A) are real and negative
B) have negative real parts
C) are rational numbers
D) None of these
Correct Answer: B
Solution :
Let \[a>0,\,\,b>0,c>0\] Given equation \[{{\operatorname{ax}}^{2}} + bx + c = 9\] we know that \[\operatorname{D} = {{b}^{2}} - 4ac\] and \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] Let \[{{\operatorname{b}}^{2}}-4ac>0,\,\,b>0\] If \[\operatorname{a} > 0, c > 0 \,then {{b}^{2}} - 4ac < {{b}^{2}}\] \[\Rightarrow \] Roots are negative Let \[{{\operatorname{b}}^{2}}-4ac<0\], then \[x=\frac{-b\pm i\sqrt{4ac-{{b}^{2}}}}{2a}\] roots are imaginary and have negative real part. \[\left( \because \,\,b>0 \right).\]
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