<s>[INST] Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$ We have a parabola and a circle with the following equations and their graph placed at the end of my question.
Parabola: $y^2 = 4x -4$
Circle: $(x-2)^2 + y^2 = 9$
My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got
$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$
$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?
What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?

update
When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:
$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$
$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$
$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.
2nd update
I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:
$(x-2)^2 +y^2=9 \implies 
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies 
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies 
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies 
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies 
(\frac{y^2}{4} +  1)^2=9 \implies 
(\frac{y^2}{4} +  1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$
 [/INST] When you are searching for intersection of two figures the intersection dot must satisfy parabola equation which indicates that
$$ y^2 = 4x - 4$$
Which means that
$$ 4x - 4 \geq 0$$
$$ x \geq 1 $$
The last inequality will exclude $x = -3$ solution.
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