Question

x= √( √ 5+1)/( √ 5-1) show that x^2-x-1=0

Answer 1

Answer:

${x}^{2} - x - 1 = 0$

Step-by-step explanation:

$x = \sqrt{ \frac{ \sqrt{5} + 1 }{ \sqrt{5} - 1} } = \sqrt{ \frac{ {( \sqrt{5} + 1 )}^{2} }{5 - 1} } \\ \\ = \sqrt{ \frac{ {( \sqrt{5} + 1 )}^{2} }{4} } \\ \\ = \frac{ \sqrt{5} + 1 }{2} \\ \\ x = \frac{ \sqrt{5} + 1}{2} \\ \\ {x}^{2} = \frac{ \sqrt{5} + 1 }{ \sqrt{5} - 1 } = \frac{ {( \sqrt{5} + 1 )}^{2} }{5 - 1} = \frac{ 5 + 1 + 2 \sqrt{5}}{4} = \frac{6 + 2 \sqrt{5} }{4} = \frac{3 + \sqrt{5} }{2} \\ \\ therefore \: \: \: \: {x}^{2} - x + 1 = \frac{ \sqrt{5} + 3}{ 2 } - \frac{ \sqrt{5} + 1}{2} - 1 \\ \\ = \frac{ \sqrt{5} + 3 - \sqrt{5} - 1 - 2}{2} \\ \\ = \frac{0}{2} \\ \\ = 0 = rhs \\ \\ therefore \: \: \: {x}^{2} - x - 1 = 0$