Question
The quadratic equations $${x^2} - 6x + a = 0\,\,{\text{and }}\,{x^2} - cx + 6 = 0$$ have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is
A.
1
B.
4
C.
3
D.
2
Answer :
2
Solution :
Let the roots of equation $${x^2} - 6x + a = 0\,$$ be $$\alpha $$ and $$4\beta $$ and that of the equation
$${x^2} - cx + 6 = 0$$ be $$\alpha $$ and $$3\beta $$ . Then
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\alpha + \,4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a \cr
& {\text{and }}\alpha {\text{ + 3}}\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6 \cr
& \Rightarrow \,\,a = 8 \cr
& \therefore \,\,{\text{The equation becomes }}\,{x^2} - 6x + 8 = 0 \cr
& \Rightarrow \,\,\left( {x - 2} \right)\left( {x - 4} \right) = 0 \cr
& \Rightarrow \,\,{\text{roots are 2 and 4}} \cr
& \Rightarrow \,\,\alpha = 2,\beta = 1 \cr
& \therefore \,\,{\text{Common root is 2}}{\text{.}} \cr} $$