Question

pla solve the above question!​

Answer 1

Question:-

If x = 4 - √15, find the value of (x + 1/x)² - (1 - 1/x)²

Given:-

• x = 4 - √15

To Find:-

• The value of $\sf{\bigg(x + \dfrac{1}{x}\bigg)^2 - \bigg(x - \dfrac{1}{x}\bigg)^2}$

Solution:-

We have the value of:-

x = 4 - √15

From here we can find the value of $\sf{\dfrac{1}{x}}$

∵ x = 4 - √15

∴ $\sf{\dfrac{1}{x} = \dfrac{1}{4 - \sqrt{15}}}$

Rationalizing the denominator,

$\sf{\dfrac{1}{x} = \dfrac{1}{4 - \sqrt{15}} \times \dfrac{4 + \sqrt{15}}{4 + \sqrt{15}}}$

= $\sf{\dfrac{1}{x} = \dfrac{4 + \sqrt{15}}{(4 - \sqrt{15})(4 + \sqrt{5})}}$

We know,

• (a - b)(a + b) = a² - b²

= $\sf{\dfrac{1}{x} = \dfrac{4 + \sqrt{15}}{(4)^2 - (\sqrt{15})^2}}$

= $\sf{\dfrac{1}{x} = \dfrac{4 + \sqrt{15}}{16 - 15}}$

= $\sf{\dfrac{1}{x} = \dfrac{4 + \sqrt{15}}{1}}$

= $\sf{\dfrac{1}{x} = 4 + \sqrt{15}}$

∴ The value of 1/x is 4 + √15

Putting the values of x and 1/x in the given equation:-

= $\sf{\bigg(x + \dfrac{1}{x}\bigg)^2 - \bigg(x - \dfrac{1}{x}\bigg)^2}$

= $\sf{[(4 - \sqrt{15})^2 + (4 + \sqrt{15})^2] - [(4 - \sqrt{15})^2 - (4 + \sqrt{15})^2}$

= Using the identities:-

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²

= [{(4)² - 2 × 4 × √15 + (√15)²} + {(4)² + 2 × 4 × √15² + (√15)²}] – [{(4² - 2 × 4 × √15 + (√15)²} - {(4)² + 2 × 4 × √15 + (√15)²}]

= [{16 - 8√15 + 15} + {16 + 8√15 + 15}] – [{16 - 8√15 + 15} - {16 + 8√5 + 15}]

= [{21 - 8√15} + {21 + 8√15}] – [{21 - 8√15} - {21 + 8√15}]

= [21 - 8√15 + 21 + 8√15] – [21 - 8√15 - 21 - 8√15]

= [42] - [-16√5]

= 42 + 16√5

$\boxed{\underline{\sf{The\:value\:of\:\bigg(x + \dfrac{1}{x}\bigg)^2 - \bigg(x - \dfrac{1}{x}\bigg)^2\:is\:42 + 16\sqrt{15}}}}$

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Answer 2

$= > {(x + \frac{1}{x}) }^{2} - {(x - \frac{1}{x}) }^{2}$

$= > {( \frac{ {x}^{2} + 1 }{x} )}^{2} - {( \frac{ {x}^{2} - 1}{x})}^{2}$

$= > \frac{ ({x}^{4} + 1 + 2 {x}^{2}) - ( {x}^{4} + 1 - 2 {x}^{2} ) }{ {x}^{2} }$

$= > \frac{4 {x}^{2} }{ {x}^{2} }$

$= > 4$

Hope this answer helps you ^_^ !