a-2-b-2-4-ab-2-a-b-

Question Number 180069 by Acem last updated on 06/Nov/22

a^{2} - b^{2} = 4, a b = 2, a + b = ?

Commented by Acem last updated on 07/Nov/22

H a v e a g o o d d a y m y b r o t h e r s

m r . W a n d m r . R a s h e e d

Answered by Rasheed.Sindhi last updated on 07/Nov/22

a^{2} - b^{2} = 4, a b = 2, a + b = ?

a^{2} - b^{2} = 4

(a - b) (a + b) = 4

{(a - b)}^{2} {(a + b)}^{2} = 4^{2}

(a^{2} + b^{2} - 2 a b) (a^{2} + b^{2} - 2 a b) = 16

{(a^{2} + b^{2})}^{2} - {(2 (2))}^{2} = 16

{(a^{2} + b^{2})}^{2} = 32

a^{2} + b^{2} = \pm 4 \sqrt{2}

{(a + b)}^{2} = a^{2} + b^{2} + 2 a b = \pm 4 \sqrt{2} + 2 (2)

= 4 \pm 4 \sqrt{2}

a + b = 2 \sqrt{1 \pm \sqrt{2}}

Commented by Acem last updated on 07/Nov/22

A m s o r r y a g a i n, t h a n k f o r y o u

Commented by Rasheed.Sindhi last updated on 07/Nov/22

N o m a t t e r, {i t}^{'} s a m i s p r i n t o n l y (n o t a

m i s t a k e); I^{'} v e e d i t e d m y a n s w e r s .

Commented by Acem last updated on 06/Nov/22

I a m v e r y s o r r y, a^{2} - b^{2} = 4, i {d o n}^{'} t k n o w w h y

t h e = 4 {d i d n}^{'} t t y p e d t h o u g h i d i d \dots s o r r y a g a i n!

A p o l o g i e s f o r t h e i n c o n v e n i e n c e

Commented by CElcedricjunior last updated on 07/Nov/22

A t t e n t i o n!!!

p o u r q u o i m e t t r e 1 \mp \sqrt{2}

d a n s l a r a c i n e

Commented by Rasheed.Sindhi last updated on 07/Nov/22

C a n y o u p l e a s e t r a n s l a t e i n E n g l i s h s i r!

Commented by Acem last updated on 08/Nov/22

P a r c e q u e s i a, b \in Z \Rightarrow a + b = \mp 2 \sqrt{1 \mp \sqrt{2}}

M a i s s i a, b \in R \Rightarrow a + b = \mp 2 \sqrt{1 + \sqrt{2}}

Answered by Rasheed.Sindhi last updated on 07/Nov/22

a^{2} - b^{2} = 4, a b = 2, a + b = ?

(a - b) (a + b) = 4

L e t a + b = t \dots . (i)

a - b = \frac{4}{t} \dots \dots (i i)

(i) + (i i) :

2 a = t + \frac{4}{t} \dots \dots (i i i)

(i) - (i i) :

2 b = t - \frac{4}{t} \dots \dots (i v)

(i i i) \times (i v) :

4 a b = t^{2} - \frac{16}{t^{2}} = 4 (2) = 8

t^{4} - 8 t^{2} - 16 = 0

t^{2} = \frac{8 \pm \sqrt{64 + 64}}{2} = \frac{8 \pm 8 \sqrt{2}}{2} = 4 \pm 4 \sqrt{5}

t = \sqrt{4 \pm 4 \sqrt{2}}

a + b = 2 \sqrt{1 \pm \sqrt{2}}

Answered by Rasheed.Sindhi last updated on 07/Nov/22

a^{2} - b^{2} = 4, a b = 2, a + b = ?

{(a^{2} - b^{2})}^{2} = {(4)}^{2}

a^{4} + b^{4} - 2 {(a b)}^{2} = 16

a^{4} + b^{4} - 2 {(2)}^{2} = 16

a^{4} + b^{4} = 24

{(a^{2} + b^{2})}^{2} - 2 {(a b)}^{2} = 24

{(a^{2} + b^{2})}^{2} - 2 {(2)}^{2} = 24

{(a^{2} + b^{2})}^{2} = 32

a^{2} + b^{2} = \pm 4 \sqrt{2}

{(a + b)}^{2} - 2 a b = \pm 4 \sqrt{5}

{(a + b)}^{2} - 2 (2) = \pm 4 \sqrt{2}

{(a + b)}^{2} = 4 \pm 4 \sqrt{2}

a + b = 2 \sqrt{1 \pm \sqrt{2}}

Commented by Rasheed.Sindhi last updated on 07/Nov/22

Y e s s i r a n d i f a + b \in R, t h e n

a + b = \pm 2 \sqrt{1 + \sqrt{2}}

⊤ han k s sir!

Commented by mr W last updated on 07/Nov/22

w h y n o t

a + b = \pm 2 \sqrt{1 \pm \sqrt{2}} ?

Commented by mr W last updated on 07/Nov/22

a g r e e!

Commented by Acem last updated on 08/Nov/22

T h a n k s f o r t h e m a n y s o l u t i o n s a n d f o r y o u r

e f f o r t s!

B y t h e w a y i t h i n k t h a t {t h e r e}^{'} s \mp b e f o r e t h e l a s t

n u m b e r, a + b = \pm 2 \sqrt{1 \pm \sqrt{2}} i f c = a + b i \in Z

^{'} S e e M r . {F r i x}^{'} s {n o t e}^{'}

Commented by Frix last updated on 08/Nov/22

typo I guess \dots

Z = {\dots, - 3, - 2, - 1, 0, 1, 2, 3, \dots}

you mean C = {a + b i ∣ a, b \in R \land i = \sqrt{- 1}}

Commented by Acem last updated on 08/Nov/22

E x a c t l y M r . F r i x, y o u a r e r i g h t

Answered by Acem last updated on 08/Nov/22

a^{2} - b^{2} = 4

a^{4} + b^{4} = 16 + 8 ∣_{+ 2 a^{2} b^{2}}

{(a^{2} + b^{2})}^{2} = 24 + 8 ∣_{+ 2 a^{2} b^{2}}

a^{2} + b^{2} = \mp 4 \sqrt{2}

{(a + b)}^{2} = \mp 4 \sqrt{2} + 4 ∣_{+ 2 a b}

a + b = \mp 2 \sqrt{1 + \sqrt{2}}; a, b \in R

a + b = \mp 2 \sqrt{1 \mp \sqrt{2}}; c \in Z : c = a + b i; a, b \in R

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