# What is a if eq sq root(x+3-4sq root(x-1))+asq root(x+3+4sq root(x-1))=4 have unic solution?

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### 1 Answer

You should simplify the expression under the square root, using the following substitution, such that:

`x - 1 = t^2 => x = t^2 + 1`

`x + 3 = t^2 + 1 + 3 => x + 3 = t^2 + 4`

Replacing the variable under the square root yields:

`sqrt(t^2 + 4 - 4sqrt(t^2)) + a*sqrt(t^2 + 4 + 4sqrt(t^2)) = 4`

`sqrt(t^2 + 4 - 4t) + a*sqrt(t^2 + 4 + 4t) = 4`

Using the special products `(a +- b)^2 = a^2 +- 2ab + b^2` yields:

`sqrt((t - 2)^2) + a*sqrt((t + 2)^2) = 4`

Since `sqrt(x^2) = |x|` yields:

`|t - 2| + a*|t + 2| = 4`

Using the absolute value definition yields:

`|t - 2| = {(t - 2, t>=2),(2 - t, t< 2):}`

`|t + 2| = {(t + 2, t>=-2),(-2 - t, t< -2):}`

You need to evaluate the equation considering the following intervals, such that:

`t in (-oo,-2)`

`2 - t + a(t + 2) = 4 => 2 - t + at + 2a = 4 => t*(a - 1) = 4 - 2 - 2a`

`t*(a - 1) = 2 - 2a => t*(a - 1) =-2(a - 1) => t = -2`

Since` t = -2 !in` ` (-oo,-2)` , the result is not valid.

`t in [-2,2)`

`2 - t + a*(t + 2) = 4 => 2 - t + at + 2a = 4 => t*(a - 1) = 2 - 2a`

`t(a - 1) = -2(a - 1) => t = -2`

Since `t = -2` in `t in [-2,2)` , the result is valid.

`t in [2,oo)`

`t - 2 + a*(t + 2) = 4 => t - 2 + at + 2a = 4 => t(a+1) = 6 - 2a`

`t(a+1) = 2(3 - a)` invalid

Hence, evaluating the solutions to equation yields `t = -2` .

You need to solve for x the equation `x = t^2 + 1` such that:

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

Replacing 5 for x in the original equation yields:

`sqrt(x + 3 - 4sqrt(x - 1)) + a*sqrt(x + 3 + 4sqrt(x - 1)) = 4`

`sqrt(5 + 3 - 4sqrt(5 - 1)) + a*sqrt(5 + 3 + 4sqrt(5 - 1)) = 4`

`0 + a*sqrt16 = 4 => 4a = 4 => a = 1`

**Hence, evaluating the value of a, under the condition specified by the problem, yields `a = 1.` **