Question #8b2a4

1 Answer
Jan 11, 2018

No point of intersection between the two equation.

Explanation:

Let's turn the equation around so that we have #y# isolated.

#x-5y=0# becomes
#-5y=-x#
#y=x/5#

For these two equation to intersect, we have to have the same #y# values.

Therefore, the equations #y=x/5# and #y=-x^2-7/4#
Therefore, we have:
#x/5=-x^2-7/4# We see that by turning this equation around, we could have a quadratic equation in the form #ax^2+bx+c=0#

#x/5=-x^2-7/4# becomes
#0=-x^2-7/4-x/5=>0=x^2+7/4+x/5#
We now solve for #x#.
#5[0=x^2+7/4+x/5]#
#0=20x^2+35+4x#
#0=20x^2+4x+35#
Now, before we even start our messy work for the quadratic formula, we first ask ourselves: Is there a real number answer for this equation?

We use the fact that when #b^2-4(a)(c)<0#, there is no real solutions.
Let's test this out.

#4^2-4(20)(35)<0#

#4^2-2800<0#
#-2784<0#

We can now see that there is no real solution.

So what does this mean?
It means that there is no point of intersection between the two equations.