Question

if (a-1/a)= root5, find the value of (a+ 1/a),(a^3+ 1/a^3)​

Answer 1

Step-by-step explanation:

$\tt \frac{a - 1}{a} = \sqrt{5}$

$\tt{1 - \frac{1}{a} = \sqrt{5} }$

$\tt{ - \frac{1}{a} = ( \sqrt{5} - 1) }$

$\tt{ - 1 = ( \sqrt{5} - 1)a}$

$\tt{a = - \frac{1}{ \sqrt{5} - 1 } }$

(1)

•Rationalize the denominator: $\tt{\frac{1}{\sqrt{5}-1} \cdot \frac{\sqrt{5}+1}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{{\sqrt{5}}^{2}-1}}$

$\tt{\frac{-\frac{\sqrt{5}+1}{{\sqrt{5}}^{2}-1}+1}{-\frac{\sqrt{5}+1}{{\sqrt{5}}^{2}-1}}}$

•Use this rule: $\tt{{\sqrt{x}}^{2}=x.}$

$\tt{\frac{-\frac{\sqrt{5}+1}{5-1}+1}{-\frac{\sqrt{5}+1}{5-1}}}$

•Invert and multiply.

$\tt{-(-\frac{\sqrt{5}+1}{4}+1)\times \frac{4}{\sqrt{5}+1}}$

•Rationalize the denominator: $\tt{(-\frac{\sqrt{5}+1}{4}+1)\times \frac{4}{\sqrt{5}+1} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{(-\frac{\sqrt{5}+1}{4}+1)\times 4(\sqrt{5}-1)}{{\sqrt{5}}^{2}-1}.}$

$\tt{-\frac{(-\frac{\sqrt{5}+1}{4}+1)\times 4(\sqrt{5}-1)}{{\sqrt{5}}^{2}-1}}$

Regroup terms.

$\tt{-\frac{4(-\frac{\sqrt{5}+1}{4}+1)(\sqrt{5}-1)}{{\sqrt{5}}^{2}-1}}$

Use this rule: $\tt{{\sqrt{x}}^{2}=x.}$

$\tt{-\frac{4(-\frac{\sqrt{5}+1}{4}+1)(\sqrt{5}-1)}{5-1}}$

$\tt{-(-\frac{\sqrt{5}+1}{4}+1)(\sqrt{5}-1)}$