How do you solve 2x - 9\sqrt { x } + 4= 0?

Aug 24, 2017

See below.

Explanation:

Move the term with the radical to the right hand side:

$2 x + 4 = 9 \sqrt{x}$
square both sides and collect like terms:
$81 x = 4 {x}^{2} + 16 x + 16$
collect and equate to 0:
$4 {x}^{2} - 65 x + 16 = 0$
Factor:
$\left(4 x - 1\right) \left(x - 16\right) = 0$
solution: $x = 16 , x = \frac{1}{4}$

Aug 24, 2017

$x = 16 \text{ }$ or $\text{ } x = \frac{1}{4}$

Explanation:

Given:

$2 x - 9 \sqrt{x} + 4 = 0$

Let $t = \sqrt{x}$ and find:

$0 = 8 \left(2 x - 9 \sqrt{x} + 4\right)$

$\textcolor{w h i t e}{0} = 8 \left(2 {t}^{2} - 9 t + 4\right)$

$\textcolor{w h i t e}{0} = 16 {t}^{2} - 72 t + 32$

$\textcolor{w h i t e}{0} = {\left(4 t\right)}^{2} - 2 \left(4 t\right) \left(9\right) + {9}^{2} - 49$

$\textcolor{w h i t e}{0} = {\left(4 t - 9\right)}^{2} - {7}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(4 t - 9\right) - 7\right) \left(\left(4 t - 9\right) + 7\right)$

$\textcolor{w h i t e}{0} = \left(4 t - 16\right) \left(4 t - 2\right)$

$\textcolor{w h i t e}{0} = 8 \left(t - 4\right) \left(2 t - 1\right)$

So:

$t = 4 \text{ }$ or $\text{ } t = \frac{1}{2}$

Then:

$x = {t}^{2} = {4}^{2} = 16 \text{ }$ or $\text{ } x = {t}^{2} = {\left(\frac{1}{2}\right)}^{2} = \frac{1}{4}$