hard question

solve in x 4 power(x-sqroot(x (power2) -5)))-12*2 power(x-1-(x power 2-5)power (.5))+ 8=0?

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You should write the equation such that:

`2^(2(x - sqrt(x^2 - 5))) - 12*2^(x - sqrt(x^2 - 5))*2^(-1) +8 = 0`

You should use the following substitution to solve the equation such that:

`2^(x - sqrt(x^2 - 5)) = t`

Changing the variable yields:

`t^2 - 12t/2 + 8 = 0 => t^2 - 6t + 8 = 0`

`t^2 - 6t +12 - 4= 0`

You need to group the terms such that:

`(t^2 - 4) + (-6t + 12) = 0`

`(t - 2)(t + 2) - 6(t - 2) = 0`

Factoring out `t - 2 ` yields:

`(t - 2)(t + 2 - 6) = 0 => (t - 2)(t - 4) = 0 => {(t - 2 = 0),(t - 4 = 0):} => {(t = 2),(t = 4):}`

You should solve for x the equations:

`{(2^(x - sqrt(x^2 - 5)) = 2),(2^(x - sqrt(x^2 - 5)) = 4):} =gt `

`x - sqrt(x^2 - 5) = 1 => x - 1 = sqrt(x^2 - 5)`

You should raise to square both sides such that:

`x^2 - 2x + 1 = x^2 - 5 => -2x + 1 = -5 => -2x = -6 => x = 3`

Since the square root exists if `x^2 - 5 >= 0 => x in (-oo , -sqrt5)U(sqrt5 , oo)` , then `x = 3` can be accepted since `xin (sqrt5 , oo)` .

`x - sqrt(x^2 - 5) =2 => x -2 = sqrt(x^2 - 5)`

You should raise to square both sides such that:

`x^2 - 4x + 4 = x^2 - 5 => -4x = -9 => x = 9/4 > sqrt 5`

**Hence, evaluating the solutions to the given equation yields `x = 3` and `x = 9/4` .**

Sir, what you have written about square root exists if x^2 - 5 > 0, why is it so ??? plz elaborate, what does U denote and what is the sign you gave after can be accepted since ..........

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