How many extraneous solutions does the equation below have? (2m)/(2m + 3) - (2m)/(2m - 3) = 1

Solution:

\(\frac{2m}{(2m+3)} - \frac{2m}{(2m-3)} = 1\)

\(\frac{2m(2m-3)- 2m(2m+3)}{(2m+3)(2m-3)} = 1\)

\(\frac{4m^{2} - 6m - 4m^{2} - 6m}{(4m^{2} - 9)} = 1\)

\(\frac{-12m}{4m^{2}-9} = 1\)

\(-12m = 4m^{2} - 9\)

4m² + 12m - 9 = 0(1)

Since (1) is a quadratic equation its roots can be calculated by using the formula which used for getting roots of the quadratic equation ax² + bx + c

\(\frac{-b \pm \sqrt{b^{2}-4ac} }{2a}\)

In equation (1) the variable is m and a = 4; b = 12; and c = -9

The roots are calculated as:

\(\frac{-12 \pm \sqrt{12^{2}-4(4)(-9)} }{2(4)}\)

= \(\frac{-12 \pm \sqrt{144 + 144} }{2(4)}\)

= \(\frac{-12 \pm \sqrt{2 \times 144} }{2(4)}\)

= \(\frac{-12 \pm 12\sqrt{2} }{8}\)

= \(\frac{-3}{2} \pm \frac{3}{2}\sqrt{2}\)

= \(\frac{-3}{2} - \frac{3}{2}\sqrt{2}\) and \(\frac{-3}{2} + \frac{3}{2}\sqrt{2}\)

First Substituting \(\frac{-3}{2} - \frac{3}{2}\sqrt{2}\) in equation (1)

\(4[\frac{-3}{2} - \frac{3}{2}\sqrt{2}]^{2} + 12[\frac{-3}{2} - \frac{3}{2}\sqrt{2}] - 9 = 0\)

\(4[\frac{9}{4} + 2(\frac{9}{4})+ (\frac{-3}{2})(\frac{-3}{2})\sqrt{2})] + 12(\frac{-3}{2}) + 12(\frac{-3}{2})\sqrt{2}) - 9 = 0\)

\(4(\frac{9}{4}) + 8(\frac{9}{4}) + 8(\frac{9}{4})\sqrt{2}-18 -18\sqrt{2} - 9 = 0\)

9 + 18 + 18√2 - 18 - 18√2 - 9 = 0

0 = 0

Hence the root -3/2 - 3(√2)/2 is not an extraneous solution.

Now Substituting \(\frac{-3}{2} + \frac{3}{2}\sqrt{2}\) in equation (1) we have

\(4[\frac{-3}{2} + \frac{3}{2}\sqrt{2}]^{2} + 12[\frac{-3}{2}+\frac{3}{2}\sqrt{2}] - 9 = 0\)

\(4[(\frac{-3}{2})^{2} + (\frac{3}{2}\sqrt{2})^{2} + 2(\frac{-3}{2})(\frac{3}{2}\sqrt{2})] + 12(\frac{-3}{2}) + 12(\frac{3}{2}\sqrt{2})- 9 = 0\)

\(4(\frac{9}{4}) + 4(\frac{9}{4})(2) -2({9})\sqrt{2} - 18 + 18\sqrt{2} - 9 = 0\)

9 + 18 - 18√2 - 18 + 18√2 -9 = 0

0 = 0 

Hence the root -3/2 + 3(√2)/2 is ALSO not an extraneous solution.

Both the roots are valid solutions to the given problem.

Thus, the equation (2m)/(2m+3)-(2m)/(2m-3) = 1 does Not have an extraneous solution.

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How many extraneous solutions does the equation below have? (2m)/(2m + 3) - (2m)/(2m - 3)=1

Summary:

The number of extraneous solutions the equation (2m)/(2m+3)-(2m)/(2m-3)=1 have is zero.