If the difference between the roots of x^2+ax+1=0 is less than √5, then the set of possibl…

Question

If the difference between the roots of $x^2+ax+1=0$ is less than $\sqrt5$, then the set of possible values of $a$ is

Accepted Answer

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 $${\left( {\alpha  - \beta } \right)^2} = {\left( {\alpha  + \beta } \right)^2} - 4\alpha \beta $$ )

$$ \Rightarrow \sqrt {{a^2} - 4}  < \sqrt 5 $$

$$ \Rightarrow {a^2} - 4 < 5$$

$$ \Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9$$

$$ \Rightarrow  - 3 < a < 3$$

$$ \Rightarrow a \in \left( { - 3,3} \right)$$

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