Question

10 8. If a = 9-4V5 , find the value of a² + 1/a squre​

Answer 1

a=9-4√5

1/a=1/9-4√5=9+4√5 (Rationalised)

a+1/a=9-4√5+9+4√5=18

square both sides

a^2+1/a^2+2=324

a^2+1/a^2=322

hope it helps...

Answer 2

$\bf{\large{\underline{\underline{Correct \: Question}}}}$

If a = 9 - 4√5, find the value of a² + 1/a²

$\bf{\large{\underline{\underline{Answer}}}}$

$\large{\boxed{ \tt {a}^{2} + \dfrac{1}{a^2} = 322}}$

$\bf{\large{\underline{\underline{Explanation}:-}}}$

Given :- a = 9 - 4√5

To find :- a² + 1/a²

Solution :-

$\tt a = 9 - 4 \sqrt{5}$

First find the value of 1/a

$\tt \dfrac{1}{a} = \dfrac{1}{9 - 4 \sqrt{5} }$

Rationalise the denominator

The rationalising factor of 9 - 4√5 is 9 + 4√5. So multiply both numerator and denominator by rationalising factor

$\tt = \dfrac{1}{9 - 4 \sqrt{5} } \times \dfrac{9 + 4 \sqrt{5} }{9 + 4 \sqrt{5} }$

$\tt = \dfrac{9 + 4 \sqrt{5} }{ {9}^{2} - {(4 \sqrt{5})}^{2} }$

[Since (x + y)(x - y) = x² - y²]

$\tt = \dfrac{9 + 4 \sqrt{5} }{81 - 16(5)}$

$\tt = \dfrac{9 + 4 \sqrt{5} }{81 - 80}$

$\tt = \dfrac{9 + 4 \sqrt{5} }{1}$

$\tt = 9 + 4 \sqrt{5}$

$\tt \dfrac{1}{a} = 9 + 4 \sqrt{5}$

Now find the value of a + 1/a

$\tt a + \dfrac{1}{a} = 9 - 4 \sqrt{5} + (9 + 4 \sqrt{5})$

$\tt a + \dfrac{1}{a} = 9\cancel{ - 4 \sqrt{5}} + 9 \cancel{ + 4 \sqrt{5}}$

$\tt a + \dfrac{1}{a} = 9+ 9$

$\tt a + \dfrac{1}{a} = 18$

Squaring on both sides

$\tt {(a + \dfrac{1}{a})}^{2} = {(18)}^{2}$

We know that (x + y)² = x² + y² + 2xy

Here x =.a, y = 1/a

By substituting the values in the identity we have

$\tt {a}^{2} + {( \dfrac{1}{a})}^{2} + 2(\cancel{a})( \dfrac{1}{\cancel{a}}) = 324$

$\tt {a}^{2} + \dfrac{1}{a^2}+ 2= 324$

$\tt {a}^{2} + \dfrac{1}{a^2} = 324 - 2$

$\tt {a}^{2} + \dfrac{1}{a^2} = 322$

$\Huge{\boxed{ \tt {a}^{2} + \dfrac{1}{a^2} = 322}}$