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 $a=1,b=-k,c=-21,d=1,g=-3k,f=35$ $\therefore$ 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) $49=\frac{28\times 28}{k^2}$ $\Rightarrow$ $k^2=\frac{28\times 28}{49}=16$ $\Rightarrow$ $k=\pm 4$