# How do you find the set of values for k which the line x+3y=k intersects the curve y^2=2x+3 at 2 real and distinct points?

Oct 4, 2015

I found $k \text{>} - 6$

#### Explanation:

We can try taking the first equation and substitute for $x$ into the second:
$x = k - 3 y$ into the second:
${y}^{2} = 2 \left(k - 3 y\right) + 3$
${y}^{2} = 2 k - 6 y + 3$
Rearranging:
${y}^{2} + 6 y - \left(2 k + 3\right) = 0$
We use the Quadratic Formula to solve for $y$:
${y}_{1 , 2} = \frac{- 6 \pm \sqrt{36 + 4 \left(2 k + 3\right)}}{2} =$
$= \frac{- 6 \pm \sqrt{36 + 8 k + 12}}{2} = \frac{- 6 \pm \sqrt{48 + 8 k}}{2}$
To be real and distinct we need the argument of the square root to be $> 0$, so:
$48 + 8 k > 0$
$k \text{>} - \frac{48}{8}$
$k \text{>} - 6$