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calculate real parameter m for which the equation has a unique solution:...

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alexa0048 | Student, Grade 11 | eNoter

Posted May 21, 2013 at 3:16 PM via web

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calculate real parameter m for which the equation has a unique solution:

4^x-(5m-2)*2^x+4m^2-3m=0

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 21, 2013 at 3:26 PM (Answer #1)

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The parameter m has to be determined for which `4^x-(5m-2)*2^x+4m^2-3m=0` has a unique solution.

`4^x-(5m-2)*2^x+4m^2-3m=0`

let `2^x = y`

=> `y^2-(5m-2)*y+4m^2-3m=0`

For a unique solution `(5m - 2)^2 = 4*(4m^2-3m)`

=> `25m^2 + 4 - 20m = 16m^2 - 12m`

=> `9m^2 - 8m + 4 = 0`

This equation does not have a real solution.

There is no real value of m for which the given equation has a unique solution.

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