Suppose a, b ∈ R. If the equation x2-(2 a+b) x+(2 a2+b2-b+1 / 2)=0 has two real roots, then

Question

Suppose a, b ∈ R. If the equation x2-(2 a+b) x+(2 a2+b2-b+1 / 2)=0 has two real roots, then (A) a=(1/2), b=-1 (B) a=-(1/2), b=1 (C) a=2, b=1 (D) a=-2, b=-1

Tardigrade Admin · Accepted Answer

As (1) has real roots, (2 a+b)2-4(2 a2+b2-b+1 / 2) ≥ 0 ⇒ 4 a2+3 b2-4 a b-4 b+2 ≤ 0 ⇒ (2 a-b)2+2(b-1)2 ≤ 0 ⇒ b=1, a=(1/2)

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

$a = \frac{1}{2}, b = - 1$

$a = - \frac{1}{2}, b = 1$

$a = 2, b = 1$

$a = - 2, b = - 1$

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

Q. Suppose a,b∈R. If the equation x2−(2a+b)x+(2a2+b2−b+1/2)=0 has two real roots, then

a=21​,b=−1

a=−21​,b=1

a=2,b=1

a=−2,b=−1