# How do you simplify sqrt(x-1) + sqrt( 2x) = 3?

May 6, 2018

$\rightarrow x = 2$

#### Explanation:

$\rightarrow \sqrt{x - 1} + \sqrt{2 x} = 3$

$\rightarrow \sqrt{x - 1} = 3 - \sqrt{2 x}$

$\rightarrow {\left[\sqrt{x - 1}\right]}^{2} = {\left[3 - \sqrt{2 x}\right]}^{2}$

$\rightarrow x - 1 = 9 - 6 \sqrt{2 x} + 2 x$

$\rightarrow 6 \sqrt{2 x} = x + 10$

$\rightarrow {\left[6 \sqrt{2 x}\right]}^{2} = {\left[x + 10\right]}^{2}$

$\rightarrow 36 \cdot \left(2 x\right) = {x}^{2} + 20 x + 100$

$\rightarrow {x}^{2} - 52 x + 100 = 0$

$\rightarrow {x}^{2} - 2 \cdot x \cdot 26 + {26}^{2} - {26}^{2} + 100 = 0$

$\rightarrow {\left(x - 26\right)}^{2} = {26}^{2} - 100 = 576$

$\rightarrow x - 26 = \sqrt{576} = \pm 24$

$\rightarrow x = 26 + 24 , 26 - 24 = 50 \mathmr{and} 2$

Putting $x = 50$ in given equation, we get,

$\rightarrow \sqrt{50 - 1} + \sqrt{2 \cdot 50} = 17 \left(r e j e c t e d\right)$

Putting $x = 2$ in given equation, we get,

$\rightarrow \sqrt{2 - 1} + \sqrt{2 \cdot 2} = 3 \left(a \mathcal{e} p t e d\right)$

So, the required value of x is $2.$