Question 601101
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I can't figure where you got *[tex \LARGE 3x^2\ +\ 6x\ +\ 0\ =\ 0] unless you were just guessing.  BTW, that is almost always a poor strategy.  In any case, toss it out and start over.


Use Pythagoras.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\ +\ b^2\ =\ c^2]



Substitute what you were given:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 1)^2\ +\ (x\ +\ 3)^2\ =\ \left(2\sqrt{5}\right)^2]


Expand the binomials and square the constant.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 2x\ +\ 1\ +\ x^2\ +\ 6x\ +\ 9\ =\ 20]


Collect terms in the LHS


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ +\ 8x\ -\ 10\ =\ 0]


Divide by 2


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ -\ 5\ =\ 0]


Solve the factorable quadratic.  Discard the negative root because it leads to the absurdity of negative measures of length.  Calculate *[tex \LARGE x\ +\ 1] and *[tex \LARGE x\ +\ 3]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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