A) Real
B) Equal
C) No real
D) Can't be determined
Correct Answer: C
Solution :
The given equation \[{{x}^{2}}+2cx+ab=0\] has real and unequal roots. \[\Rightarrow \] \[D={{(2c)}^{2}}-4ab>0\] \[\Rightarrow \] \[4{{c}^{2}}-4ab>0\] \[\Rightarrow \] \[{{c}^{2}}-ab>0\] Now, the equation \[{{x}^{2}}-2(a+b)x+{{a}^{2}}+{{b}^{2}}+2{{c}^{2}}=0\] \[\therefore \] \[D={{(-2(a+b))}^{2}}-4({{a}^{2}}+{{b}^{2}}+2{{c}^{2}})\] \[=4{{a}^{2}}+4{{b}^{2}}+8ab-4{{a}^{2}}-4{{b}^{2}}-8{{c}^{2}}\] \[=8ab-8{{c}^{2}}=8\,(ab-{{c}^{2}})<0\] \[(\therefore \,\,{{c}^{2}}-ab>0)\] Hence, the equation has no real roots.You need to login to perform this action.
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