# How do you solve x = sqrt(-3x + 10)  and find any extraneous solutions?

Jul 9, 2016

$x = - 5 \text{ and "x=+2" }$ are solutions to ${x}^{2} + 3 x - 10 = 0$

Extraneous $\underline{\text{values}}$ of $x > \frac{10}{3}$

For $\forall x \notin \left\{- 5 , 2\right\} \text{ " =>" } {x}^{2} + 3 x - 10 \ne 0$

#### Explanation:

$\textcolor{b l u e}{\text{Determine general solution}}$

Square both sides

${x}^{2} = - 3 x + 10 \text{ "larr" you now have a quadratic}$

${x}^{2} + 3 x - 10 = 0$

Notice that $2 \times 5 = 10 \text{ and } 5 - 2 = + 3$

Thus the factorization is$\text{ } \left(x + 5\right) \left(x - 2\right) = 0$

So $x = - 5 \text{ and } x = + 2$ are general solutions
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine extraneous values of } x}$

For the number to remain in the $\underline{\text{'Real'}}$ domain you may can not permit the square root to be that of a negative number.

So $- 3 x + 10 \ge 0$

Multiply by (-1) giving

$+ 3 x - 10 \le 0$
Notice that the inequality sign has turned the other way. This happens when you multiply by (-1)

$3 x \le 10$
$\implies x \le \frac{10}{3}$ as a required condition
$\implies x > \frac{10}{3}$ takes the solution out of the Real domain and into the Complex number domain