Question

If `a` and `b` are non-zero integers, is it possible for `x^3-ax^2+(a^2-b)x+a(2b-a^2)=0` to have more than one Real root?

See https://socratic.org/s/aB5FuCr4 to understand where this cubic comes from if it helps.

Answer 1

Yes, for example with `a=1` and `b=20` we get the cubic:

`x^3-x^2-20x+39=0`

which has three real roots.

Explanation:

`x^3-ax^2+(a^2-b)x+a(2b-a^2)=0`

The discriminant `Delta` of a cubic in the form:

`Ax^3+Bx^2+Cx+D`

is given by the formula:

`Delta = B^2C^2-4AC^3-4B^3D-27A^2D^2+18ABCD`

In our example:

`{ (A=1), (B=-a), (C=a^2-b), (D=a(2b-a^2)) :}`

So:

`Delta = (-a)^2(a^2-b)^2-4(a^2-b)^3-4(-a)^3(a(2b-a^2))-27(a(2b-a^2))^2+18(-a)(a^2-b)(a(2b-a^2))`

`color(white)(Delta) =-16a^6 + 72a^4b - 83a^2b^2 + 4b^3`

Note that:

• The sign of `a` does not affect the discriminant.

• The term of highest degree in `a` is `-16a^6`. So when `a` is large compared with `b`, the discriminant will be negative and the cubic will have only one real root.

• The term of highest degree in `b` is `4b^3`. So when `b` is large and positive compared with `a`, the discriminant will be positive and the cubic will have three real roots.

So try `a=1` and `b=20`

Then `Delta = -16+72(20)-83(400)+4(8000) = 224 > 0`

So our example cubic is:

`x^3-x^2-20x+39=0`

which does indeed have three real roots.

graph{x^3-x^2-20x+39 [-10, 10, -10, 75]}