Question

If 2(a2+b2) = (a-b)2,then show that a+b=0 plz answer my question I will mark u as BRANLIEST​

Answer 1

Given,

2(a² + b²) = (a - b)²

→ 2a² + 2b² = a² + b² - 2ab

→ 2a² - a² + 2b² - b² = - 2ab

→ a² + b² = - 2ab. .... (i)

Now,

(a + b)² = a² + b² + 2ab.

But, from (i) we know that a² + b² = - 2ab

→ (a + b)²

= a² + b² + 2ab

= - 2ab + 2ab

= 0

Hence, (a + b)² = 0

→ (a + b)² = 0²

→ a + b = 0

Hence Proved.

Answer 2

$\bold{\underline{\underline{Answer:}}}$

Given :

• 2 (a² + b²) = (a-b) ²

To show (prove) :

• a + b = 0

Solution :

$\rightarrow$ $\bold{2(a^2+b^2)=(a-b)^2}$

Expand :- (a-b)² using the identity.

=> (a-b) ² = a² + b² - 2ab

$\rightarrow$ $\bold{2a^2+2b^2=a^+b^2-2ab}$

$\rightarrow$ $\bold{2a^2-a^2=b^2-2b^2-2ab}$

$\rightarrow$ $\bold{a^2=-b^2-2ab}$

$\rightarrow$ $\bold{a^2+b^2= -2ab}$ ---> (1)

Now, to the LHS we have a² + b² i.e

(a+b)² = a² + b² + 2ab

From equation (1) we have,

a² + b² = - 2ab

$\rightarrow$ $\bold{(a+b)^2}$

$\rightarrow$ $\bold{a^2+b^2+2ab}$

$\rightarrow$ $\bold{-2ab+2ab}$

$\rightarrow$ $\bold{0}$

•°• (a + b) ² = 0

$\rightarrow$ $\bold{a+b={\sqrt{0}}}$

$\rightarrow$ $\bold{a+b=0}$

Hence proved.