If the equation $x ^ { 2 } + 5 b x + 8 c = 0 ,$ does not have two distinct real roots, then minimum value of $5 b +8c$ is
- $-1$
- $-2$
- $1$
- $2$
For the equation x^2 + 5bx + 8c = 0 to not have two distinct real roots, the discriminant D must be <= 0. D = (5b)^2 - 4(1)(8c) = 25b^2 - 32c <= 0. This implies 32c >= 25b^2. We want the minimum value of 5b + 8c. Since 8c >= (25/4)b^2, we minimize f(b) = 5b + (25/4)b^2. Derivative f'(b) = 5 + (25/2)b = 0, so b = -10/25 = -0.4. Min value = 5(-0.4) + (25/4)(-0.4)^2 = -2 + (25/4)(0.16) = -2 + 1 = -1.
Since the quadratic equation does not have two distinct real roots, its discriminant must be less than or equal to zero. Using the discriminant formula, we write $(5b)^2 - 4(1)(8c) \le 0$, which simplifies to $25b^2 \le 32c$, or $c \ge 25b^2 / 32$. We want to minimize the linear expression $5b + 8c$, so we substitute the inequality to get $5b + 8(25b^2 / 32)$. This simplifies to the function $f(b) = 5b + 25b^2 / 4$. To find the minimum of this upward-facing parabola, we take the derivative $f'(b) = 5 + 25b / 2$ and set it to zero, yielding $b = -2/5$. Substituting $b = -2/5$ back into the function gives $5(-2/5) + 25(-2/5)^2 / 4$, which evaluates to $-2 + 1$, so the minimum value is $-1$.