If a, b ∈(1, 2, 3) and the equation ax2 + bx + 1 = 0 has real roots, then

Question

If a, b ∈(1, 2, 3) and the equation ax2 + bx + 1 = 0 has real roots, then (A) a > b (B) a le b (C) number of possible ordered pairs (a, b) is 3 (D) a < b

Tardigrade Admin · Accepted Answer

We have, a x2+b x+1=0 For real roots, D ≥ 0 ∴ b2-4 a ≥ 0 ⇒ b2 ≥ 4 a ∴ (a, b)=(1,2),(1,3),(2,3) ∴ Number of ordered pairs (a, b)=3 and a is always less than b.

Q. If $a, b \in (1, 2, 3)$ and the equation $a x^{2} + b x + 1 = 0$ has real roots, then

$a > b$

$a \leq b$

number of possible ordered pairs $(a, b)$ is $3$

$a < b$

We have,
$a x^{2} + b x + 1 = 0$
For real roots, $D \geq 0$
$∴ b^{2} - 4 a \geq 0$
$\Rightarrow b^{2} \geq 4 a$
$∴ (a, b) = (1, 2), (1, 3), (2, 3)$
$∴$ Number of ordered pairs $(a, b) = 3$
and $a$ is always less than $b$ .

Q. If a,b∈(1,2,3) and the equation ax2+bx+1=0 has real roots, then

a>b

a≤b

number of possible ordered pairs (a,b) is 3

a<b