Question

If a= 3+2√2 find
1) a²+1/a²
2) a³+1/a³

Answer 1

solution:-

a = 3 + 2√2
1/a = 1/(3 + 2√2)
= [1/(3 + 2√2)] x [3 - 2√2][3 - 2√2]
= (3 - 2√2)

(a6 + a4 + a2 +1) / a3
= (a3 + a + 1/a + 1/a3)
= (3 + 2√2)3 + (3 + 2√2) + (3 - 2√2) + (3 - 2√2)3
= (3 + 2√2)3 + (3 - 2√2)3 + (3 + 2√2) + (3 - 2√2)
=[2(3)3 + 6(3)(2√2)2] + [6]
= [54 + 144] + 6
= 198 + 6
= 204.

Answer 2

$\sf{a = 3 + 2 \sqrt{2}}$

$\sf{\frac{1}{a} = \frac{1}{3 + 2 \sqrt{2} } }$

$\sf{rationalise \: the \: denominator}$

$\sf{\frac{1}{a} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } }$

$\sf{\frac{1}{a} = \frac{3 - 2 \sqrt{2} }{ {3}^{2} - {(2 \sqrt{2} )}^{2} } }$

$\sf{\frac{1}{a} = \frac{3 - 2 \sqrt{2} }{9 - 8} }$

$\sf{\frac{1}{a} = {3 - 2 \sqrt{2} } }$

$\sf{a + \frac{1}{a} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} }$

$\sf{a + \frac{1}{a} = 6}$

$1) \: \sf{(a + \frac{1}{a})}^{2} = {a}^{2} + \frac{1}{ {a}^{2} } + 2$

$\sf{{6}^{2} = {a}^{2} + \frac{1}{ {a}^{2} } + 2}$

$\sf{{a}^{2} + \frac{1}{ {a}^{2} } = 36 - 2}$

$\fbox{ \sf{{a}^{2} + \frac{1}{ {a}^{2} } = 34}}$

$2) \: \sf{ {(a + \frac{1}{a})}^{3} = {a}^{3} + \frac{1}{ {a}^{3} } + 3(a + \frac{1}{a} )}$

$\sf{ {6}^{3} = {a}^{3} + \frac{1}{ {a}^{3} } + 3(6)}$

$\sf{{a}^{3} + \frac{1}{ {a}^{3} } = 216 - 18}$

$\fbox{\sf{{a}^{3} + \frac{1}{ {a}^{3} } = 198}}$