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Question

If a + b + c > $fraction numerator 9 c over denominator 4 end fraction$ and equation ax² + 2bx – 5c = 0 has non–real complex roots, then

a > 0, c > 0
a > 0, c < 0
a < 0, c < 0
a < 0, c > 0

The correct answer is: a > 0, c < 0

Given, 4a + 4b – 5c > 0
Let f(x) = ax² + 2bx – 5c, then
f(2) = 4a + 4b – 5c > 0
Since equation f(x) = 0 has imaginary roots, therefore f(x) will have same sign as that of a for all x Î R. Since f(2) > 0

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