Question

Question #0e900

Answer 1

In case A: Either `k=-4` with repeated root `x = 1` or `k=0` with roots `x = -1+-sqrt(2)`.

In case B: `k = -3/5` with roots `x = 1/3` and `x = -1`

In case C: `x = -k +-sqrt(2-3k)`

Explanation:

In case A, since the roots are reciprocals of one another, we must have:

`(2k+2) = +-(k-2)`

If `(2k+2) = (k-2)` then `k = -4` and `x = 1`

If `(2k+2) = -(k-2)` then `k = 0` and `x = -1+-sqrt(2)`

In case B, first divide through by `k` to get:

`x^2-(1+k)/kx +(3k+2)/k = 0`

The sum of the roots is `(1+k)/k`

The product of the roots is `(3k+2)/k`

So we are told:

`(1+k)/k = 2 xx (3k+2)/k`

Multiply through by `k` to get:

`1+k = 2(3k+2) = 6k+4`

So `5k = -3` and `k = -3/5`

So the quadratic is:

`-3/5x^2-2/5x+1/5 = 0`

Multiply by `-5` to get:

`3x^2+2x-1 = 0`

That is:

`(3x-1)(x+1) = 0`

So `x=1/3` or `x=-1`

In case C,

`x = -k +-sqrt(2-3k)`

We seem to be missing some additional constraint.