Question

If x=2 root 2 + root 7 then (x-1/x)^2 =

Answer 1

Answer:

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Answer 2

Answer:

$\large{\underline{\boxed{\sf (x - \dfrac{1}{x})^2 = 28}}}$

Step-by-step explanation:

Given that ;

x = 2√2 + √7

Now,

$\implies \tt \frac{1}{x} = \frac{1}{2 \sqrt{2} + \sqrt{7} } \\ \\ \implies \tt \frac{1}{x} = \frac{1}{2 \sqrt{2} + \sqrt{7} } \times \frac{2 \sqrt{2} - \sqrt{7} }{2 \sqrt{2} - \sqrt{7} } \\ \\ \implies \tt \frac{1}{x} = \frac{2 \sqrt{2} - \sqrt{7} }{(2 \sqrt{2}) {}^{2} - ( \sqrt{7}) {}^{2}}$

[Because, (a + b)(a - b) = a² - b²]

$\implies \tt \frac{1}{x} = \frac{2 \sqrt{2} - \sqrt{7} }{8 - 7} \\ \\ \implies \tt \frac{1}{x} = \frac{2 \sqrt{2} - \sqrt{7} }{1} \\ \\ \implies \tt \frac{1}{x} = 2 \sqrt{2} - \sqrt{7}$

So,

$\implies \tt x - \frac{1}{x} = 2 \sqrt{2} + \sqrt{7} - (2 \sqrt{2} - \sqrt{7} ) \\ \\ \implies \tt x - \frac{1}{x} = 2 \sqrt{2} + \sqrt{7} - 2 \sqrt{2} + \sqrt{7} \\ \\ \implies \tt x - \frac{1}{x} = \sqrt{7} + \sqrt{7} \\ \\ \implies \tt x - \frac{1}{x} = 2 \sqrt{7}$

On squaring both sides,

$\implies \tt (x - \frac{1}{x}) {}^{2} = (2 \sqrt{7} ) {}^{2} \\ \\ \boxed{ \red{ \therefore \: \bf \left( x - \frac{1}{x} \right )^{2} = 28}}$