JEE MAIN - Mathematics (2002 - No. 41)
Difference between the corresponding roots of $${x^2} + ax + b = 0$$ and $${x^2} + bx + a = 0$$ is same and $$a \ne b,$$ then
$$a + b + 4 = 0$$
$$a + b - 4 = 0$$
$$a - b - 4 = 0$$
$$a - b + 4 = 0$$
Explanation
Let $$\alpha ,\beta $$ and $$\gamma ,\delta $$ be the roots of the equations $${x^2} + ax + b = 0$$
and $${x^2} + bx + a = 0$$ respectively.
$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$
and $$\gamma + \delta = - b,\gamma \delta = a.$$
Given $$\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|$$
$$ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}$$
$$ \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta $$
$$ \Rightarrow {a^2} - 4b = {b^2} - 4a$$
$$ \Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$
$$ \Rightarrow a + b + 4 = 0$$
( as $$a \ne b$$ )
and $${x^2} + bx + a = 0$$ respectively.
$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$
and $$\gamma + \delta = - b,\gamma \delta = a.$$
Given $$\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|$$
$$ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}$$
$$ \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta $$
$$ \Rightarrow {a^2} - 4b = {b^2} - 4a$$
$$ \Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$
$$ \Rightarrow a + b + 4 = 0$$
( as $$a \ne b$$ )
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