Question

a square - 4a -1=0 find a - 1/a , and a + 1/a​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{a-\frac{1}{a}=4}}}$

$\green{\tt{\therefore{a+\frac{1}{a}=2\sqrt{5}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies {a}^{2} - 4a - 1 = 0 \\ \\\red{\underline \bold{To \: Find:}} \\ \tt: \implies a - \frac{1}{a} = ? \\ \\ \tt: \implies a - \frac{1}{a} = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {a}^{2} - 4a - 1 = 0 \\ \\ \tt: \implies a = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \tt: \implies a = \frac{ - ( - 4) \pm \sqrt{ { (- 4)}^{2} - 4 \times 1 \times ( - 1) } }{2 \times 1} \\ \\ \tt: \implies a = \frac{4 \pm \sqrt{16 +4} }{2} \\ \\ \tt: \implies a = \frac{4 \pm \sqrt{20} }{2} \\ \\ \tt: \implies a = \frac{4 \pm2 \sqrt{5} }{2} \\ \\ \green{\tt: \implies a = 2\pm \sqrt{5} } \\ \\ \bold{For \: finding \: value} \\ \tt: \implies a - \frac{1}{a} \\ \\ \tt: \implies 2 + \sqrt{5} - \frac{1}{2 - \sqrt{5} } \\ \\ \tt: \implies 2 + \sqrt{5} - \frac{1}{2 - \sqrt{5} } \times \frac{2 + \sqrt{5} }{2 + \sqrt{5} } \\ \\ \tt: \implies 2 + \sqrt{5} - \frac{2 + \sqrt{5} }{4 - 5} \\ \\ \tt: \implies 2 + \sqrt{5} -( - 2 + \sqrt{5} ) \\ \\ \tt: \implies 2 + \sqrt{5} + 2 - \sqrt{5} \\ \\ \green{\tt: \implies a - \frac{1}{a} = 4} \\ \\ \bold{Similarly : } \\ \tt: \implies a + \frac{1}{a} \\ \\ \tt: \implies 2 + \sqrt{5} + \frac{1}{2 - \sqrt{5} } \\ \\ \tt: \implies 2 + \sqrt{5} + \frac{1}{2 - \sqrt{5} } \times \frac{2 + \sqrt{5} }{2 + \sqrt{5} } \\ \\ \tt: \implies 2 + \sqrt{5} + \frac{2 + \sqrt{5} }{4 - 5} \\ \\ \tt: \implies 2 + \sqrt{5} + ( - 2 + \sqrt{5} ) \\ \\ \tt: \implies 2 + \sqrt{5 } - 2 + \sqrt{5} \\ \\ \green{\tt: \implies a + \frac{1}{a} = 2\sqrt{5}}$

Answer 2

Answer:

Given :-

a²-4a - 1=0

To find :-

$a - \frac{1}{a} \: = \: ? \\ a + \frac{1}{a} \: = \: ?$

Solution :-

$a \: = \frac{ - b + - \sqrt{ {b}^{2} - 4ac} }{2a}$

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

$a = \frac{4 + - \sqrt{16 + 4} }{2}$

$a \: = \frac{4 + - \sqrt{20} }{2}$

$a \: = \frac{4 + - 2 \sqrt{5} }{2}$

$a = 2 + \sqrt{5} \: or \: 2 - \sqrt{5}$

Now for finding the value,

$a - \frac{1}{a}$

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

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

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

$2 + \sqrt{5} - ( - 2 + \sqrt{5} )$

=$4$

Therefore,

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

And,

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

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

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

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

$2 + \sqrt{5} - (2 - \sqrt{5} )$

=$2 \sqrt{5}$

Therefore,

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