Question

if a=2+√5 and b=1÷a find a^2+b^2

Answer 1

Answer :

48

Step-by-step explanation :

• a = 2 + √5

• b = 1 / a

Find the value of b :

b = 1 / 2 + √5

b = 1 / 2 + √5 × 2 - √5 / 2 - √5

b = 2 - √5 / 2² - (√5)²

b = 2 - √5 / 4 - 5

b = - 2 + √5

b = √5 - 2

Find a + b :

( a + b ) = 2 + √5 + √5 - 2

( a + b ) = √5 + √5

( a + b ) = 2 √ 5

Find a² + b² :

( a + b ) = 2 √ 5

On squaring both sides ;

( a + b )² = ( 2√5 )²

a² + b² + 2 = 20

a² + b² = 20 - 2

a² + b² = 18

Hence, the answer is 18.

Answer 2

Answer


$a = 2 + \sqrt{5}$

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


so,

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


$b = \frac{1 \times( 2 - \sqrt{5} )}{2 + \sqrt{5} \times (2 - \sqrt{5}) }$


$b = \frac{2 - \sqrt{5} }{ {2}^{2} - 5}$



$b = \frac{2 - \sqrt{5} }{4 - 5}$



$b = \frac{2 - \sqrt{5} }{ - 1}$

$b = \sqrt{5 } - 2$


${a}^{2} + {b}^{2}$



so put value of a and b in eq.



${(2 + \sqrt{5} )}^{2} + { (\sqrt{5 } - 2)}^{2}$
we. get



${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2b$

$(4 + 5 + 2(4)( \sqrt{5} )\: \: + (4 + 5 +( - 2 \times 4 \times \sqrt{5} ))$



$(4 + 5 + 8 \sqrt{5} ) + (4 + 5 - 8 \sqrt{5} )$


$9 + 9$



18 ans