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Mar 17, 2016
 #1
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Mar 17, 2016
 #2
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Mar 17, 2016
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Mar 17, 2016
 #2
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(2M)/(2M+3)-(2M)/(2M-3)=1 How many extraneous solutions does the equation have.

 

\(\small{ \begin{array}{rcll} \frac{2M}{2M+3}-\frac{2M}{2M-3} &=& 1 \quad & | \quad :2M \\ \frac{1}{2M+3}-\frac{1}{2M-3} &=& \frac{1}{2M} \quad & | \quad \cdot (2M+3) \cdot (2M-3) \\ \frac{(2M+3) \cdot (2M-3) }{2M+3}-\frac{(2M+3) \cdot (2M-3) }{2M-3} &=& \frac{(2M+3) \cdot (2M-3) }{2M} \\ (2M-3) - (2M+3) &=& \frac{(2M+3) \cdot (2M-3) }{2M} \\ 2M-3 - 2M-3 &=& \frac{(2M+3) \cdot (2M-3) }{2M} \\ -6 &=& \frac{(2M+3) \cdot (2M-3) }{2M} \quad & | \quad (2M+3) \cdot (2M-3) = (2M)^2-3^2 = 4M^2-9 \\ -6 &=& \frac{4M^2-9}{2M} \quad & | \quad \cdot 2M \\ -6\cdot 2M &=& 4M^2-9\\ 4M^2-9 &=& -6\cdot 2M\\ 4M^2-9 &=& -12M\quad & | \quad +12M \\ 4M^2+12M-9 &=& 0\\ \end{array} }\)

 

\(\boxed{~ \begin{array}{rcll} a\cdot M^2+b\cdot M+c &=& 0 \\ M &=& \dfrac{-b \pm \sqrt{b^2-4ac} }{2a} \end{array} ~}\)

 

\(\small{ \begin{array}{rcll} 4M^2+12M-9 &=& 0 \quad | \quad a=4 \quad b = 12 \quad c = -9 \\ M &=& \dfrac{-12 \pm \sqrt{12^2-4\cdot 4 \cdot (-9) } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm \sqrt{144+144 } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm \sqrt{ 2\cdot 144 } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm \sqrt{ 2\cdot 12^2 } }{2\cdot 4} \\ M &=& \dfrac{-12 \pm 12\cdot\sqrt{ 2} }{2\cdot 4} \\ M &=& \dfrac{-3 \pm 3\cdot\sqrt{ 2} }{2 } \\\\ M_1 &=& \dfrac{-3 + 3\cdot\sqrt{ 2} }{2 } \\ \mathbf{M_1} & \mathbf{=} & \mathbf{0.62132034356} \\\\ M_2 &=& \dfrac{-3 - 3\cdot\sqrt{ 2} }{2 } \\ \mathbf{M_2} &\mathbf{=} & \mathbf{-3.62132034356 } \end{array} }\)

 

laugh

Mar 17, 2016
 #1
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use http://www.cymath.com/  it will explain you step by step

Mar 17, 2016
 #5
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Mar 17, 2016

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