A) \[k=\pm 4\]
B) \[k=\pm 1\]
C) \[k=\pm 3\]
D) \[k=0\]
Correct Answer: A
Solution :
Key Idea: The equations \[a{{x}^{2}}+bx+c=0\] ...(i) and \[d{{x}^{2}}+gx+f=0\] ...(ii) have a common roots, if \[{{(dc-af)}^{2}}=(bf-cg)(ag-bd)\] Given quadratic equations are \[{{x}^{2}}-kx-21=0\] ...(iii) and \[{{x}^{2}}-3kx+35=0\] ...(iv) Now, on comparing Eqs. (iii) and Eqs. (iv) with (i) and (ii), we get \[a=1,b=-k,\text{ }c=-\text{ }21,\text{ }d=1,\text{ }g=-3\text{ }k,\text{ }f=35\] \[\therefore \]For common roots \[{{(-21-35)}^{2}}=(-35k-63k)(-3k+k)\] \[\Rightarrow \] \[{{(-56)}^{2}}=(-98k)(-2k)\] \[\Rightarrow \] \[{{k}^{2}}=\frac{56\times 56}{98\times 2}=16\] \[\Rightarrow \] \[k=\pm 4\] Alternative Method Let a be the common root to the equations \[{{x}^{2}}-kx-21=0\]and \[{{x}^{2}}-3kx+35=0\] \[\Rightarrow \] \[{{\alpha }^{2}}-k\alpha -21=0\] ...(i) and \[{{\alpha }^{2}}-3k\alpha +35=0\] ...(ii) Now, by cross multiplication method \[\frac{{{\alpha }^{2}}}{(-35k-63k)}=\frac{\alpha }{(-21-35)}=\frac{1}{(-3k+k)}\] \[\Rightarrow \] \[\frac{{{\alpha }^{2}}}{-98k}=\frac{\alpha }{-56}=\frac{-1}{2k}\] \[\Rightarrow \] \[\frac{\alpha }{-56}=\frac{-1}{2k}\] \[\Rightarrow \] \[\alpha =\frac{28}{k}\] ?. (iii) Also, \[\frac{{{\alpha }^{2}}}{-98k}=\frac{-1}{2k}\] \[\Rightarrow \] \[{{\alpha }^{2}}=\frac{98}{2}=49\] ?. (iv) From Eqs. (iii) and (iv) \[49=\frac{28\times 28}{{{k}^{2}}}\] \[\Rightarrow \] \[{{k}^{2}}=\frac{28\times 28}{49}=16\] \[\Rightarrow \] \[k=\pm 4\]
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