m = 3(1-\sqrt{2}) \quad \Rightarrow \quad 2m + 3 = 3(1-\sqrt{2}) + 3 \neq 0 \quad \text{and} \quad 2m - 3 = 3(1-\sqrt{2}) - 3 \neq 0
\][/tex]
### Conclusion
Since neither solution makes the denominator zero, there are no extraneous solutions.
Final Answer:
[tex]\[
0
\][/tex]
