The condition that the equation 1/x + 1/x+b = 1/m + 1/m+b has real roots that are equal in magnitude but opposite in sign

Question

The condition that the equation 1/x + 1/x+b = 1/m + 1/m+b has real roots that are equal in magnitude but opposite in sign

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Hadley 4 weeks 2021-08-21T23:16:57+00:00 1 Answer 0 views 0

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    2021-08-21T23:18:43+00:00

    Answer:

    The required condition is

     {b}^{2}  = 2 {m}^{2}

    Step-by-step explanation:

     \frac{1}{x}  +  \frac{1}{x + b}  =  \frac{1}{m}  +  \frac{1}{m + b}

     \frac{(x + b) + x}{x( x+ b)}  =  \frac{(m + b) + m}{m(m+ b)}

     \frac{2x + b}{x(x + b)}  = \frac{2m + b}{m(m + b)}

    (2x + b)m(m + b) = (2m + b)x(x + b)

    2m(m + b)x + mb(m + b) = (2m + b) {x}^{2}  + (2m + b)bx

    (2m + b) {x}^{2}  + ((2m + b)b - 2m(m + b))x - mb(m + b) = 0

    It is given that roots are equal and opposite ,

    or \:  \alpha  =  -  \beta

    or \:  \alpha  +  \beta  = 0

    or \:   - \frac{coeff. \: of \: x}{coeff. \: of  \: {x}^{2}  }  = 0

     \frac{(2m + b)b - 2m(m + b)}{2m + b}  = 0

    2mb +  {b}^{2}  - 2 {m}^{2}  - 2mb = 0

     {b}^{2}  - 2 {m}^{2}  = 0

     {b}^{2}  = 2 {m}^{2}

    This is the required condition.

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