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

1 Answer

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]}