(2M)/(2M+3)-(2M)/(2M-3)=1 How many extraneous solutions does the equation have.
Solve for x over the real numbers:
(2 x)/(2 x+3)-(2 x)/(2 x-3) = 1
Bring (2 x)/(2 x+3)-(2 x)/(2 x-3) together using the common denominator (2 x-3) (2 x+3):
-(12 x)/((2 x-3) (2 x+3)) = 1
Multiply both sides by (2 x-3) (2 x+3):
-12 x = (2 x-3) (2 x+3)
Expand out terms of the right hand side:
-12 x = 4 x^2-9
Subtract 4 x^2-9 from both sides:
-4 x^2-12 x+9 = 0
Divide both sides by -4:
x^2+3 x-9/4 = 0
Add 9/4 to both sides:
x^2+3 x = 9/4
Add 9/4 to both sides:
x^2+3 x+9/4 = 9/2
Write the left hand side as a square:
(x+3/2)^2 = 9/2
Take the square root of both sides:
x+3/2 = 3/sqrt(2) or x+3/2 = -3/sqrt(2)
Subtract 3/2 from both sides:
x = 3/sqrt(2)-3/2 or x+3/2 = -3/sqrt(2)
Subtract 3/2 from both sides:
Answer: | x = 3/sqrt(2)-3/2 or x = -3/2-3/sqrt(2)
P.S. I changed your variable M into x for some technical reasons.
+26387 (2M)/(2M+3)-(2M)/(2M-3)=1 How many extraneous solutions does the equation have.
\(\small{ \begin{array}{rcll} \frac{2M}{2M+3}-\frac{2M}{2M-3} &=& 1 \quad & | \quad :2M \\ \frac{1}{2M+3}-\frac{1}{2M-3} &=& \frac{1}{2M} \quad & | \quad \cdot (2M+3) \cdot (2M-3) \\ \frac{(2M+3) \cdot (2M-3) }{2M+3}-\frac{(2M+3) \cdot (2M-3) }{2M-3} &=& \frac{(2M+3) \cdot (2M-3) }{2M} \\ (2M-3) - (2M+3) &=& \frac{(2M+3) \cdot (2M-3) }{2M} \\ 2M-3 - 2M-3 &=& \frac{(2M+3) \cdot (2M-3) }{2M} \\ -6 &=& \frac{(2M+3) \cdot (2M-3) }{2M} \quad & | \quad (2M+3) \cdot (2M-3) = (2M)^2-3^2 = 4M^2-9 \\ -6 &=& \frac{4M^2-9}{2M} \quad & | \quad \cdot 2M \\ -6\cdot 2M &=& 4M^2-9\\ 4M^2-9 &=& -6\cdot 2M\\ 4M^2-9 &=& -12M\quad & | \quad +12M \\ 4M^2+12M-9 &=& 0\\ \end{array} }\)
\(\boxed{~ \begin{array}{rcll} a\cdot M^2+b\cdot M+c &=& 0 \\ M &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)
\(\small{ \begin{array}{rcll} 4M^2+12M-9 &=& 0 \quad | \quad a=4 \quad b = 12 \quad c = -9 \\ M &=& \dfrac{-12 \pm \sqrt{12^2-4\cdot 4 \cdot (-9) } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm \sqrt{144+144 } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm \sqrt{ 2\cdot 144 } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm \sqrt{ 2\cdot 12^2 } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm 12\cdot\sqrt{ 2} }{2\cdot 4} \\ M &=& \dfrac{-3 \pm 3\cdot\sqrt{ 2} }{2 } \\\\ M_1 &=& \dfrac{-3 + 3\cdot\sqrt{ 2} }{2 } \\ \mathbf{M_1} & \mathbf{=} & \mathbf{0.62132034356} \\\\ M_2 &=& \dfrac{-3 - 3\cdot\sqrt{ 2} }{2 } \\ \mathbf{M_2} &\mathbf{=} & \mathbf{-3.62132034356 } \end{array} }\)
