Hi, could anyone help me solve this equation?
3x = sqrt.5 (x + 2) ----- square root applies to 5 only
in the form a + b sqrt.5
Thanks.
$\displaystyle 3x = \sqrt{5} \;(x+2)$
$\displaystyle 3x = x\sqrt{5} \;+2\sqrt{5}$
$\displaystyle 3x - x\sqrt{5} =2\sqrt{5}$
$\displaystyle x(3 - \sqrt{5}) =2\sqrt{5}$
$\displaystyle x=\frac {2\sqrt{5}}{3 - \sqrt{5}}$
Now Rationalize the denominator,
$\displaystyle x=\frac {2\sqrt{5}}{3 - \sqrt{5}}\times \frac {3+ \sqrt{5}}{3 + \sqrt{5}} $
$\displaystyle x=\frac {6\sqrt{5} - 2\times 5}{9-5}$
$\displaystyle x=\frac {-10+6\sqrt{5}}{4}$
$\displaystyle x=\frac {-10}{4}+ \frac{6}{4}\sqrt{5}$
$\displaystyle x=\frac {-5}{2}+ \frac{3}{2}\sqrt{5}$