Question

Find the value of k if f(x) has three distinct real roots ?

`x^3-3x+k=0`

Answer 1

See explanation below

Explanation:

If equation has three real roots all of them are distinct, then

`f(x)=(x-a)(x-b)(x-c)` where `a,b, c` are roots of `f(x)`

Developing:

`f(x)=(x^2-bx-ax+ab)(x-c)=x^3-cx^2-bx^2+bcx-ax^2+acx+abx-abc=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc`

We know that two polynomial expresions are equal if and only if his coefficients are equal. Then (Cardano formulae)

`a+b+c=0`
`ab+ac+bc=-3`
`-abc=k`

Now, if we take a look to `x^3-3x` we see that has 3 roots if we re-write in form `x(x^2-3)` ; `x=0, x=+-sqrt3`

Adding +k we are traslating graph up or down if k is positive or negative respectively and this fact we can eliminate 2 real roots adding 2 complex roots.(if a polinomial has a complex root, then his conjugate is also root). Hope this helps

Answer 2

`-2 \le k \le 2`

Explanation:

Let's start by taking a look at `x^3-3x`: it has three roots, and local minimum and a maximum:

graph{x^3-3x [-2 2 -3 3]}

The minimum and maximum are found by posing

`f'(x)=3x^2-3 = 0 \iff x^2-1=0 \iff x=\pm 1`

With corresponding values `f(1)=1-3=-2` for the minimum and `f(-1) = -1+3=2` for the maximum.

Adding `k` to the function translates the graph upwards (`k>0`) or downwards (`k<0`). You can see that the function has three zeroes as long as the maximum is above the `x` axis, and the minimum is below.

If we translate the function up three units, for example, the minimum would be above the `x` axis, and we would lose solutions:

graph{x^3-3x+3 [-5 5 -1 6]}

Similarly, we can't translate the graph down more than two units: see this example with `k=-3`

graph{x^3-3x-3 [-5 5 -6 1]}

While any `k` between `-2` and `2` is ok:

graph{x^3-3x+1 [-3 3 -2 4]}