Question

if (a2-4a-1)=0 find the values of a+1/a​

Answer 1

Answer:

the values are (7+4√5)2+√5 and (7-4√5)/2-√5

Step-by-step explanation:

so the given equation is a quadratic and it has roots :

  (4±√16+4)/2              [USING QUADRATIC FORMULA]

2±√5 are the values of a

so taking a=2+√5 first a+1/a is

2+√5+1/(2+√5)=(2+5+4√5)/2+√5 which gives us

(7+4√5)2+√5

now taking a=2-√5 we get a+1/a as

2-√5+1/(2-√5)=(2+5-4√5)/2-√5 which gives us

(7-4√5)/2-√5

I REALLY HOPE THIS HELPS YOU

Answer 2

Answer:

${a}^{2} - 4a - 1 = 0$

${a}^{2} - 4a = 1$

${a}^{2} - 4a + 4 = 1 + 4 = 5$

${(a - 2)}^{2} = 5$

$a - 2 = \sqrt{5}$

$a = 2 + \sqrt{5}$

$a + \frac{1}{a}$

$(2 + \sqrt{5} ) + \frac{1}{(2 + \sqrt{5}) }$

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

$\frac{4 + 5 + 4 \sqrt{5} + 1 }{2 + \sqrt{5} }$

$\frac{10 + 4 \sqrt{5} }{2 + \sqrt{5} }$

$\frac{(10 + 4 \sqrt{5})(2 - \sqrt{5}) }{(2 + \sqrt{5})(2 - \sqrt{5}) }$

$\frac{20 - 10 \sqrt{5} + 8 \sqrt{5} - 20 }{4 - 5}$

$\frac{ - 2 \sqrt{5} }{ - 1}$

$2 \sqrt{5}$