Question

If a=1-√2 find value of a+1/a and a^2+1/a^2

Answer 1

Hiiii friend,


A = 1-✓2


Therefore,


1/A = 1/1-✓2

Rationalizing the denominator 1-✓2


=> 1/A = 1/1-✓2 × 1+✓2/1+✓2 = (1+✓2)/(1-✓2)(1+✓2)


=> 1/A = (1-✓2)/(1)² - (✓2)² = (1-✓2)/1-2 = (1+✓2)/-1 = (1+✓2)



Therefore,

(A+1/A) = (1-✓2)(1+✓2) = 2


=> (A+1/A)² = (2)² = 4


=> (A²+1/A²) + 2 × A × 1/A = 4


=> (A²+1/A²) +2 = 4


=> (A²+1/A²) = 4-2


=> (A²+1/A²) = 2




HOPE IT WILL HELP YOU...... :-)

Answer 2

Solution:-

given by:-

$a = 1 - \sqrt{2} \\ a + \frac{1}{a} \\ 1 - \sqrt{2} + \frac{1}{1 - \sqrt{2} } \\ \frac{ 1 + 2- 2 \sqrt{2} + 1}{1 - \sqrt{2} } \\ \frac{4 - 2 \sqrt{2} }{1 - \sqrt{2} } \times \frac{1 + \sqrt{2} }{1 + \sqrt{2} } \\ \frac{4 - 2 \sqrt{2} + 4 \sqrt{2} - 4 }{1 - 2} \\( - 2 \sqrt{2} )ans \\ now \: \\ {x}^{2} + \frac{1}{ {x}^{2} } \\ 3 - 2 \sqrt{2} + \frac{1}{3 - 2\sqrt{2} } \\ \frac{9 + 8 - 12 \sqrt{2} + 1}{3 - 2 \sqrt{2} } \\ \frac{18 - 12 \sqrt{2} }{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \frac{54 - 36 \sqrt{2} + 36 \sqrt{2} - 24 \sqrt{2} }{9 - 8} \\ (54 - 24 \sqrt{2} )ans$