If $x^{2} + \frac{1}{x^{2}} = 47$ $x^2+1/x^2=47$ , then $\sqrt{x} + \frac{1}{\sqrt{x}} =$ $sqrtx+1/sqrtx=$ ?

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If $x^{2} + \frac{1}{x^{2}} = 47$ $x^2+1/x^2=47$ , then $\sqrt{x} + \frac{1}{\sqrt{x}} =$ $sqrtx+1/sqrtx=$ ?

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Answer 1 · 2018-07-01T14:39:22

The answer is $= 3$ $=3$

Explanation:

If

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

${(x + \frac{1}{x})}^{2} = x^{2} + \frac{1}{x^{2}} + 2 \cdot x \cdot \frac{1}{x} = x^{2} + \frac{1}{x^{2}} + 2$ $(x+1/x)^2=x^2+1/x^2+2*x*1/x=x^2+1/x^2+2$

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

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

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

${(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2} = {(\sqrt{x})}^{2} + {(\frac{1}{\sqrt{x}})}^{2} + 2 \cdot \sqrt{x} \cdot \frac{1}{\sqrt{x}}$ $(sqrtx+1/sqrtx)^2=(sqrtx)^2+(1/sqrtx)^2+2*sqrtx*1/sqrtx$

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

$= 7 + 2 = 9$ $=7+2=9$

Therefore,

${(\sqrt{x} + \frac{1}{\sqrt{x}})}^{2} = 9$ $(sqrtx+1/sqrtx)^2=9$

$\Rightarrow$ $=>$ , $\sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{9} = 3$ $sqrtx+1/sqrtx=sqrt9=3$

Answer 2 · 2018-07-01T18:43:20

$3$ $3$

Explanation:

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

Hence,

$y^{4} = {(\sqrt{x} + \frac{1}{\sqrt{x}})}^{4}$ $y^4=(sqrtx+1/sqrtx)^4$

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

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

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

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

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

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

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

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

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If $x^{2} + \frac{1}{x^{2}} = 47$ $x^2+1/x^2=47$ , then $\sqrt{x} + \frac{1}{\sqrt{x}} =$ $sqrtx+1/sqrtx=$ ?

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If x^2+1/x^2=47x2+1x2=47x^2+1/x^2=47, then sqrtx+1/sqrtx=√x+1√x=sqrtx+1/sqrtx=?

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If $x^{2} + \frac{1}{x^{2}} = 47$ $x^2+1/x^2=47$ , then $\sqrt{x} + \frac{1}{\sqrt{x}} =$ $sqrtx+1/sqrtx=$ ?