If the roots of the equation x^{2}-2ax+a^{2}+a-3 =0 are less than 3 then find the set of all possible values of a. (ans : -infinity, 2)

**Given** equation is:

*x*^{2 }- 2*ax* + *a*^{2 }+ *a* - 3 = 0

⇒ *x*^{2 }- 2*ax* + (*a*^{2 }+ *a* - 3) = 0

Let *p* and *q* be the roots of the given equation.

So, sum of roots = *p +q* = 2*a* and product of roots = (*a*^{2 }+ *a* - 3)

It is given that roots of equation are less than 3.

So, sum of roots < 6 (As 3 + 3 = 6)

2*a* < 6

⇒ *a* < 3

So, for any integral value of *p* and *q* less than 3, value of *a* is always less than 3.

Hence, the solution set of *a *will always be (2, -∞).

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