# How do you solve  x - 3 = sqrt(x - 1) and find any extraneous solutions?

Mar 12, 2017

$x = 5 , \text{ extraneous solution } x = 2$

#### Explanation:

$\textcolor{b l u e}{\text{square both sides}}$

${\left(x - 3\right)}^{2} = {\left(\sqrt{x - 1}\right)}^{2}$

$\Rightarrow {x}^{2} - 6 x + 9 = x - 1$

Equate the quadratic to zero

$\Rightarrow {x}^{2} - 7 x + 10 = 0$

Factorising the quadratic gives.

$\left(x - 2\right) \left(x - 5\right) = 0$

$\Rightarrow x = 5 \text{ or } x = 2$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the equation and if the left side equals the right side then they are the solutions.

• x=5to5-3=sqrt(5-1)to2=2rArr" a solution"

• x=2to2-3=sqrt1to-1=1rArr" not a solution"

$\Rightarrow x = 5 \text{ is the solution, but "x=2" is extraneous}$