Number of roots (Part - 2)

For the equation above, something that you could do intuitively is try to turn it simply into x^2 + ax + 1 = 0 since the only problem with |x| is that the inside x can be positive or negative but both turn into the positive form anyways.

So if we were to actually turn the equation into x^2 + ax + 1 = 0, then for one solution of x in that equation, it will give us two solutions for the real equation that we were given at the start, because for example if you were to put x = -1 into x^2 + a|x| + 1 = 0 you will get the positive output of it in all of the terms with x in it. The x^2 will give you a positive output for both the negative and positive version of the input. This is also the same for the |x|. So in total or rather at the same time, when you put those terms together you will both get the positive outputs of them. The constant 1 in the equation also doesn't change.

All in all, for any one solution when solving this equation: x^2 + ax + 1 = 0, you will get two solutions for the real equation given: x^2 + a|x| + 1 = 0.

When actually solving x^2 + ax + 1 = 0, it's actually just solving a quadratic equation with either 1, 2 or no solutions at all.

If you were to have no solutions, then there are just no solutions. But the thing is we only care about if it is even possible to have solutions or not, so we cannot find values of a in this condition.

If you were to have 1 solution then that means both solutions are the same which means that the discriminant is 0, which is when you calculate it, a = -2 or 2. Since they are the same solution, inputting them to the original equation will again give you two solutions.

If you were to have 2 solutions then that means they are distinct and when you input each of them into the original equation you will get the negative and positive versions for each of them which give you a total of 4 solutions.

So there can only be pairs containing two real solutions when solving for x^2 + a|x| + 1 = 0, in other words, you can only get even numbers of solutions because of the absolute value that is attached to the x. Therefore it is not possible to get 3 solutions or roots to the original equation, but only 2, 4 or no solutions at all for any value of a.

A quadratic equation can have at most 2 roots, unless you would like to argue with Gauss.

equation will have 3 roots if one of root had been 0 which is not.