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+1 vote
29.5k views
in Complex number and Quadratic equations by (52.7k points)
edited by

If a + b + c > 9c/4 and equation ax2 + 2bx − 5c = 0 has non-real complex roots, then

(A)  a > 0, c > 0

(B)  a > 0, c < 0

(C)  a < 0, c < 0 

(D)  a < 0, c > 0

1 Answer

+2 votes
by (46.6k points)
selected by
 
Best answer

Correct option   (B) 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 ∈ R. Since f(2) > 0.

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