# Solve for x in RR the equation sqrt(x+3-4sqrt(x-1))+sqrt(x+8-6sqrt(x-1))=1 ?

Sep 19, 2016

$x \in \left[5 , 10\right]$

#### Explanation:

Let $u = x - 1$. We can then rewrite the left hand side of the equation as

$\sqrt{u + 4 - 4 \sqrt{u}} + \sqrt{u + 9 - 6 \sqrt{u}}$

$= \sqrt{{\left(\sqrt{u} - 2\right)}^{2}} + \sqrt{{\left(\sqrt{u} - 3\right)}^{2}}$

$= | \sqrt{u} - 2 | + | \sqrt{u} - 3 |$

Note the presence of $\sqrt{u}$ in the equation and that we are only looking for real values, so we have the restriction $u \ge 0$. With that, we will now consider all remaining cases:

Case 1: $0 \le u \le 4$

$| \sqrt{u} - 2 | + | \sqrt{u} - 3 | = 1$

$\implies 2 - \sqrt{u} + 3 - \sqrt{2} = 1$

$\implies - 2 \sqrt{u} = - 4$

$\implies \sqrt{u} = 2$

$\implies u = 4$

Thus $u = 4$ is the only solution in the interval $\left[0 , 4\right]$

Case 2: $4 \le u \le 9$

$| \sqrt{u} - 2 | + | \sqrt{u} - 3 | = 1$

$\implies \sqrt{u} - 2 + 3 - \sqrt{u} = 1$

$\implies 1 = 1$

As this is a tautology, every value in $\left[4 , 9\right]$ is a solution.

Case 3: $u \ge 9$

$| \sqrt{u} - 2 | + | \sqrt{u} - 3 | = 1$

$\implies \sqrt{u} - 2 + \sqrt{u} - 3 = 1$

$\implies 2 \sqrt{u} = 6$

$\implies \sqrt{u} = 3$

$\implies u = 9$

Thus $u = 9$ is the only solution in the interval $\left[9 , \infty\right)$

Taken together, we have $\left[4 , 9\right]$ as the solution set for real values of $u$. Substituting in $x = u + 1$, we arrive at the final solution set $x \in \left[5 , 10\right]$

Looking at the graph of the left hand side, this matches with what we would expect: