# How do you solve (x+5)^(1/2) - (5-2x)^(1/4) = 0 and find any extraneous solutions?

Mar 30, 2018

$x = - 2$

And, $- 10$ is the extraneous root.

#### Explanation:

We have,

${\left(x + 5\right)}^{\frac{1}{2}} - {\left(5 - 2 x\right)}^{\frac{1}{4}} = 0$

$\Rightarrow {\left(x + 5\right)}^{\frac{1}{2}} = {\left(5 - 2 x\right)}^{\frac{1}{4}}$ [Transpose ${\left(5 - 2 x\right)}^{\frac{1}{4}}$ to R.H.S]

$\Rightarrow {\left({\left(x + 5\right)}^{\frac{1}{2}}\right)}^{2} = {\left({\left(5 - 2 x\right)}^{\frac{1}{4}}\right)}^{2}$ [Squaring both sides]

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

$\Rightarrow {\left(x + 5\right)}^{2} = 5 - 2 x$ [Squaring Again]

$\Rightarrow {x}^{2} + 10 x + 25 = 5 - 2 x$ [Expanded the L.H.S]

$\Rightarrow {x}^{2} + 10 x + 2 x + 25 - 5 = \cancel{5} \cancel{- 2 x} \cancel{+ 2 x} \cancel{- 5}$ [Add $2 x - 5$ to both sides]

$\Rightarrow {x}^{2} + 12 x + 20 = 0$

$\Rightarrow {x}^{2} + \left(10 + 2\right) x + 20 = 0$ [Break up $12$ as $10 + 2$]

$\Rightarrow {x}^{2} + 10 x + 2 x + 20 = 0$

$\Rightarrow x \left(x + 10\right) + 2 \left(x + 10\right) = 0$

$\Rightarrow \left(x + 10\right) \left(x + 2\right) = 0$

So, $x = - 10 , - 2.$

Now, If we substitute the respective values for $x$, we can find the values are extraneous roots or not.

Checking for $- 10$:-

L.H.S. :- ${\left(- 10 + 5\right)}^{\frac{1}{2}} = \sqrt{- 5} = 5 i$.

And, R.H.S :- ${\left(5 - 2 \cdot - 10\right)}^{\frac{1}{4}} = {\left(25\right)}^{\frac{1}{4}} = \sqrt{5}$

So, L.H.S $\ne$ R.H.S.

So, $- 10$ is an extraneous root of this equation.

Checking For $- 2$:-

L.H.S. :- ${\left(- 2 + 5\right)}^{\frac{1}{2}} = \sqrt{3}$.

And, R.H.S :- ${\left(5 - 2 \cdot - 2\right)}^{\frac{1}{4}} = {\left(9\right)}^{\frac{1}{4}} = \sqrt{3}$

So, L.H.S $=$ R.H.S.

So, $- 2$ is not an extraneous root of this equation, but it is a legit root.

Hope This Helps.