Question

2(a^2+b^2)=(a+b)^2show that a=b​

Answer 1

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Answer 2

Given :-

• 2(a² + b²) = (a+b)²

Aim :-

• To show that a = b

2(a² + b²) = (a+b)²

Let us expand on both the sides,

• (a+b)² = a² + 2ab + b² [By applying identity]

→ 2a² + 2b² = a² + 2ab + b²

Let us transpose each term to the LHS (Left hand side of the equation) one by one.

Transposing a²

→ 2a² + 2b² - a² = 2ab + b²

Subtracting,

→ 2b² + a² = 2ab + b²

Transposing b²,

→ 2b² + a² - b² = 2ab

Subtracting,

→ a² + b² = 2ab

Transposing 2ab,

→ a² + b² - 2ab = 0

The LHS (Left hand side) of the equation formed is an identity.

• a² + b² - 2ab = (a-b)²

By expressing a² + b² - 2ab as (a-b)²,

(a-b)² = 0

Transposing the power,

(a-b) = √0

$\sqrt[n]{0} \: \sf{will \: always \: be \: zero}$.

Hence,

(a - b) = 0

Transposing (-b) to the RHS (Right hand side) we get :-

a = b

Hence proved.

Identities :-

• (a+b)² = a² + 2ab + b²

• (a-b)² = a² - 2ab + b²

• (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca

• (x+a)(x+b) = x² + x(a+b) + ab

• a²-b² = (a+b)(a-b)

• (a+b)³ = a³ + 3a²b + 3ab² + b³

• (a-b)³ = a³ - 3a²b + 3ab² - b³

• a³+b³ = (a+b)(a² - ab + b²)

• a³-b³ = (a-b)(a² + ab + b²)

• a³+b³+c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)