Question 197253
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1.) *[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2-4a+3=0]


This factors because *[tex \LARGE -3 \times -1 = 3] and *[tex \LARGE (-3) + (-1) = -4], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2 - 4a + 3 = (a - 3)(a - 1) = 0]


Use the Zero Product Rule:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a - 3 = 0 \ \ \Rightarrow\ \ a = 3]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a - 1 = 0 \ \ \Rightarrow\ \ a = 1]


2.) *[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 4x - 10] does not factor.


You can tell whether a quadratic will factor by computing the discriminant. The discriminant is the expression under the radical in the quadratic formula, namely *[tex \LARGE b^2 - 4ac].  If the result is not a perfect square, then the quadratic does not factor over the rationals.  Here you have -4 squared is 16 and 4 times 1 times -10 = -40 and 16 - (-40) is 56 which is not a perfect square.


Since the quadratic does not factor, you can either complete the square or use the much easier process of using the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-b \pm sqrt{b^2 - 4ac}}{2a} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-4) \pm sqrt{(-4)^2 - 4(1)(-10)}}{2(1)} = \frac{4 \pm \sqrt{56}}{2} = \frac{4 \pm 2\sqrt{14}}{2} = 2 \pm \sqrt {14}]


The other two do not factor either, proof of which is left as an exercise for the student.  You do them the same way -- use the quadratic formula.


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