How do you show that kx^2+2x-(k-2) = 0 has at least one real root for any value of k ?

May 29, 2015

$k {x}^{2} + 2 x - \left(k - 2\right)$ has discriminant $\Delta$ given by the formula

$\Delta = {2}^{2} - \left(4 k \cdot - \left(k - 2\right)\right)$

$= 4 + 4 {k}^{2} - 8 k$

$= 4 \left({k}^{2} - 2 k + 1\right)$

$= 4 {\left(k - 1\right)}^{2} \ge 0$

If $k = 1$ then $\Delta = 0$ and $k {x}^{2} + 2 x - \left(k - 2\right) = 0$ has one repeated real root $\left(x = - 1\right)$.

If $k \ne 1$ then $\Delta > 0$ and $k {x}^{2} + 2 x - \left(k - 2\right) = 0$ has 2 distinct real roots.

May 29, 2015

More simply, $x = - 1$ is a root for all values of $k$

Substituting $x = - 1$, we get:

$k {x}^{2} + 2 x - \left(k - 2\right) = k {\left(- 1\right)}^{2} + 2 \left(- 1\right) - \left(k - 2\right)$

$= k - 2 - \left(k - 2\right) = 0$

So $\left(x + 1\right)$ is one factor of $k {x}^{2} + 2 x - \left(k - 2\right)$, the other being $\left(k x - k + 2\right)$

$\left(x + 1\right) \left(k x - k + 2\right)$

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

$= k {x}^{2} - k x + 2 x + k x - k + 2$

$= k {x}^{2} + 2 x - \left(k - 2\right)$

The other root of $k {x}^{2} + 2 x - \left(k - 2\right) = 0$ is $x = \frac{k - 2}{k}$ unless $k = 0$.