Question

Suppose $a, b \in R$. If the equation $ x^2-(2 a+b) x+\left(2 a^2+b^2-b+1 / 2\right)=0 $ has two real roots, then

• A. $a=\frac{1}{2}, b=-1$
• B. $a=-\frac{1}{2}, b=1$
• C. $a=2, b=1$
• D. $a=-2, b=-1$

Answer 1

Answer: (A)

As (1) has real roots,
$ (2 a+b)^2-4\left(2 a^2+b^2-b+1 / 2\right) \geq 0$
$\Rightarrow 4 a^2+3 b^2-4 a b-4 b+2 \leq 0 $
$\Rightarrow (2 a-b)^2+2(b-1)^2 \leq 0 \Rightarrow b=1, a=\frac{1}{2}$