# What is sqrt(4x-3)= 2+sqrt(2x-5)?

Jul 11, 2018

$x = \left\{3 , 7\right\}$

#### Explanation:

Given:

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

Square both sides:

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

ACTUALLY square them:

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

Group like terms:

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

Square both sides AGAIN:

$4 {x}^{2} - 8 x + 4 = 16 \left(2 x - 5\right)$

Multiply:

$4 {x}^{2} - 8 x + 4 = 32 x - 80$

Group like terms:

$4 {x}^{2} - 40 x + 84 = 0$

Factor out $4$:

$4 \left({x}^{2} - 10 x + 21\right) = 0$

Then

$4 \left({x}^{2} - 3 x - 7 x + 21\right) = 0$

$4 \left[x \left(x - 3\right) - 7 \left(x - 3\right)\right] = 0$

So

$4 \left(x - 3\right) \left(x - 7\right) = 0$

Jul 11, 2018

${x}_{1} = 3$ and ${x}_{2} = 7$

#### Explanation:

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

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

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

$4 x - 3 + 2 x - 5 - 2 \sqrt{8 {x}^{2} - 26 x + 15} = 4$

$6 x - 12 = 2 \sqrt{8 {x}^{2} - 26 x + 15}$

$3 x - 6 = \sqrt{8 {x}^{2} - 26 x + 15}$

${\left(3 x - 6\right)}^{2} = 8 {x}^{2} - 26 x + 15$

$9 {x}^{2} - 36 x + 36 = 8 {x}^{2} - 26 x + 15$

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

$\left(x - 3\right) \cdot \left(x - 7\right) = 0$

Hence ${x}_{1} = 3$ and ${x}_{2} = 7$