Question

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

• A. there is no solution pair (a, b)
• B. there are infinitely many solution pairs (a, b)
• C. there are exactly two solution pairs (a, b)
• D. there is exactly one solution pair (a, b)

Answer 1

Answer: (D)

Given,

$(1+a+b{)}^2=3(1+a^2+b^2)$

$1+a^2+b^2+2a+2b+2ab$

$=3+3a^2+3b^2$

$\Rightarrow 2a^2-2b^2-2a-2b-2ab+2=0$

$\Rightarrow (a^2-2a+1)+(b^2-2b+1)$

$+(a^2+b^2-2ab)=0$

$\Rightarrow (a-1{)}^2+(b-1{)}^2+(a-b{)}^2=0$

$\therefore a-1=0,b-1=0,a-b=0$

$\Rightarrow a=1,b=1,a=b$

$\therefore a=b=1$

Exactly one pair.