A) \[k=\pm 4\]
B) \[k=\pm 1\]
C) \[k=\pm 3\]
D) \[k=0\]
Correct Answer: A
Solution :
Let \[\alpha \] be the common root to the equations: \[{{x}^{2}}-kx-21=0\] and \[{{x}^{2}}-3kx+35=0\] \[\therefore \,\,\,\,\,\,'\alpha '\] satisfies both the equations \[\therefore \,\,\,\,\,\,{{\alpha }^{2}}-k\alpha -21=0\] ...(i) and \[\therefore \,\,\,\,\,\,{{\alpha }^{2}}-3\,\,k\alpha +35=0\] ...(ii) From (i) and (ii), \[{{\alpha }^{2}}-21=\frac{{{\alpha }^{2}}+35}{3}\] \[\Rightarrow \,\,\,3{{\alpha }^{2}}-63={{\alpha }^{2}}+35\] \[\Rightarrow \,\,\,{{\alpha }^{2}}=49\Rightarrow \alpha =\pm 7\] Now, again by eliminating \[{{\alpha }^{2}}\] from (i) and (ii), we get \[k\alpha +21=3k\alpha -35\] \[\Rightarrow \,\,\,2k\alpha =56\Rightarrow k=\frac{56}{2\alpha }\] When \[\alpha =7\] then \[k=4\] When \[\alpha =-7\] then \[k=-4\] Hence, \[k=\pm 4\]
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