Question 250260
<font face="Garamond" size="+2">


Let *[tex \Large x] represent the smaller number.  Then the larger number is *[tex \Large x\ +\ 3]


The square of the larger number:  *[tex \Large x^2\ +\ 6x\ + 9]


Ten times the smaller number: *[tex \Large 10x].  Nine more than that is *[tex \Large 10x\ +\ 9]


Equate the two expressions (the sentence has the verb "is" in it)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ + 9\ =\ 10x\ +\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ =\ 10x]


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


Solve for *[tex \LARGE x].  Remember, this is a quadratic, so there will be two roots to the equation.  Check them against the parameters of the problem to determine if none, one, or both of the roots are solutions to the problem.


========================


The area of a rectangle is the length times the width.  The perimeter of a rectangle is two times the width plus two times the length.


Let *[tex \Large w] represent the width.  Then the length is *[tex \Large w\ +\ 3]


The area function in terms of the width is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ w^2\ +\ 3w]


The perimeter function in terms of the width is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(w)\ =\ 2(w\ +\ 3)\ +\ 2w\ =\ 4w\ +\ 6]


Twice the perimeter is then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2P(w)\ =\ 8w\ +\ 12]


But this is 6 more than the Area, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(w)\ =\ 2P(w)\ -\ 6]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ 3w\ =\ 8w\ +\ 12\ -\ 6\ =\ 8w\ +\ 6]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 5w\ -\ 6\ =\ 0]


Factor and solve for *[tex \LARGE w].  Discard the negative root -- width cannot be a negative number.  The positive root is the width.  Three more than that is the length.



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