Question

How do you solve `y=x^2+5x+4, y=-x-8` using substitution?

Answer 1

There are no points of intersection.

Explanation:

We have the equations;

`y=x^2+5x+4`

`y=-x-8`

I'm going to substitute the second equation into the first equation:

`-x-8=x^2+5x+4`

and now solve for `x`:

`x^2+6x+12=0`

I don't see any factors, so I'll shift over to the quadratic formula, with the general formula of:

`x = (-b \pm sqrt(b^2-4ac)) / (2a)`

where we have `ax^2+bx+c=0`

In our case we have `a=1, b=6, c=12`, and so:

`x = (-6 \pm sqrt(6^2-4(1)(12))) / (2(1))`

`x = (-6 \pm sqrt(36-48)) / (2)`

`x = (-6 \pm sqrt(-12)) / (2)`

It's at this point that we stop - the points of intersection should be Real numbers (`RR`) and we have imaginary ones. Looking at the graphs of both equations, we can see there are no points of intersection:

graph{(y-x^2-5x-4)(y+x+8)=0 [-25.66, 25.66, -12.83, 12.83]}