Question 200602
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If you think the solution to an equation is a particular number, then the quickest way to make sure is to substitute that number back into the equation and see if you have a true statement.


You think that one of the roots is 3:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{3}\ =^?\ \frac{4}{3 + 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{3}\ = \frac{4}{4}].


That one checks.  Now let's try 4:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{3}\ =^?\ \frac{4}{4 + 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4}{3}\ \neq \frac{4}{5}].


Oops.  I guess we have to go back to the beginning.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{3}\ = \frac{4}{x + 1}]


Step 1:  Cross multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x + 1) = 12]


Step 2:  Standard form of the quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 + x - 12 = 0]


Step 3:  Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3 \times 4 = -12] and *[tex \LARGE -3 + 4 = 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 3)(x + 4) = 0]


Step 4:  Zero Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 3 = 0\ \ \Rightarrow\ \ x = 3] or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + 4 = 0\ \ \Rightarrow\ \ x = - 4]


We have already checked 3, so lets try -4:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-4}{3}\ =^?\ \frac{4}{-4 + 1}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{-4}{3}\ = \frac{4}{-3}]. Checks.


What's the lesson?  Slow and steady, step by step.  And never stop at "I think..." You aren't done until you are sure.


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