Answer:
I have provided two solutions both concluding the same thing.
There is no solution.
Simplest solution says this is a parabola opened up at vertex (-2,5) which means it never crosses the x-axis. There is no solution because thee curve does not touch the x-axis.
Step-by-step explanation:
Harder solution (algebraic solution):
The vertex form a parabola is [tex]y=a(x-h)^2+k[/tex]
We are given the vertex (h,k) is (-2,5)
So we have [tex]y=a(x+2)^2+5[/tex]
Now we also know [tex]a>0[/tex] since the parabola opens up.
That is all we know about a.
Let's see what [tex]y=a(x+2)^2+5[/tex] is in standard form
[tex]y=a(x+2)(x+2)+5\\y=a(x^2+4x+4)+5\\y=ax^2+4ax+4a+5\\\\[/tex]
So we are asked to solve for the solution set of
[tex]0=ax^2+4ax+4a+5\\A=a\\B=4a\\C=4a+5\\\\[/tex]
Plug into quadratic formula
I'm going to write the quadratic formula with the capital letters to be less confusing:
[tex]x=\frac{-B \pm \sqrt{B^2-4AC}}{2A}\\x=\frac{-4a \pm \sqrt{16a^2-4(a)(4a+5)}}{2a}\\x=\frac{-4a \pm \sqrt{16a^2-16a^2-20a}}{2a}\\x=\frac{-4a \pm \sqrt{-20a}}{2a}\\x=\frac{-4a \pm \sqrt{4} \sqrt{-5a}}{2a}\\x=\frac{-4a \pm 2 \sqrt{-5a}}{2a}\\x=\frac{-4a}{2a} \pm \frac{\sqrt{-5a}}{a}\\[/tex]
This says [tex]a[/tex] has to be negative... The inside of the square root... So there was no real solution.\\
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Simpler solution (graph/visual)
You could have also drawn a parabola open up with vertex at (-2,5) and we should have seen that it was impossible it cross the x-axis.
