# How do you find all values of k such that 2x^2-3x+5k=0 has two solutions?

Feb 16, 2017

For $x \in \mathbb{C} : k \ne \frac{9}{40}$

For $x \in \mathbb{R} : k < \frac{9}{40}$

#### Explanation:

$2 {x}^{2} - 3 x + 5 k = 0$

The discriminant here is: ${\left(- 3\right)}^{2} - 4 \times 2 \times 5 k$

$= 9 - 40 k$

Assume $k \in \mathbb{Q}$

For the equation to have two roots (real and complex) the discriminant must not equal zero.

$\to 9 - 40 k \ne 0$

$k \ne \frac{9}{40}$

If the roots are to be real only the discriminant must be greater than zero.

$\to 9 - 40 k > 0$

$\therefore 40 k < 9$

$k < \frac{9}{40}$