# If x^2+1/x^2=47, then sqrtx+1/sqrtx=?

Jul 1, 2018

The answer is $= 3$

#### Explanation:

If

${x}^{2} + \frac{1}{x} ^ 2 = 47$

${\left(x + \frac{1}{x}\right)}^{2} = {x}^{2} + \frac{1}{x} ^ 2 + 2 \cdot x \cdot \frac{1}{x} = {x}^{2} + \frac{1}{x} ^ 2 + 2$

Substituting the value of ${x}^{2} + \frac{1}{x} ^ 2 = 47$

${\left(x + \frac{1}{x}\right)}^{2} = 47 + 2 = 49$

$x + \frac{1}{x} = \sqrt{49} = 7$

${\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)}^{2} = {\left(\sqrt{x}\right)}^{2} + {\left(\frac{1}{\sqrt{x}}\right)}^{2} + 2 \cdot \sqrt{x} \cdot \frac{1}{\sqrt{x}}$

$= x + \frac{1}{x} + 2$

$= 7 + 2 = 9$

Therefore,

${\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)}^{2} = 9$

$\implies$, $\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{9} = 3$

Jul 1, 2018

$3$

#### Explanation:

Set $y = \sqrt{x} + \frac{1}{\sqrt{x}}$

Hence,

${y}^{4} = {\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)}^{4}$

${y}^{4} = {x}^{2} + 4 x + 6 + \frac{4}{x} + \frac{1}{x} ^ 2$

${y}^{4} = {x}^{2} + \frac{1}{x} ^ 2 + 4 \cdot \left(x + \frac{1}{x}\right) + 6$

${y}^{4} = 47 + 4 \cdot \left({\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)}^{2} - 2\right) + 6$

${y}^{4} = 4 \cdot \left({y}^{2} - 2\right) + 53$

${y}^{4} = 4 {y}^{2} + 45$

${y}^{4} - 4 {y}^{2} - 45 = 0$

$\left({y}^{2} + 5\right) \cdot \left({y}^{2} - 9\right) = 0$

Thus, ${y}^{2} = 9$, so $y = \sqrt{x} + \frac{1}{\sqrt{x}} = 3$