Question

If 2x = 3+√7, find 4x2+1/x2​

Answer 1

2x = 3+√7

=> x = (3+√7)/2

4x²+1/x²

= 4{(3+√7)/2}² + 1/{(3+√7)/2}²

= 4{(16+6√7)/4} + 1/{(16+6√7)/4}

= (16+6√7) + 4/(16+6√7)

= (512+192√7)/(16+6√7)

= {32(16+6√7)}/(16+6√7)

= 32 .

Hope it's helpful to you.

Answer 2

⇨GIVEN:-

$\tt ↠ \red{2x = 3 + \sqrt{7} }$

⇨FIND:-

$\tt ↠ \blue{ {4x}^{2} + \frac{1}{ {x}^{2} } }$

⇨SOLUTION:-

$\tt⤳ \green{2x = 3 + \sqrt{7} }.....(i)$

$\tt➾ \frac{1}{x} = \frac{2}{3 + \sqrt{7} } \\ \tt \gray{❅ rationalise \: the \: denominator} \\ \tt ➾ \frac{2}{3 + \sqrt{7} } \times \frac{3 - \sqrt{7} }{3 - \sqrt{7} } \\۞\tt by \: using \: identity \: (a + b)(a - b) = {a}^{2} - {b}^{2} \: we \: have\\ \tt \frac{2(3 - \sqrt{7} )}{ {3}^{2} - ( \sqrt{7} {)}^{2} } = \frac{2(3 - \sqrt{7} )}{ 9 - 7 } \\ \tt➾ \frac{ \cancel2(3 - \sqrt{7} )}{ \cancel2} \\ \tt ➾ \pink{ \frac{1}{x} = 3 - \sqrt{7}} .....(ii)$

adding eq(i) and (ii) we have,

$\tt⪼2x + \frac{1}{x} = 3 \cancel{+ \sqrt{7} } + 3 \cancel{- \sqrt{7} }$

$\tt⪼2x + \frac{1}{x} = 3 + 3 \\ \tt⪼2x + \frac{1}{x} = 6$

now, squaring both sides we have,

$\tt \implies {(2x + \frac{1}{x}) }^{2} = {6}^{2}$

$\tt ✼by \: using {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: we \: have$

$\tt \implies {(2x)}^{2} + \frac{(1 {)}^{2} }{ {x}^{2} } + 2 \times 2 \cancel x \times \frac{1}{ \cancel x } = 36$

$\tt \implies 4 {x}^{2} + \frac{1}{ {x}^{2} } + 4 = 36$

$\tt \implies 4 {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 4$

$\tt \implies 4 {x}^{2} + \frac{1}{ {x}^{2} } = 32$

$\tt Hence, \huge\boxed{ \bold{ 4 {x}^{2} + \frac{1}{ {x}^{2} } = 32}}$