Question 985157
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Let the required integer be represented by *[tex \Large x].  10 added to the integer's square is *[tex \Large x^2\ +\ 10].  40 more than the integer is *[tex \Large x\ +\ 40], and the problem says that these two results are equal.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 10\ =\ x\ +\ 40]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ x\ -\ 30\ =\ 0]


When you solve this quadratic, you will obtain two roots.  I'll leave it up to you to decide whether both of the roots satisfy the original problem conditions.


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

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