Question 191107
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By the numbers:


<b>Step 1:</b> Variables on the left, constants on the right.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2-4x=-13 ]


<b>Step 2:</b> Divide the coefficient on the 1st order term by 2, square the result, add that result to both sides of the equation (-4 divided by 2 is -2, -2 squared is 4, so add 4 to each side):



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2-4x + 4=-13+4 = -9 ]


<b>Step 3:</b> Now that we have 'completed the square' by creating a perfect square trinomial on the left, factor it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 2)^2 = -9 ]


<b>Step 4:</b> Take the square root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 2 = \pm sqrt{-9}]


Clang! Buzz-Buzz-Buzz! Warning, Will Robinson!  We have a negative radicand.


<b>Step 5:</b>  Factor the -1 out of the radicand:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 2 = \pm sqrt{(-1)(9)}]


<b>Step 6:</b> Use the definition of the imaginary number *[tex \LARGE i],namely *[tex \LARGE i^2 = -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 2 = \pm 3i]


<b>Step 7:</b> Add 2 to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 2 \pm 3i]



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