# How do you solve (5-x)^(1/2) = x + 1 and find any extraneous solutions?

Aug 28, 2017

See a solution process below:

#### Explanation:

First, we can square both sides of the equation to eliminate the fractional exponent while keeping the equation balanced:

${\left({\left(5 - x\right)}^{\frac{1}{2}}\right)}^{2} = {\left(x + 1\right)}^{2}$

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

$5 - x = {x}^{2} + 2 x + 1$

Next, we can convert the equation into standard form:

$5 - x - \textcolor{red}{5} + \textcolor{b l u e}{x} = {x}^{2} + 2 x + 1 - \textcolor{red}{5} + \textcolor{b l u e}{x}$

$5 - \textcolor{red}{5} - x + \textcolor{b l u e}{x} = {x}^{2} + 2 x + \textcolor{b l u e}{x} + 1 - \textcolor{red}{5}$

$0 - 0 = {x}^{2} + 3 x - 4$

$0 = {x}^{2} + 3 x - 4$

${x}^{2} + 3 x - 4 = 0$

Now, we can use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{3}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 4}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{{\textcolor{b l u e}{3}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{- 4}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{- \textcolor{b l u e}{3} \pm \sqrt{9 - \left(- 16\right)}}{2}$

$x = \frac{- \textcolor{b l u e}{3} - \sqrt{9 - \left(- 16\right)}}{2}$ and $x = \frac{- \textcolor{b l u e}{3} + \sqrt{9 - \left(- 16\right)}}{2}$

$x = \frac{- \textcolor{b l u e}{3} - \sqrt{9 + 16}}{2}$ and $x = \frac{- \textcolor{b l u e}{3} + \sqrt{9 + 16}}{2}$

$x = \frac{- \textcolor{b l u e}{3} - \sqrt{25}}{2}$ and $x = \frac{- \textcolor{b l u e}{3} + \sqrt{25}}{2}$

$x = \frac{- \textcolor{b l u e}{3} - 5}{2}$ and $x = \frac{- \textcolor{b l u e}{3} + 5}{2}$

$x = - \frac{8}{2}$ and $x = \frac{2}{2}$

$x = - 4$ and $x = 1$

Substituting both solutions into the original equation gives:

${\left(5 - \left(- 4\right)\right)}^{\frac{1}{2}} = - 4 + 1$ and ${\left(5 - 1\right)}^{\frac{1}{2}} = 1 + 1$

${\left(5 + 4\right)}^{\frac{1}{2}} = - 3$ and ${\left(4\right)}^{\frac{1}{2}} = 2$

${\left(9\right)}^{\frac{1}{2}} = - 3$ and $2 = 2$

$3 \ne - 3$ and $2 = 2$

Therefore the solution $x = - 4$ is extraneous.

Aug 28, 2017

color(magenta)(x=1 or x=-4

#### Explanation:

${\left(5 - x\right)}^{\frac{1}{2}} = x + 1$

$\therefore \sqrt{5 - x} = x + 1$

square both sides

$\therefore {\left(\sqrt{5 - x}\right)}^{2} = {\left(x + 1\right)}^{2}$

$\therefore 5 - x = {x}^{2} + 2 x + 1$

$\therefore {x}^{2} + 2 x + 1 = 5 - x$

$\therefore {x}^{2} + 2 x + x + 1 - 5 = 0$

$\therefore {x}^{2} + 3 x - 4 = 0$

$\therefore \left(x - 1\right) \left(x + 4\right) = 0$

:.color(magenta)(x=1 or x=-4