How do you solve sqrt(4x - 3) = 2 + sqrt(2x - 5)?

Jun 5, 2016

x=7

Explanation:

First let's fix the domain:

$4 x - 3 \ge 0$ and $2 x - 5 \ge 0$

$x \ge \frac{3}{4}$ and $x \ge \frac{5}{2}$

that's $x \ge \frac{5}{2}$

Then let's square both parts:

${\left(\sqrt{4 x - 3}\right)}^{2}$ and ${\left(2 + \sqrt{2 x - 5}\right)}^{2}$

$4 x - 3 = 4 + 4 \sqrt{2 x - 5} + 2 x - 5$

let's put the irrational term on the right and the remaining terms on the left:

$4 x - 3 - 4 - 2 x + 5 = 4 \sqrt{2 x - 5}$

$2 x - 2 = 4 \sqrt{2 x - 5}$

let's divide all terms by 2:

$x - 1 = 2 \sqrt{2 x - 5}$

let's again square both parts:

${\left(x - 1\right)}^{2} = {\left(2 \sqrt{2 x - 5}\right)}^{2}$

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

${x}^{2} - 2 x + 1 = 8 x - 20$

${x}^{2} - 10 x + 21 = 0$

whose solutions are x=2 and x=7

but only the second one belongs to the domain