Consider the equation (1+a+b)^2=3left(1+a^2+b | vedclass

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Question

(d)

Given,

$(1+a+b)^2=3\left(1+a^2+b^2\right)$

$1+a^2+b^2+2 a+2 b+2 a b$

$=3+3 a^2+3 b^2$

$\Rightarrow 2 a^2+2 b^2-2 a-2 b-2 a b+2=0$

Vedclass · Accepted Answer

there is exactly one solution pair $(a, b)$

Vedclass · Answer

(d)

Given,

$(1+a+b)^2=3\left(1+a^2+b^2ight)$

$1+a^2+b^2+2 a+2 b+2 a b$

$=3+3 a^2+3 b^2$

$\Rightarrow 2 a^2+2 b^2-2 a-2 b-2 a b+2=0$

$\Rightarrow \quad\left(a^2-2 a+1ight)+\left(b^2-2 b+1ight)$

$+\left(a^2+b^2-2 a bight)=0$

$\Rightarrow \quad(a-1)^2+(b-1)^2+(a-b)^2=0$

$	herefore \quad a-1=0, \quad b-1=0, a-b=0$

$\Rightarrow \quad a=1, b=1, a=b$

$	herefore \quad a=b=1$

Exactly one pair.

Consider the equation $(1+a+b)^2=3\left(1+a^2+b^{2})\right.$ where $a, b$ are real numbers. Then,

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