# Question #0e900

Jun 22, 2015

In case A: Either $k = - 4$ with repeated root $x = 1$ or $k = 0$ with roots $x = - 1 \pm \sqrt{2}$.

In case B: $k = - \frac{3}{5}$ with roots $x = \frac{1}{3}$ and $x = - 1$

In case C: $x = - k \pm \sqrt{2 - 3 k}$

#### Explanation:

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

$\left(2 k + 2\right) = \pm \left(k - 2\right)$

If $\left(2 k + 2\right) = \left(k - 2\right)$ then $k = - 4$ and $x = 1$

If $\left(2 k + 2\right) = - \left(k - 2\right)$ then $k = 0$ and $x = - 1 \pm \sqrt{2}$

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

${x}^{2} - \frac{1 + k}{k} x + \frac{3 k + 2}{k} = 0$

The sum of the roots is $\frac{1 + k}{k}$

The product of the roots is $\frac{3 k + 2}{k}$

So we are told:

$\frac{1 + k}{k} = 2 \times \frac{3 k + 2}{k}$

Multiply through by $k$ to get:

$1 + k = 2 \left(3 k + 2\right) = 6 k + 4$

So $5 k = - 3$ and $k = - \frac{3}{5}$

$- \frac{3}{5} {x}^{2} - \frac{2}{5} x + \frac{1}{5} = 0$

Multiply by $- 5$ to get:

$3 {x}^{2} + 2 x - 1 = 0$

That is:

$\left(3 x - 1\right) \left(x + 1\right) = 0$

So $x = \frac{1}{3}$ or $x = - 1$

In case C,

$x = - k \pm \sqrt{2 - 3 k}$

We seem to be missing some additional constraint.