JEE MAIN - Mathematics (2007 - No. 22)
If the difference between the roots of the equation $${x^2} + ax + 1 = 0$$ is less than $$\sqrt 5 ,$$ then the set of possible values of $$a$$ is
$$\left( {3,\infty } \right)$$
$$\left( { - \infty , - 3} \right)$$
$$\left( { - 3,3} \right)$$
$$\left( { - 3,\infty } \right)$$
Explanation
Let $$\alpha $$ and $$\beta $$ are roots of the equation $${x^2} + ax + 1 = 0$$
So, $$\alpha + \beta = - a$$ and $$\alpha \beta = 1$$
given $$\left| {\alpha - \beta } \right| < \sqrt 5 $$
$$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 $$
(as $${\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta $$ )
$$ \Rightarrow \sqrt {{a^2} - 4} < \sqrt 5 $$
$$ \Rightarrow {a^2} - 4 < 5$$
$$ \Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9$$
$$ \Rightarrow - 3 < a < 3$$
$$ \Rightarrow a \in \left( { - 3,3} \right)$$
So, $$\alpha + \beta = - a$$ and $$\alpha \beta = 1$$
given $$\left| {\alpha - \beta } \right| < \sqrt 5 $$
$$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 $$
(as $${\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta $$ )
$$ \Rightarrow \sqrt {{a^2} - 4} < \sqrt 5 $$
$$ \Rightarrow {a^2} - 4 < 5$$
$$ \Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9$$
$$ \Rightarrow - 3 < a < 3$$
$$ \Rightarrow a \in \left( { - 3,3} \right)$$
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