2-2-

2-2-

Question Number 1181 by 22 last updated on 11/Jul/15

\sqrt{2 + \sqrt{2}} = ?

Answered by Rasheed Ahmad last updated on 24/Jul/15

I n c e r t a i n c a s e s \sqrt{a + b \sqrt{c}} c a n b e

s i m p l i f i e d i n t o p + q \sqrt{c} f o r m (n o t

i n a l l c a s e s) . T h e p r o c e d u r e i s a s

f o l l o w s

L e t \sqrt{a + b \sqrt{c}} = p + q \sqrt{c}

a + b \sqrt{c} = {(p + q \sqrt{c})}^{2}

= p^{2} + q^{2} c + 2 p q \sqrt{c}

B y c o m p a r i n g t h e t h e c o e f f i c i e n t s s s

o f \sqrt{c} a n d t h e t e r m s n o t c o n t a i n i n g

\sqrt{c} w e h a v e

a = p^{2} + q^{2} c a n d b = 2 p q

I f t h e s e t w o s i m u l t a n e o u s e q n s

h a v e s o l u t i o n f o r r a t i o n a l p a n d

q t h e g i v e n e x p r e s s i o n i s

t r a n s f o r m a b l e i n t o t h e f o r m

p + q \sqrt{c} o t h e r w i s e n o t .

N o w i n t h i s p a r t i c u l a r s i t u a t i o n

a = 2 b = 1 a n d c = 2

H e n c e p^{2} + 2 q^{2} = 2 a n d 2 p q = 1

F r o m t h e l a t t e r e q n p = \frac{1}{2 q} a n d b y

s u b s t i t u t i n g t h i s i n t h e f o r m e r

e q n w e g e t

8 q^{4} - 8 q^{2} + 1 = 0

U n f o r t i n u a t e l y t h i s i s n o t

s o l v e a b l e f o r r a t i o n a l q .

S o \sqrt{2 + \sqrt{2}} c a n b e r e p r e s e n t e d

i n o n l y i n d e c i m a l f o r m w h i c h i s

a p p r o x i m a t e l y 1.847759 \dots