Question

if x=√6+√5,then the value of x²+1/x²-2 is​

Answer 1

STEP-BY-STEP EXPLANATION:

.

$\bf x = \sqrt{6} + \sqrt{5} \: \: \: ...(given)$

$:\implies\tt {x}^{2} = ( \sqrt{6} + \sqrt{5})^{2}$

$:\implies\tt {x}^{2} = { (\sqrt{6} )}^{2} + 2( \sqrt{6} )( \sqrt{5} ) + {( \sqrt{5}) }^{2}$

$:\implies\tt {x}^{2} = 6 + 2 \sqrt{30} + 5$

$:\implies\tt \bf {x}^{2} = 11 + 2 \sqrt{30}$

$:\implies\tt \frac{1}{ {x}^{2} } = \frac{1}{11 + 2 \sqrt{30} }$

$:\implies\tt \frac{1}{ {x}^{2} } = \frac{1}{11 + 2 \sqrt{30} } \times \frac{11 - 2 \sqrt{30} }{11 - 2 \sqrt{30} }$

$:\implies\tt \frac{1}{ {x}^{2} } = \frac{11 - 2 \sqrt{30} }{ {11}^{2} - {(2 \sqrt{30}) }^{2} }$

$:\implies\tt \frac{1}{ {x}^{2} } = \frac{11 - 2 \sqrt{30} }{121 - 120}$

$:\implies\tt \bf \frac{1}{x ^{2} } = 11 - 2 \sqrt{30}$

$:\implies\tt {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 11 + 2 \sqrt{30} + 11 - 2 \sqrt{30} - 2$

$:\implies\tt {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 22 - 2$

$\bf Hence, \: {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 20$

REQUIRED ANSWER,

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• $\bf {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 20$

Answer 2

Answer: 20.

Explanation:

x = √6 + √5

∴ 1/x = 1/(√6 + √5) = 1(√6 - √5)/(√6 + √5)(√6 - √5)

[Multiplying numerator & denominator by its conjugate.]

= 1/x = √6 - √5

Now, finding x² + 1/x² - 2:-

x² + 1/x² - 2 = (x)² + (1/x)² - 2(x)(1/x)

= (x - 1/x)²

= [(√6 + √5) - (√6 - √5)]²

= (2√5)²

= 20.