Question

If x=6−√35, find the value of x²+1÷x².

Answer 1

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Answer 2

Answer:

The value of $x^{2} + \frac{1}{x^{2} } = 142$

Step-by-step explanation:

Given $x=6-\sqrt{35}$,

We can write  $\frac{1}{x} = \frac{1}{6-\sqrt{35} }$

Rationalizing the numerator and the denominator,

$\frac{1}{x}= \frac{1}{6-\sqrt{35}}\times \frac{6+\sqrt{35}}{6+\sqrt{35}}$

$\frac{1}{x}= \frac{1.(6+\sqrt{35})}{(6-\sqrt{35})(6+\sqrt{35})}$

$\frac{1}{x}= \frac{6+\sqrt{35}}{(6)^{2}-(\sqrt{35})^{2}}$  $[\because (a-b)(a+b)=a^{2}-b^{2}]$

$\frac{1}{x}=\frac{6+\sqrt{35}}{36-35}$

$\frac{1}{x} = 6+\sqrt{35}$

Now finding the value of $x^{2} +\frac{1}{x^{2} }$,

$x^{2} +\frac{1}{x^{2} } = (6-\sqrt{35} )^{2} + (6+\sqrt{35} )^{2}$

$x^{2} +\frac{1}{x^{2} } = 6^{2} +(\sqrt{35})^{2} -2(6)(\sqrt{35} ) +6^{2} +(\sqrt{35}^{2}) +2(6)(\sqrt{35})$

$x^{2} + \frac{1}{x^{2} } = 36+35+36+35$

$x^{2} + \frac{1}{x^{2} } = 142$