Given that ax​2 + bx + c has always the same sign as c if

a) b>4ac   b) c>0  c) c<0   d) none of these 

Dear Student,
Please find below the solution to the asked query:

fx=ax2+bx+cCase i c>0If c>0, then we wantax2+bx+c>0 for all xRwhich is possible when opening is upward a>0 and discrimimant is negativei.e. graph does not cut x-axis.So we havea>0andb2-4ac<0Case iiIf c<0, we wantax2+bx+c<0which is possible when opening is downward a<0 and discriminant is negativei.e. graph does not cut x-axis.a<0andb2-4ac<0Hence option d is correct.

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