Question

If
$a = 5 + 2 \sqrt{6}$
and
$b = \frac{1}{a}$
then find the value of
${a}^{2} + {b}^{2}$
​

Answer 1

Heya......❤✌

Given, a= 5+2√6

Hence,

1/a = 1/(5+2√6) × (5-2√6) /(5-2√6)

= (5-2√6)/(25-24)

= 5-2√6

Therefore,

a^2 + b^2

= a^2 + 1/a^2

= (5+2√6) ^2 + (5-2√6)^2

= 25+24+20√6 +25+24-20√6

= 98

Hope it’s helpful....... ☺

Answer 2

Given : a = 5 + 2√6 and b = 1/a

To find : The value of a² + b²

solution : a = 5 + 2√6

b = 1/a = 1/(5 + 2√6)

= 1/(5 + 2√6) × (5 - 2√6)/(5 - 2√6)

= (5 - 2√6)/[(5 + 2√6)(5 - 2√6)]

= (5 - 2√6)/[(5)² - (2√6)²] [ using formula, a² - b² = (a + b)( a - b)]

= (5 - 2√6)/(25 - 24)

= (5 - 2√6)

so, b = (5 - 2√6)

now a² + b² = (a + b)² - 2ab

= [(5 + 2√6) + (5 - 2√6)]² - 2[(5 + 2√6)(5 - 2√6)]

= [10]² - 2[(5)² - (2√6)²]

= 100 - 2[25 - 24]

= 100 - 2 × 1

= 98

Therefore the value of a² + b² = 98.