Question

If a square-3a-1=0, find the value of a square+1/a square​

Answer 1

★ᏀᏆᏙᎬΝ:-

$\sf a^2 - 3a - 1 = 0$

★Ꭲᝪ ᖴᏆᑎᗞ:-

$\sf a^2 + \dfrac{1}{a^2}$

★ꌗꂦ꒒ꀎ꓄ꀤꂦꈤ:-

$\sf a^2 - 3a - 1 = 0$

Dividing the equation by a,

$\sf \dfrac{a^2}{a} - \dfrac{3a}{a} - \dfrac{1}{a} = \dfrac{0}{a}$

$\sf \implies a- 3 - \dfrac{1}{a} = 0$

$\sf \implies a - \dfrac{1}{a} = 3$

$\qquad\qquad\qquad\bigstar\boxed{\bf a - \dfrac{1}{a} = 3}\bigstar$

__________________________

We know that,

$\qquad\qquad\qquad\bigstar\boxed{\sf (a-b)^2 = a^2+b^2-2ab}\bigstar$

Using this identity,

$\sf \implies \Bigg ( a - \dfrac{1}{a} \Bigg )^2 = (a)^2 + \Bigg ( \dfrac{1}{a} \Bigg )^2 - 2 \times a \times \dfrac{1}{a}$

$\sf \implies ( 3 )^2 = a^2 + \dfrac{1}{a^2}- 2$

$\sf \implies 9=a^2 + \dfrac{1}{a^2} - 2$

$\sf \implies 9+2=a^2 + \dfrac{1}{a^2}$

$\sf \implies 11=a^2 + \dfrac{1}{a^2}$

$\qquad\qquad\qquad\bigstar\boxed{\bf a^2 +\dfrac{1}{a^2}= 11}\bigstar$

$\mathcal{HOPE \ IT \ HELPS}$