Question

If the roots of the equation x²-2ax+a²+a-3 =0 are less than 3 then find the set of all possible values of a. (ans : -infinity, 2)

Answer 1

Given equation is:

x² - 2ax + a² + a - 3 = 0

⇒ x² - 2ax + (a² + a - 3) = 0

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

So, sum of roots = p +q = 2a and product of roots = (a² + a - 3)

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

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

2a < 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, -∞).