If the difference between the roots of the equation $x^2+a\cdot x+1=0$ is less than $\sqrt{5}$, then find the set of possible values of a.

Question

If the difference between the roots of the equation $x^2+a\cdot x+1=0$ is less than $\sqrt{5}$, then find the set of possible values of a.

My attempt is as follows:

$$\left|\frac{\sqrt{D}}{1}\right|<\sqrt{5}$$

Squaring both sides

$$D<5$$ $$a^2-4<5$$ $$a^2-9<0$$ $$(a-3)(a+3)<0$$ $$a\in \left(-3,3\right)$$

But for $a\in \left(-2,2\right)$, D would be negative and $\sqrt{D}$ would be a complex number and in the original equation we are comparing $\left|\sqrt{D}\right|<\sqrt{5}$ and how can we compare a complex number with real no as $\sqrt{D}$ would be complex number if $a\in \left(-2,2\right)$.

So $a\in (-3,-2] \cup [2,-3)$, but my answer is not matching as actual answer is $a\in (-3,3)$.

What am I missing here?

Answer 1

For real values of a I think your answer is right.

Assume other case when it got complex roots then difference is x! . If we take it's magnitude which is $ |x!|=x$ . So here magnitude of difference is $ <\sqrt5$. Now if question is about magnitude of difference then answer [-3,3] is right.

Finally problem is question , To get answer [-3,3] question should have been magnitude of difference of roots is less then $\sqrt{5}$ so that it includes complex roots.

Answer 2

Edited to add: I misread the question -- see comments below.

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You have $$\sqrt{a^2-4}<\sqrt 5$$ This inequality only makes sense if $a^2-4\ge 0$. But you ignored that subtlety when you squared both sides.

For an even simpler example: suppose you have the inequality $$\sqrt x < 1$$ Squaring both sides will give you $$x < 1$$ But I hope it is obvious that the correct solution is $$0\le x < 1$$