Question 55801

Apr 25, 2016

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

Explanation:

$x = a$

${x}^{2} = {a}^{2}$

$4 \sqrt{x - 4} = b$

$16 x - 64 = {b}^{2}$

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

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = a + \cancel{b} + 2 \cdot \sqrt{a + b} \cdot \sqrt{a - b} + a - \cancel{b}$

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = 2 a + 2 \sqrt{\left(a + b\right) \left(a - b\right)}$

$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = 2 a + 2 \sqrt{{a}^{2} - {b}^{2}}$

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = 2 x + 2 \sqrt{{x}^{2} - 16 x + 64}$

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = 2 x + 2 \sqrt{{\left(x - 8\right)}^{2}}$

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = 2 x + 2 \left(x - 8\right)$

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = 2 x + 2 x - 16$

${\left(\sqrt{a + b} + \sqrt{a - b}\right)}^{2} = 4 x - 16$

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

Apr 25, 2016

Slightly different approach in the beginning to that of Ali Ergin but we end up at the same point part way through.

$4 x - 64$

Explanation:

Let $a = 4 \sqrt{x - 4}$ giving

${\left(\sqrt{x + a} + \sqrt{x - a} \textcolor{w h i t e}{.}\right)}^{2}$

${\left(\sqrt{x + a} \textcolor{w h i t e}{.}\right)}^{2} + 2 \sqrt{x + a} \sqrt{x - a} + {\left(\sqrt{x - a} \textcolor{w h i t e}{.}\right)}^{2}$

$x + \cancel{a} + 2 \sqrt{x + a} \sqrt{x - a} + x - \cancel{a}$

$2 x + 2 \sqrt{x + a} \sqrt{x - a}$

$2 x + 2 \sqrt{\left(x + a\right) \left(x - a\right)}$

$2 x + 2 \sqrt{{x}^{2} - {a}^{2}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

But $a = 4 \sqrt{x - 4} \implies {a}^{2} = 16 \left(x - 4\right) = 16 x - 64$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So by substitution for a giving

$2 x + 2 \sqrt{{x}^{2} - \left(16 x - 64\right)}$

2x+2sqrt(x^2-16x+64))#

Note that $\left(- 8\right) \times \left(- 8\right) = + 64 \text{ and } - 8 - 8 = - 16$

$2 x + 2 \sqrt{{\left(x - 8\right)}^{2}}$

$2 x + 2 \left(x - 8\right)$

$4 x - 64$