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4-2.Quadratic Equations and Inequations
normal
Consider the equation $(1+a+b)^2=3\left(1+a^2+b^{2})\right.$ 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 $(a, b)$
(KVPY-2016)
Solution
(d)
Given,
$(1+a+b)^2=3\left(1+a^2+b^2\right)$
$1+a^2+b^2+2 a+2 b+2 a b$
$=3+3 a^2+3 b^2$
$\Rightarrow 2 a^2+2 b^2-2 a-2 b-2 a b+2=0$
$\Rightarrow \quad\left(a^2-2 a+1\right)+\left(b^2-2 b+1\right)$
$+\left(a^2+b^2-2 a b\right)=0$
$\Rightarrow \quad(a-1)^2+(b-1)^2+(a-b)^2=0$
$\therefore \quad a-1=0, \quad b-1=0, a-b=0$
$\Rightarrow \quad a=1, b=1, a=b$
$\therefore \quad a=b=1$
Exactly one pair.
Standard 11
Mathematics
