If $a+b+c>\dfrac{9c}{4}$ and quadratic equation $ax^{2}+2bx-5c=0$ has non-real, then
- $a>0, c>0$
- $a>0, c<0$
- $a<0,c<0$
- $a<0, c>0$
For non-real roots, the discriminant D = (2b)^2 - 4(a)(-5c) < 0, so 4b^2 + 20ac < 0, or b^2 + 5ac < 0. Given a+b+c > 9c/4, we analyze the signs. If a > 0 and c < 0, the product ac is negative, which supports the condition for non-real roots.
Given the quadratic equation ax^2 + 2bx - 5c = 0 has non-real roots, its discriminant must be strictly negative: (2b)^2 - 4(a)(-5c) < 0, which simplifies to b^2 + 5ac < 0. This inequality implies that the product ac must be negative, meaning a and c have opposite signs. The condition a + b + c > 9c/4 implies a + b > 5c/4. For the parabola to not cross the x-axis while maintaining the stated inequalities, a must be positive, which forces c to be negative. Therefore, a > 0 and c < 0.