Question

If $a,bϵ\{1,2,3\}$ and the equation $a{x}^2+bx+1=0$ has real roots, then

• A. $a>b$
• B. $a\le b$
• C. Number of possible ordered pairs $(a,b)$ is $3$
• D. $a<b$

Answer 1

Answer: (C), (D)

The correct options are
C Number of possible ordered pairs $(a,b)$ is $3$
D $a<b$

Given : $a,b\in \{1,2,3\}$

$a{x}^2+bx+1=0$ ....... $\left(i\right)$

Roots of this equation are

$x=\frac{-b\pm \sqrt{b^2-4a}}{2a}$

Now for $x$ to be real, $b^2-4a$ should be greater than or equal to zero i.e. $b^2-4a\ge 0$

$⟹b^2\ge 4a$ ..... $\left(i\right)$

For $b=1$, $\left(ii\right)$ is not true

If $b=2⟹a\le 1⟹a=1$

$\therefore (1,2)$ is the one pair

If $b=3⟹a\le 2.25⟹a=1,2$

$\therefore (1,3),(2,3)$ are another pairs.

Thus, the possible ordered pairs are $(1,2),(1,3),(2,3)$ and $a<b$

Hence, the no. of possible ordered pairs $(a,b)$ is $3$ and $a<b$.