Question

How many extraneous solutions does the equation below have?

Answer 1

Answer:

The equation has:

              zero extraneous solution.

( Both are true solution )

Step-by-step explanation:

We are given a algebraic expression in terms of variable m as:

            $\dfrac{2m}{2m+3}-\dfrac{2m}{2m-3}=1$

On taking least common multiple we get:

     $\dfrac{2m\times ( 2m-3)-2m\times (2m+3)}{(2m+3)(2m-3)}=1$

i.e.

$\dfrac{4m^2-6m-4m^2-6m}{4m^2-9}=1\\\\\\\dfrac{-12m}{4m^2-9}=1\\\\\\-12m=4m^2-9\\\\\\4m^2+12m-9=0$

Hence, on solving for m we get:

 $m=\dfrac{-3\pm 3\sqrt{2}}{2}$

• when $m=\dfrac{-3+3\sqrt{2}}{2}$

Then,  

$\dfrac{2\times \dfrac{-3+3\sqrt{2}}{2}}{2\times \dfrac{-3+3\sqrt{2}}{2}+3}-\dfrac{2\times \dfrac{-3+3\sqrt{2}}{2}}{2\times \dfrac{-3+3\sqrt{2}}{2}-3}=1\\\\\\\dfrac{-3+3\sqrt{2}}{-3+3\sqrt{2}+3}-\dfrac{-3+3\sqrt{2}}{-3+3\sqrt{2}-3}=1\\\\\\\\\dfrac{-3+3\sqrt{2}}{3\sqrt{2}}-\dfrac{-3+3\sqrt{2}}{-6+3\sqrt{2}}=1\\\\\\\dfrac{-1+\sqrt{2}}{\sqrt{2}}-\dfrac{-1+\sqrt{2}}{-2+\sqrt{2}}=1\\\\\\(\sqrt{2}-1)(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}-2})=1\\\\\\(\sqrt{2}-1)(\dfrac{\sqrt{2}-2-\sqrt{2}}{(\sqrt{2}-2)})=1$

$\dfrac{-2\times (\sqrt{2}-1)}{(\sqrt{2}-2)\times \sqrt{2})}=1$

$\dfrac{-\sqrt{2}(\sqrt{2}-1)}{(\sqrt{2}-2)}=1\\\\\\\dfrac{\sqrt{2}-2}{\sqrt{2}-2}=1\\\\\\1=1$

Hence, the solution is a true solution.

• Similarly we may check for:

 $m=\dfrac{-3-3\sqrt{2}}{2}$

It will also be a true solution.

Answer 2

Two. m=1.5 and m=-1.5