Question

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?

Answer 1

I found `k">"-6`

Explanation:

We can try taking the first equation and substitute for `x` into the second:
`x=k-3y` into the second:
`y^2=2(k-3y)+3`
`y^2=2k-6y+3`
Rearranging:
`y^2+6y-(2k+3)=0`
We use the Quadratic Formula to solve for `y`:
`y_(1,2)=(-6+-sqrt(36+4(2k+3)))/2=`
`=(-6+-sqrt(36+8k+12))/2=(-6+-sqrt(48+8k))/2`
To be real and distinct we need the argument of the square root to be `>0`, so:
`48+8k>0`
`k">"-48/8`
`k">"-6`