Question

if x=3 + root 3 find the value of x square +9/x square
Please need answer fast

Answer 1

Step-by-step explanation:

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Answer 2

The correct answer is $\frac{30+9\sqrt{3}}{2}$.

Given: x = $3+\sqrt{3}$.

To Find: The value of $x^{2} +\frac{9}{x^{2} }$.

Solution:

x = $3+\sqrt{3}$

$x^{2} =(3+\sqrt{3})^{2}$

$x^{2} =9+3+6\sqrt{3}$

$x^{2} =12+6\sqrt{3}$

$\frac{1}{x^{2} } =\frac{1}{12+6\sqrt{3} }$

$\frac{1}{x^{2} } =\frac{1}{12+6\sqrt{3} }*\frac{12-6\sqrt{3} }{12-6\sqrt{3} }$

$\frac{1}{x^{2} } =\frac{12-6\sqrt{3}}{12^{2} -(6\sqrt{3} )^{2} }$

$\frac{1}{x^{2} } =\frac{12-6\sqrt{3}}{144-108 }$

$\frac{1}{x^{2} } =\frac{12-6\sqrt{3}}{36 }$

$\frac{1}{x^{2} } =\frac{2-\sqrt{3}}{6 }$

Put the value of $x^{2}$ and $\frac{1}{x^{2} }$ in the equation.

$x^{2} +\frac{9}{x^{2} }$

= $12+6\sqrt{3}+9(\frac{2-\sqrt{3} }{6} )$

= $12+6\sqrt{3}+3(\frac{2-\sqrt{3} }{2} )$

= $12+6\sqrt{3}+\frac{6-3\sqrt{3} }{2}$

= $\frac{24+12\sqrt{3}+ 6-3\sqrt{3} }{2}$

= $\frac{30+9\sqrt{3}}{2}$

Hence, the answer is $\frac{30+9\sqrt{3}}{2}$.

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