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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 (A) 4a + c > 2b (B) 4a + c Please give the solution of this question.

 


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.


 


 

Grade:12

1 Answers

mayank singh
16 Points
11 years ago

ans is B

puting x=-1

a-b+c

but

a+c<b or a+c-b<0

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

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