Question 278740
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Close, but not quite.


You have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ -\ 4t\ =\ 7]


You want to add a value to both sides so that the LHS is a perfect square trinomial.  Here's where the *[tex \Large \frac{1}{2}] comes in.  Divide the coefficient on the first degree term by 2.  -4 divided by 2 is -2.  Then square this value, -2 squared is 4.  Now, add 4 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t^2\ -\ 4t\ +\ 4\ =\ 11]


Next, factor the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(t\ -\ 2\right)^2\ =\ 11]


Take the square root of both sides, remembering to consider both positive and negative roots in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ -\ 2\ =\ \pm\sqrt{11}]


Finally add 2 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ 2\ \pm\ \sqrt{11}]


Now you can use a calculator and round off, or, if you are really supposed to just make an estimate, proceed thus:


3 squared is 9, 4 squared is 16.  16 - 9 is 7 and 16 minus 11 is 5  2/7 is about .3, so root 11 is very close to 3.3.  2 plus 3.3 is 5.3 and 2 minus 3.3 is -1.3.


The calculator says:  5.3166, -1.3166



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