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 :
$$k = \pm 4$$
Solution :
Let $$\alpha $$ be the common root to the equations :
$$\eqalign{
& {x^2} - kx - 21 = 0\,\,\,{\text{and }}{x^2} - 3kx + 35 = 0 \cr
& \therefore '\alpha '{\text{ satisfies both the equations}} \cr
& \therefore {\alpha ^2} - k\alpha - 21 = 0\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr
& {\text{and }}{\alpha ^2} - 3k\alpha + 35 = 0\,\,\,\,.....\left( {{\text{ii}}} \right) \cr
& {\text{From }}\left( {\text{i}} \right)\,\,{\text{and }}\left( {{\text{ii}}} \right), \cr
& {\alpha ^2} - 21 = \frac{{{\alpha ^2} + 35}}{3} \cr
& \Rightarrow 3{\alpha ^2} - 63 = {\alpha ^2} + 35 \cr
& \Rightarrow {\alpha ^2} = 49 \cr
& \Rightarrow \alpha = \pm 7 \cr} $$
Now, again by eliminating $${\alpha ^2}$$ from (i) and (ii), we get
$$\eqalign{
& k\alpha + 21 = 3k\alpha - 35 \cr
& \Rightarrow 2k\alpha = 56 \cr
& \Rightarrow k = \frac{{56}}{{2\alpha }} \cr
& {\text{When, }}\alpha = 7{\text{ then }}k = 4 \cr
& {\text{When, }}\alpha = - 7{\text{ then }}k = - 4 \cr
& {\text{Hence, }}k = \pm 4 \cr} $$