prove-that-a-a-2-b-2-a-a-2-b-2-a-b-a-b-a-b-C-

Question Number 192129 by universe last updated on 08/May/23

p r o v e t h a t

∣ a + \sqrt{a^{2} - b^{2}} ∣ + ∣ a - \sqrt{a^{2} - b^{2}} ∣ = ∣ a + b ∣ + ∣ a - b ∣

a, b \in C

Answered by AST last updated on 08/May/23

S q u a r i n g b o t h s i d e s

(∣ x ∣^{2} = x \bar{x}; ∣ x y ∣=∣ x ∣∣ y ∣; \overset{_____}{x + y} = \bar{x} + \bar{y})

{L H S}^{2} =

(a + \sqrt{a^{2} - b^{2}}) (\bar{a} + \overset{}{\sqrt{a^{2} - b^{2}}}) + (a - \sqrt{a^{2} - b^{2}}) (\bar{a} - \overset{}{\sqrt{a^{2} - b^{2}}})

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

= 2 ∣ a ∣^{2} + 2 ∣ a^{2} - b^{2} ∣ + 2 ∣ b ∣^{2}

{R H S}^{2} = (a + b) (\bar{a} + \bar{b}) + (a - b) (\bar{a} - \bar{b}) + 2 ∣ a^{2} - b^{2} ∣

= 2 ∣ a ∣^{2} + 2 ∣ a^{2} - b^{2} ∣ + 2 ∣ b ∣^{2}

S i n c e L H S a n d R H S w e r e b o t h p o s i t i v e b e f o r e

s q u a r i n g b o t h s i d e s

{L H S}^{2} = {R H S}^{2} \Rightarrow L H S = R H S

Commented by York12 last updated on 08/May/23

s i r I w a n n a a s k y o u [s e v e r a l q u e s i o n s,

I a m a h i g h s c h o o l s t u d e n t

a n d I w a n n a a s k a b o u t b o o k s r e c o m m e n d a t i o n d

s o s i r t h a t i s m y t e l e g r a m : b e n g u b l e r

Commented by AST last updated on 09/May/23

I {d o n}^{'} t u s e t e l e g r a m .

Commented by York12 last updated on 09/May/23

s o s i r h o w c a n I r e a c h y o u o u t

Answered by universe last updated on 09/May/23

\frac{1}{2} {∣ a + b + a - b + 2 \sqrt{(a + b) (a - b)} ∣ + ∣ a + b + a - b - 2 \sqrt{(a + b) (a - b)} ∣}

l e t a + b = x

a n d a - b = y

\frac{1}{2} {∣ {(\sqrt{x})}^{2} + {(\sqrt{y})}^{2} + 2 \sqrt{x y} ∣ + ∣ {(\sqrt{x})}^{2} + {(\sqrt{y})}^{2} - 2 \sqrt{x y} ∣}

\frac{1}{2} {∣ \sqrt{x} + \sqrt{y} ∣^{2} + ∣ \sqrt{x} - \sqrt{y} ∣^{2}}

l e t \sqrt{x} = u a n d \sqrt{y} = v

\frac{1}{2} {(u + v) (\bar{u} + \bar{v}) + (u - v) (\bar{u} - \bar{v})}

\frac{1}{2} {u \bar{u} + u \bar{v} + v \bar{u} + v \bar{v} + u \bar{u} - u \bar{v} - v \bar{u} + v \bar{v}}

u \bar{u} + v \bar{v} \Rightarrow∣ u^{2} ∣ + ∣ v^{2} ∣ \Rightarrow∣ {(\sqrt{x})}^{2} ∣ + ∣ {(\sqrt{y})}^{2} ∣

∣ x ∣ + ∣ y ∣ \Rightarrow ∣ a + b ∣ + ∣ a - b ∣

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