Question

Tell ans of 2 plz and 8

Answer 1

Step-by-step explanation:

3. Given,

x = 3 + √8

Then,

$\frac{1}{x} = \frac{1}{3 + \sqrt{8} }$

Rationalizing the denominator we get:

$\frac{1}{x} = \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{ {(3)}^{2} - {( \sqrt{8} )}^{2} } \\ \\ \frac{1}{x} = \frac{3 - \sqrt{8} }{9 - 8} \\ \\ \bf \frac{1}{x} = 3 - \sqrt{8}$

Now,

$(i) \: x + \frac{1}{x} = 3 + \sqrt{8} + 3 - \sqrt{8} \\ \\ x + \frac{1}{x} = 3 + 3 \\ \\ x + \frac{1}{x} = 6 \\ \\ \bf squaring \: both \: the \: sides \: we \: get \\ \\ {(x + \frac{1}{x} )}^{2} = {(6)}^{2} \\ \\ {(x)}^{2} + {( \frac{1}{x}) }^{2} + 2 \times x \times \frac{1}{x} = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36 \\ \\ \bf {x}^{2} + \frac{1}{ {x}^{2} } = 34$

(ii)

Squaring the above equation again we get:

${( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = {(34)}^{2} \\ \\ {( {x}^{2} )}^{2} + {( \frac{1}{ {x}^{2} } )}^{2} + 2 . {x}^{2} . \frac{1}{ {x}^{2} } = 1156 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 1156 \\ \\ \bf {x}^{4} + \frac{1}{ {x}^{4} } = 1154$

8.

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$

L.H.S.

Rationalizing the denominator we get:

$= \frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \\ = \frac{5(7 - 4 \sqrt{3}) + 2 \sqrt{3}(7 - 4 \sqrt{3} ) }{ {(7)}^{2} - {(4 \sqrt{3}) }^{2} } \\ \\ = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24 }{49 - 48} \\ \\ = 11 - 6 \sqrt{3}$

Equating L.H.S and R.H.S we get,

$11 - 6 \sqrt{3} = a + b \sqrt{3} \\ \\ \bf a = 11 \: \: and \: \: b = - 6$