Question 324747
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Let *[tex \Large x] represent the number.  The square of the number is *[tex \Large x^2].  Five more than that is *[tex \Large x^2\ + 5].  Four times the number is *[tex \Large 4x].  Two more than that is *[tex \Large 4x\ +\ 2].  Finally "is" means equals, so:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ +\ 3\ =\ 0]


Just factor and solve.  Remember to check both answers in the original equation.



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