Question 969322
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Whatever gave you the idea that *[tex \Large b\ =\ 4]?


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5(4)^2\ +\ 20(4)\ =\ 80\ +\ 80\ =\ 160\ \not =\ 0]


So 4 is most certainly not an element of the solution set to the equation


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5b^2\ +\ 20b\ =\ 0]


Divide both sides by 5


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ +\ 4b\ =\ 0]


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


So *[tex \Large b\ =\ 0] or *[tex \Large b\ =\ -4]


Check your work


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5(0)^2\ +\ 20(0)\ =\ 0\ +\ 0\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5(-4)^2\ +\ 20(-4)\ =\ 80\ -\ 80\ =\ 0]


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

*[tex \Large \ \
*[tex \Large