Question 524816
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Using the quadratic formula on this problem is a lot like killing a housefly with a 12 gauge shotgun, but we aim to please...


Presuming that the ground in the area of the bottom of the house wall and the bottom of the ladder is perfectly perpendicular to the wall of the house (mostly because unless this is true you cannot solve the problem with the information given), use Pythagoras.  Let *[tex \Large x] represent the distance from the ground to the top of the ladder measured along the wall.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ c^2\ -\ b^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ (20)^2\ -\ (15)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 175]


Put it into standard quadratic form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 175\ =\ 0]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 0\ \ ], and *[tex \LARGE c\ =\ -175]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-(0)\ \pm\ \sqrt{(0)^2\ -\ 4(1)(-175)}}{2(1)}]


Do the arithmetic and discard the negative root since distance is a positive number.


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