# Q. If the roots of the equation $x^2 - 2ax + a^2 + a - 3 = 0$ are real and less than 3, then

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Solution:

## Let $f (x) = x^2 - 2ax + a^2 + a - 3$ Since, both root are less than 3. $\Rightarrow \alpha < 3, \beta < 3$ $\Rightarrow$ Sum , S = $\alpha + \beta < 6$ $\Rightarrow \frac{ \alpha + \beta }{ 2} < 3$ $\Rightarrow \frac{ 2a}{ 2} < 3$ $\Rightarrow a < 3$ Again, product, P = $\alpha \ \beta$ $\Rightarrow P < 9 \Rightarrow \alpha \beta < 9$ $\Rightarrow \alpha^2 + \alpha - 12 = 0$ $\Rightarrow ( \alpha - 3 ) \ (\alpha + 4 ) < 0$ $\Rightarrow - 4 < a < 3$ $\hspace16mm$ ...(iii) Again, D = $B^2 - 4 AC \ge 0$ $\Rightarrow ( - 2a)^2 - 4 . 1 (a^2 + a - 3 ) \ge 0$ $\Rightarrow 4a^2 - 4a^2 - 4a + 12 \ge 0$ $\Rightarrow - 4 a + 12 \ge 0 \Rightarrow a \le 3$ $\hspace19mm$ ...(ii) Again a f (3) > 0 $\Rightarrow 1 [ ( 3)^2 - 2a \ (3) + a^2 + a - 3 ] > 0$ $\Rightarrow 9 - 6 a + a^2 + a - 3 > 0 \Rightarrow a \le 3$ $\hspace19mm$ ...(iii) $\Rightarrow (a - 2) \ ( a - 3) > 0$ $\therefore$ $a \ \in \ ( - \infty, 2 ) \cup (3, \infty)$ $\hspace19mm$ ...(iv) From Eqs. (i), (ii), (iii) and (iv), we get $a \ \in (4, 2)$ Therefore, (a) is the answer NOTE There is correction in answer a < 2 should be - 4 < a < 2

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