WBJEE · Maths · Quadratic Equation
If \(a, b \in\{1,2,3\}\) and the equation \(a x^{2}+b x+1=0\) has real roots, then
- A \(a>b\)
- B \(a \leq b\)
- C number of possible ordered pairs \((a, b)\) is 3
- D \(a < b\)
Answer & Solution
Correct Answer
(D) \(a < b\)
Step-by-step Solution
Detailed explanation
We have, \[ a x^{2}+b x+1=0 \] For real roots, \(D \geq 0\) \[ \therefore \quad b^{2}-4 a \geq 0 \] \(\Rightarrow \quad b^{2} \geq 4 a\) \(\therefore(a, b)=(1,2),(1,3),(2,3)\) \(\therefore\) Number of ordered pairs \((a, b)=3\) and \(a\) is always less than \(b\).
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