A) Both the equations always have real roots
B) At least one equation always has real roots
C) Both the equation always have non real roots
D) At least one equation always has real and equal roots
Correct Answer: B
Solution :
Let the discriminant of the equation \[{{\operatorname{x}}^{2}}+px+q=0\,\,\,by\,\,{{D}_{1}}\], then \[{{D}_{1}}={{p}^{2}}-4q\] and the discriminant \[{{D}_{2}}\] of the equation \[{{\operatorname{x}}^{2}}+rx+s=0\,\,is\,\,D={{r}^{2}}-4s\] \[\therefore \,\,\,{{D}_{1}}+{{D}_{2}}={{p}^{2}}+{{r}^{2}}-4(q+s)={{p}^{2}}+{{r}^{2}}-2\,pr\] [from the given relation] \[\therefore \,\,\,{{D}_{1}}+{{D}_{2}} ={{\left( p\,-r \right)}^{2}} \ge 0\] Clearly at least one of \[{{\operatorname{D}}_{1}}\,\,and\,\,{{D}_{2}}\] must be non-negative consequently at least one of the equation has real roots.
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