Question

If `x^3+ax^2-3x+b = 0` has a root `2-i`, then what are the possible real values of `a` and `b` ?

Answer 1

`a=-2" "` and `" "b = 10`

Explanation:

Given:

`x^3+ax^2-3x+b = 0`

Find real values of `a, b` such that `2-i` is a root.

If the coefficients of the polynomial are real, then any non-real roots will occur in complex conjugate pairs. So `2+i` is also a root and we have factors:

`(x-2-i)(x-2+i) = (x-2)^2-i^2`

`color(white)((x-2-i)(x-2+i)) = x^2-4x+4+1`

`color(white)((x-2-i)(x-2+i)) = x^2-4x+5`

The remaining factor must take the form `(x+c)` for some constant `c` to be determined, since the leading coefficient of `x^3` in the product is `1`.

So we have:

`x^3+ax^2-3x+b = (x+c)(x^2-4x+5)`

`color(white)(x^3+ax^2-3x+b) = x^3+(c-4)x^2+(5-4c)x+5c`

Equating the coefficients of `x` we find:

`-3 = 5-4c`

and hence:

`c = 2`

Equating the other coefficients, we find:

`{ (a = c-4 = 2-4 = -2), (b = 5c = 5 * 2 = 10) :}`