Question

how many extraneous solutions does the equation have?
\\(\frac{2m}{2m + 3} - \frac{2m}{2m - 3} = 1\\)
3
1
2
0

Explanation:

Step1: Find the domain restrictions

The denominators are \(2m + 3\) and \(2m - 3\). Set them equal to zero to find values that make the equation undefined:

• \(2m+3 = 0\) gives \(m=-\frac{3}{2}\)
• \(2m - 3=0\) gives \(m=\frac{3}{2}\)

Step2: Solve the equation

First, find a common denominator \((2m + 3)(2m - 3)\) and multiply both sides by it:

$$ 2m(2m - 3)-2m(2m + 3)=(2m + 3)(2m - 3) $$

Expand each term:

$$ 4m^{2}-6m-4m^{2}-6m = 4m^{2}-9 $$

Simplify the left side:

$$ -12m=4m^{2}-9 $$

Rearrange into standard quadratic form:

$$ 4m^{2}+12m - 9 = 0 $$

Use the quadratic formula \(m=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 4\), \(b = 12\), \(c=-9\):

$$ m=\frac{-12\pm\sqrt{144+144}}{8}=\frac{-12\pm\sqrt{288}}{8}=\frac{-12\pm12\sqrt{2}}{8}=\frac{-3\pm3\sqrt{2}}{2} $$

Step3: Check for extraneous solutions

The solutions we found \(\frac{-3 + 3\sqrt{2}}{2}\) and \(\frac{-3 - 3\sqrt{2}}{2}\) are not equal to \(\pm\frac{3}{2}\) (the domain restrictions). So there are no extraneous solutions.

Answer:

0