Question-164549

Question Number 164549 by wwww last updated on 18/Jan/22

Answered by alephzero last updated on 18/Jan/22

(a) a - \frac{1}{a} = 4

\frac{a^{2} - 1}{a} = 4

\frac{a^{2} - 1 - 4 a}{a} = 0

\Rightarrow a^{2} - 4 a - 1 = 0

a = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2 \sqrt{5}}{2} =

= 2 \pm \sqrt{5}

Let a + \frac{1}{a} = A

\Rightarrow A_{1} = 2 + \sqrt{5} + \frac{1}{2 + \sqrt{5}} = 2 + \sqrt{5} - 2 + \sqrt{5} =

= 2 \sqrt{5}

A_{2} = 2 - \sqrt{5} + \frac{1}{2 - \sqrt{5}} = 2 - \sqrt{5} - 2 - \sqrt{5} =

= - 2 \sqrt{5}

\Rightarrow a + \frac{1}{a} = \pm 2 \sqrt{5}

(b) a^{3} + \frac{1}{a^{3}} = B

a_{1}^{3} = 38 + 17 \sqrt{5} \land a_{2}^{3} = 38 - 17 \sqrt{5}

\Rightarrow B_{1} = 38 + 17 \sqrt{5} + \frac{1}{38 + 17 \sqrt{5}} = 34 \sqrt{5}

B_{2} = 38 - 17 \sqrt{5} + \frac{1}{38 - 17 \sqrt{5}} = - 34 \sqrt{5}

Answered by Rasheed.Sindhi last updated on 18/Jan/22

a - \frac{1}{a} = 4

(a) : {(a - \frac{1}{a} = 4)}^{2}

a^{2} + \frac{1}{a^{2}} - 2 = 16

{(a + \frac{1}{a})}^{2} = 20

a + \frac{1}{a} = \pm 2 \sqrt{5}

(b) : {(a + \frac{1}{a})}^{3} = {(\pm 2 \sqrt{5})}^{3}

a^{3} + \frac{1}{a^{3}} + 3 (a + \frac{1}{a}) = \pm 40 \sqrt{5}

a^{3} + \frac{1}{a^{3}} + 3 (\pm 2 \sqrt{5}) = \pm 40 \sqrt{5}

a^{3} + \frac{1}{a^{3}} = \pm 40 \sqrt{5} \mp 6 \sqrt{5} = \pm 34 \sqrt{5}

(c) : {(a - \frac{1}{a} = 4)}^{3}

a^{3} - \frac{1}{a^{3}} - 3 (a - \frac{1}{a}) = 64

a^{3} - \frac{1}{a^{3}} - 3 (4) = 64

a^{3} - \frac{1}{a^{3}} = 76

(a^{3} - \frac{1}{a^{3}}) (a^{3} + \frac{1}{a^{3}}) = (76) (\pm 34 \sqrt{5})

a^{6} - \frac{1}{a^{6}} = \pm 2584 \sqrt{5}