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Grade: 12

                        

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.

8 years ago

Answers : (1)

mayank singh
16 Points
							

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

8 years ago
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