Question 12

# If both the roots of the equation $$x^{2}-4ax+3-3a+4a^{2}=0$$ exceed 2, then the value of a:

Solution

It is given that both the roots of given equation exceeds 2. Therefore, at x = 2, equation will be greater than zero.
$$4 - 8a + 3 - 3a + 4a^2 > 0$$
$$4a^2-11a+7>0$$
(4a-7)(a-1) > 0
a < 1 and a > 7/4