Consider a quadratic polynomial $f(x) = x^{2} - 4ax + 5a^{2} - 6a.$ $p :$ denotes smallest positive integral value of $a$, for which $f(x)$ is positive for every real $x$, and $q :$ denotes largest distance between the roots of the equation $f(x)= 0$ The value of $p + q$ is

For the polynomial $f(x) = x^2 - 4ax + 5a^2 - 6a$ to be positive for every real $x$, its discriminant must be strictly less than zero. Setting the discriminant $(-4a)^2 - 4(1)(5a^2 - 6a) < 0$ yields $16a^2 - 20a^2 + 24a < 0$, which simplifies to $-4a^2 + 24a < 0$ or $a(a - 6) > 0$. This inequality holds for $a < 0$ or $a > 6$, making the smallest positive integer $a = 7$, so $p = 7$. The distance between the roots of a quadratic $Ax^2 + Bx + C = 0$ is given by $\sqrt{B^2 - 4AC} / |A|$, which here evaluates to $\sqrt{24a - 4a^2} / 1$. The maximum of this distance occurs at the vertex of the parabola $a = -24 / (2(-4)) = 3$, where the maximum distance $q$ is $\sqrt{24(3) - 4(3)^2} = 6$. Therefore, the value of $p + q$ is $7 + 6 = 13$.

• Thanks
• Wrong explanation