# How do you solve for x in the simplest radical form for (x^2+2x+4)/x =(2x)/1?

Apr 14, 2015

If $\frac{a}{b} = \frac{c}{d} \to a \cdot d = b \cdot c$ this is called cross-product.

Applying this, we have:

$\left({x}^{2} + 2 x + 4\right) \cdot 1 = x \cdot 2 x \to$

${x}^{2} + 2 x + 4 = 2 {x}^{2} \to$

Subtract $2 x$ on both sides and multiply by $- 1$

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

There are several ways to solve this. I will rewrite for the square:

$\left({x}^{2} - 2 x + 1\right) - 5 = 0 \to$
${\left(x - 1\right)}^{2} - 5 = 0 \to$
${\left(x - 1\right)}^{2} = 5 \to$
$x - 1 = + \sqrt{5} \mathmr{and} - \sqrt{5} \to$

Solution:

$x = 1 + \sqrt{5} \mathmr{and} x = 1 - \sqrt{5}$