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Algebraic Identities

If x=√6 +√5 ,...

Question

If $x = \sqrt{6} + \sqrt{5}$ , then $x^{2} + \frac{1}{x^{2}} - 2$ .

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Solution

Given: $x = \sqrt{6} + \sqrt{5}$

$\frac{1}{x} = \frac{1}{\sqrt{6} + \sqrt{5}}$

Rationalising the denominator,

$\frac{1}{x} = \frac{(\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5}) (\sqrt{6} - \sqrt{5})}$

$\frac{1}{x} = \frac{(\sqrt{6} - \sqrt{5})}{(\sqrt{6})^{2} - (\sqrt{5})^{2}}$

$\frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{6 - 5}$

$\frac{1}{x} = \frac{\sqrt{6} - \sqrt{5}}{1}$

$\frac{1}{x} = \sqrt{6} - \sqrt{5}$

Now, $x^{2} + \frac{1}{x^{2}} - 2$

$= (\sqrt{6} + \sqrt{5})^{2} + (\sqrt{6} - \sqrt{5})^{2} - 2$

$= (\sqrt{6})^{2} + (\sqrt{5})^{2} + 2 \sqrt{30} + (\sqrt{6})^{2} + (\sqrt{5})^{2} - 2 \sqrt{30} - 2$

$= 6 + 5 + 2 \sqrt{30} + 6 + 5 - 2 \sqrt{30} - 2$

$= 22 - 2$

$= 20$

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Similar questions

x = \sqrt{6} + \sqrt{5}, then x^{2} + \frac{1}{x^{2}} - 2 =_________.

x = \sqrt{6} + \sqrt{5}

x^{2} + \frac{1}{x^{2}} - 2 =

2 \sqrt{6}

2 \sqrt{5}

x = \sqrt{6} + \sqrt{5}

x^{2} + \frac{1}{x^{2}} - 2 =

x = \sqrt{6} - \sqrt{5}

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

?

x = 5 - 2 \sqrt{6}

x^{2} + \frac{1}{x^{2}}

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