Question

If a= 5+2√6 and b =1/a then what will be the value of (square + b square)?​

Answer 1

Step-by-step explanation:

Given, a=5+2√6 , b=1/a

therefore, a²=(5+2√6)²

a²=5²+2×5×2√6+(2√6)²

a²=25+20√6+24

a²=49+20√6

and, b²=(1/a)² =1/a²

therefore,

a²+b²=49+20√6+1/a²

(49a²+20√6a²+1)

= _________________

a²

That's your answer. I hope it will help you.

Answer 2

Answer:

The value of ${a^2} + {b^2} = 98.$

Step-by-step explanation:

The value of $a$ is shown below.

$a = 5 + 2\sqrt 6$

The value of $b$ is shown below.

$b = \frac{1}{a}$

Substitute $5 + 2\sqrt 6$ for $a$ in above equation as follows:

$b = \tex\frac{1}{{5 + 2\sqrt 6 }}$

Rationalize the value of $b$ as follows:

$b = \frac{1}{{5 + 2\sqrt 6 }} \times \frac{{5 - 2\sqrt 6 }}{{5 - 2\sqrt 6 }} \\= \frac{{5 - 2\sqrt 6 }}{{\left( {5 + 2\sqrt 6 } \right)\left( {5 - 2\sqrt 6 } \right)}} \\ = \frac{{5 - 2\sqrt 6 }}{{{5^2} - {{\left( {2\sqrt 6 } \right)}^2}}} \\= \frac{{5 - 2\sqrt 6 }}{{25 - 4 \times 6}} \\=\frac{{5 - 2\sqrt 6 }}{{25 - 24}} \\= \frac{{5 - 2\sqrt 6 }}{1} \\= 5 - 2\sqrt 6$

The value of ${a^2} + {b^2}$is shown below.

 ${a^2} + {b^2} = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \\= {5^2} + 2\left( 5 \right)\left( {2\sqrt 6 } \right) + {\left( {2\sqrt 6 } \right)^2} + {5^2} - 2\left( 5 \right)\left( {2\sqrt 6 } \right) + {\left( {2\sqrt 6 } \right)^2} \\= 25 + 4\left( 6 \right) + 25 + 4\left( 6 \right) \\= 25 + 24 + 25 + 24 \\= 49 + 49 \\= 98$  

Therefore, the value of ${a^2} + {b^2} = 98$.