# How do you solve sqrt(x)= 2x?

Feb 27, 2017

See the entire solution process below:

#### Explanation:

First, square each side of the equation:

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

$x = 4 {x}^{2}$

Next, subtract $\textcolor{red}{x}$ from each side of the equation:

$x - \textcolor{red}{x} = 4 {x}^{2} - \textcolor{red}{x}$

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

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

Then, factor an $x$ out of each term on the left side of the equation:

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

Now, solve each term for $0$ to find all the solutions to the problem:

Solution 1)

$x = 0$

Solution 2)

$4 x - 1 = 0$

$4 x - 1 + \textcolor{red}{1} = 0 + \textcolor{red}{1}$

$4 x - 0 = 1$

$4 x = 1$

$\frac{4 x}{\textcolor{red}{4}} = \frac{1}{\textcolor{red}{4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} x}{\cancel{\textcolor{red}{4}}} = \frac{1}{4}$

$x = \frac{1}{4}$

The solution is: $x = 0$ and $x = \frac{1}{4}$