A) Are real and negative
B) Have negative real parts
C) Are rational numbers
D) None of these
Correct Answer: B
Solution :
The roots of the equations are given by \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] (i) Let \[{{b}^{2}}-4ac>0,b>0\] Now if \[a>0,c>0,{{b}^{2}}-4ac<{{b}^{2}}\] Þ the roots are negative. (ii) Let \[{{b}^{2}}-4ac<0,\] then the roots are given by \[x=\frac{-b\pm i\sqrt{(4ac-{{b}^{2}})}}{2a},\,\,\,\,\,\,(i=\sqrt{-1})\] Which are imaginary and have negative real part\[(\because \,\,b>0)\] \ In each case, the roots have negative real part.
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