Question

The values of k for which the equations $ {{x}^{2}}-kx-21=0 $ and $ {{x}^{2}}-3kx+35=0 $ will have a common roots are

• A. $ k=\pm 4 $
• B. $ k=\pm 1 $
• C. $ k=\pm 3 $
• D. $ k=0 $

Answer 1

Answer: (A)

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 $