Question 246960
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Your error is that you did not put the quadratic into standard form first.


Standard form is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ +\ c\ =\ 0]


But your


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


is actually in 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2\ +\ bx\ =\ -c\]


form.


Either way, your 5 in the quadratic formula has to be -5, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(4)\ \pm\ \sqrt{(4)^2\ -\ 4(1)(-5)}}{2(1)}\ =\ \frac{-(4)\ \pm\ \sqrt{36}}{2}\ =\ \frac{-4\ \pm\ 6}{2}\ =\ -2\ \pm\ 3]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -2\ +\ 3\ =\ 1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -2\ -\ 3\ =\ -5]



Check:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-5)^2\ +\ 4(-5)\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25\ -\ 20\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (1)^2\ +\ 4(1)\ =\ 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ +\ 4\ =\ 5]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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