AIEEE 2007 (Offline)Question 1
If the difference between the roots of the equation $x^{2}+ax+1=0$ is less than $\sqrt5$, then the set of possible values of $a$ is
Quadratic Equation · Relation Between Roots and Coefficients
Solution
Concept. $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$. Formulas used. (1) $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$Step 1 - Use (1). $(\alpha-\beta)^2=a^2-4$. Step 2 - Condition. $a^2-4<5$, so $a^2<9$ and $a\in(-3,3)$. Answer: (C)