Real and distinct roots of a cubic equation

Question

The real values of $a$ for which the equation $x^3-3x+a=0$ has three real and distinct roots is

Answer 1

The discriminant of the polynomial is

$$108-27a^2$$

If the discriminant is positive, we have three distint real roots.

So, a must satisfy the condition |a|<2.

Answer 2

If you want them to be real and distinct...

Natural choice would be to consider :$$(x-l)(x-m)(x-n)=(x^2-(l+m)x+lm)(x-n)=x^3-(l+m+n)x^2+(lm+n(l+m))x-lmn$$

Equating the coefficients we would have : $$l+m+n=0$$

$$lm+n(l+m)=-3\Rightarrow lm-n^2=-3\Rightarrow -lmn =n(3-n^2)$$

We need to find bound for $-lmn=n(3-n^2)$...

Can you do it by using derivative?

Answer 3

The derivative cancels at $x=1$ and $x=-1$. To these correspond a maximum value of $a+2$ and a minimum value of $a-2$. In order to have three real roots, you need three $x$ intercepts; this means that you must have $a+2>0$ and $a-2<0$. So, the condition is $|a|<2$.

Answer 4

Here is a graphical approach. Note that '+ a' shifts the graph up or down.

1) Disregard '+ a'.

2) Graph $ f(x) = x^3 - 3x $

http://www.wolframalpha.com/input/?i=plot%28x%5E3+-+3x%29#

3) from the graph we see that we must find the y values at the local extrema. You did that already getting y = 2 and -2

4) It should be obvious now by inspection that a must be between -2 and 2

Final answer ,

-2 < a < 2