Question 348713
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ \sqrt{4\ -\ 3x}\ =\ -8]


Add *[tex \Large -x] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\sqrt{4\ -\ 3x}\ =\ -x\ -\ 8]


Multiply both sides by -1


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{4\ -\ 3x}\ =\ x\ +\ 8]


Square both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\ -\ 3x\ =\ x^2\ +\ 16x\ +\ 64]


Put it in standard form by collecting like terms on the left:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ x^2\ +\ 19x\ +\ 60\ =\ 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 4)(x\ +\ 15)\ =\ 0]


Hence, *[tex \LARGE x\ =\ -4] or *[tex \LARGE x\ =\ -15]


Check the answers:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -4\ -\ \sqrt{4\ -\ 3(-4)}\ =\ -8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{16)}\ =\ 4]:  TRUE


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -15\ -\ \sqrt{4\ -\ 3(-15)}\ =\ -8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{49}\ =\ 7\ \neq\ -7]:  FALSE. Exclude the extraneous root introduced by squaring the original equation.


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