Does the parabola y= -1/2x^2 - 5x - 10 ever intersect the line y= 2?

Apr 14, 2018

$\text{ Yes, at the points } \left(- 5 - \sqrt{17} , 2\right) \mathmr{and} \left(- 5 + \sqrt{17} , 2\right)$.

Explanation:

If it does, then the $x \text{-co-ordinates}$ of their points of

intersection (if any) must satisfy the following eqn. :

$2 = - \frac{1}{2} {x}^{2} - 5 x - 2$.

To get rid of fractions, multiplying by $2$, we have,

$4 = - {x}^{2} - 10 x - 4 , \mathmr{and} , {x}^{2} + 10 x + 8 = 0$.

$\therefore {x}^{2} + 2 \cdot x \cdot 5 + {5}^{2} - 25 + 8 = 0 , i . e . ,$.

${\left(x + 5\right)}^{2} = 25 - 8 = 17$.

$\therefore x + 5 = \pm \sqrt{17}$.

$\therefore x = - 5 \pm \sqrt{17}$.

The corresponding $y \text{-co-ordinate}$ is already known to be $2$.

Accordingly, the given curves do intersect each other at the

points $\left(- 5 - \sqrt{17} , 2\right) \mathmr{and} \left(- 5 + \sqrt{17} , 2\right)$.