JEE MAIN - Mathematics (2008 - No. 22)
The quadratic equations $${x^2} - 6x + a = 0$$ 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
1
4
3
2
Explanation
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
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a$$
and $$\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 3\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6$$
$$ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 8$$
$$\therefore$$ The equation becomes
$${x^2} - 6x + 8 = 0$$
$$ \Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0$$
$$ \Rightarrow $$ roots are $$2$$ and $$4$$
$$ \Rightarrow \alpha = 2,\beta = 1$$
$$\therefore$$ Common root is $$2.$$
and $$4$$ $$\beta $$ and that of the equation
$${x^2} - cx + 6 = 0$$ be $$\alpha $$ and $$3\beta .$$ Then
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 4\beta = 6;\,\,\,\,\,\,\,4\alpha \beta = a$$
and $$\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha + 3\beta = c;\,\,\,\,\,\,\,3\alpha \beta = 6$$
$$ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 8$$
$$\therefore$$ The equation becomes
$${x^2} - 6x + 8 = 0$$
$$ \Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0$$
$$ \Rightarrow $$ roots are $$2$$ and $$4$$
$$ \Rightarrow \alpha = 2,\beta = 1$$
$$\therefore$$ Common root is $$2.$$
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