How many extraneous solutions does the equation below have?
$\frac{2 m}{2 m+3}-\frac{2 m}{2 m-3}=1$
To find the number of extraneous solutions, we first identify the values of \(m\) that are not allowed. An extraneous solution arises when a potential solution makes the denominators in the original equation equal to zero. The denominators are \(2m+3\) and \(2m-3\). Setting these to zero gives:
Thus, any solution resulting in \(m = \pm \frac{3}{2}\) would be an extraneous solution.
Next, we solve the equation \(\frac{2m}{2m+3} - \frac{2m}{2m-3} = 1\) by clearing the denominators. Multiplying the entire equation by the common denominator \((2m+3)(2m-3)\) yields:
\[2m(2m-3) - 2m(2m+3) = (2m+3)(2m-3)\]
Expanding the left and right sides:
\[(4m^2 - 6m) - (4m^2 + 6m) = 4m^2 - 9\]
Simplifying the left side:
\[4m^2 - 6m - 4m^2 - 6m = 4m^2 - 9\]
\[-12m = 4m^2 - 9\]
Rearranging into a standard quadratic form:
\[4m^2 + 12m - 9 = 0\]
We use the quadratic formula to solve \(4m^2 + 12m - 9 = 0\), where \(a=4, b=12, c=-9\):
\[m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-12 \pm \sqrt{12^2 - 4(4)(-9)}}{2(4)}\]
\[m = \frac{-12 \pm \sqrt{144 + 144}}{8} = \frac{-12 \pm \sqrt{288}}{8} = \frac{-12 \pm 12\sqrt{2}}{8}\]
Dividing both terms in the numerator by 4:
\[m = \frac{-3 \pm 3\sqrt{2}}{2}\]
The potential solutions are \(m = \frac{-3 + 3\sqrt{2}}{2}\) and \(m = \frac{-3 - 3\sqrt{2}}{2}\). Neither of these values equals \(\frac{3}{2}\) or \(-\frac{3}{2}\). Therefore, there are no extraneous solutions.
Extraneous solutions often appear in physics when calculating the velocity of objects or wave interference patterns. If a formula derived for a physical system predicts a speed that makes a division by zero in the governing model (such as at a critical resonant frequency), that mathematical result is considered extraneous, as it cannot exist in reality. Identifying these helps scientists avoid impossible predictions in their models.