Question

if alpha and beta are the roots of x^2-x-1=0 then the value of alpha square/beta+beta square/alpha​

Answer 1

Answer:

$\alpha \: and \: \beta \: are \: the \: roots \: of \\ the \: equation \\ {x}^{2} - x - 1 = 0 \\ then \: the \: value \: of \: \frac{ { \alpha }^{2} }{ \frac{ {( \beta + \beta )}^{2} }{ \alpha } } \\ = \frac{ \alpha }{ {2 \beta }^{2} }$

first we have to find the roots

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

here ,a=1, b=-1,c=-1

we know that

quadratic equation formula

$\frac{ -b + or - \sqrt{ {b}^{2} - 4ac} }{2a} \\ = \frac{ - ( - 1) + or - \sqrt{ { (- 1)}^{2} - 4(1)( - 1)} }{2(1)} \\ = \frac{1 + or - \sqrt{1 + 4} }{2} \\ \\ = \frac{1 + \sqrt{5} }{2} (or) \frac{1 - \sqrt{5} }{2}$

now,

$\alpha = \frac{1 + \sqrt{5} }{2} \\ \beta = \frac{1 - \sqrt{5} }{2} \\ then \\ \: \frac{ \alpha }{ {2 \beta }^{2} } = \frac{ \frac{1 + \sqrt{5} }{2} }{ \frac{1 - \sqrt{5} }{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1 + \sqrt{5} }{1 - \sqrt{5} }$

rationalize the denominator

$= \frac{1 + \sqrt{5} }{1 - \sqrt{5} } \times \frac{1 + \sqrt{5} }{1 + \sqrt{5} } \\ = \frac{ {(1 + \sqrt{5} )}^{2} }{ {(1)}^{2} - {( \sqrt{5}) }^{2} } \\ = \frac{1 + 2 \sqrt{5} + {( \sqrt{5} )}^{2} }{1 - 5} \\ = \frac{1 + 2 \sqrt{5} + 5 }{ - 4} \\ = \frac{6 + 2 \sqrt{5} }{ - 4} \\ = \frac{2(3 + \sqrt{5} )}{ - 4} \\ = \frac{ - 3 - \sqrt{5} }{2}$

hope this helps you