Question

if a^2 - 3a - 1 = 0, find the value of a^2 + 1/a^2​

Answer 1

$a ^{2} - 3a - 1 = 0$

${a}^{2} - 1 = 3a$

$a - \frac{1}{a} = 3$

Squaring on both sides

$(a - \frac{1}{a} ) ^{2} = (3)^{2}$

${a}^{2} + ( \frac{1}{a} )^{2} - 2(a)( \frac{1}{a} ) = 9$

${a}^{2} + ( \frac{1}{a} ) ^{2} = 9 + 2$

${a}^{2} + ( \frac{1}{a}) ^{2} = 11$

Answer 2

Answer:

Step-by-step explanation:

Here, we are given :

$a^{2} - 3a - 1 = 0$

What we actually need to find ?

• Value(a) = ?
• $a^{2} + \frac{1}{a^{2}}$

Solution:

Since,

$a^{2} -3a -1 = 0$

Therefore by factorisation, we get

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

$a = \frac{-3 + \sqrt{13}}{2}, \frac{-3 - \sqrt{13}}{2}$

So, by moving forward...

$a^{2} + \frac{1}{a^{2}}\\= {(\frac{-3 + \sqrt{13}}{2})}^2 + {(\frac{2}{-3 + \sqrt{13}})}^2\\= (-3-\sqrt{13} + \frac{1}{-3-\sqrt{13}}) - 2 \\= \frac{22 - 6 \sqrt{13}}{-3-\sqrt{13}} \: or \: \frac{ 6 \sqrt{13}- 22}{3+\sqrt{13}}$

Answer appears bit complex in the end (-_-;)

Mark me brainliest if it helped you, atleast

Happy Learning