Coefficients of the equation, Mathematics

If coefficients of the equation ax2 + bx + c = 0, a ¹ 0 are real and roots of the equation are non-real complex and a + c < b, then

(A) 4a + c > 2b (B) 4a + c < 2b (C) 4a + c = 2b (D) none of these

Please give the solution of this question.

Solution": Let assume x=-1

a-b+c but

a+c

hence f(x)<0 for all real values of x

therefore

putting x=-2

we get f(X)=4a+c-2b<0

or 4a+c<2b


 

Posted Date: 9/19/2012 6:39:41 AM | Location : United States







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