The value of $P$ for which both the roots of the equation $\displaystyle 4x^{2}-20Px+(25P^{2}+15P-66)= 0$ are less than $2$, lies in
- $\displaystyle \left ( \frac{4}{5},2 \right )$
- $\displaystyle (2,\infty )$
- $\displaystyle \left (-1, -\frac{4}{5} \right )$
- $\displaystyle ( -\infty,-1)$
For roots to be less than 2, the vertex -b/2a must be < 2, f(2) > 0, and D >= 0. Solving these inequalities for P leads to P < -1.
For both roots to be less than 2, the value of the polynomial at x = 2 must be strictly positive, the vertex must be less than 2, and the discriminant must be positive. First, the discriminant condition requires (-20P)^2 - 16(25P^2 + 15P - 66) > 0, which simplifies to 400P^2 - 400P^2 - 240P + 1056 > 0, yielding P < 22/5. Second, the vertex condition requires -(-20P) / 8 < 2, which gives 5P/2 < 2 and means P < 4/5. Third, the polynomial at x = 2 gives 16 - 40P + 25P^2 + 15P - 66 > 0, simplifying to 25P^2 - 25P - 50 > 0. Dividing by 25 gives P^2 - P - 2 > 0, so (P - 2)(P + 1) > 0, which holds true when P < -1 or P > 2. The intersection of P < -1, P < 4/5, and P < 22/5 is P < -1. The correct interval is from negative infinity to -1.