Question 199094
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 10x - 39 = 0]


<b>Factoring</b>


Notice that *[tex \Large 3 \times -13 = -39] and *[tex \Large 3 + (-13) = -10], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 10x - 39 = (x + 3)(x - 13) = 0]


Now use the Zero Product Rule ( *[tex \Large ab = 0] if and only if *[tex \Large a = 0] or *[tex \Large b = 0])


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x + 3)(x - 13) = 0] means:


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


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 13)= 0 \ \ \Rightarrow\ \ x = 13]


<b>Complete the Square</b>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 10x - 39 = 0]


Add the inverse of the constant term to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 10x = 39]


Divide the coefficient on the *[tex \Large x] term by 2, square the result, and add that result to both sides (-10 divided by 2 is -5, -5 squared is 25):


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 10x + 25 = 64]


The left-hand side is now a perfect square trinomial, so factor it:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 5)^2 = 64]


Take the square root of both sides, considering both the positive and negative roots:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - 5 = \pm\sqrt{64} = \pm 8]


Add the inverse of the constant term on the left to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 5 \pm 8]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = -3]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 13]


<b>Quadratic Formula</b>


The equation


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 10x - 39 = 0]


is in the form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ax^2 + bx + c = 0]


Where *[tex \Large a = 1], *[tex \Large b = -10], and *[tex \Large c = -39]


So:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-10) \pm sqrt{(-10)^2 - 4(1)(-39)}}{2(1)} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{10 \pm sqrt{100 + 156}}{2} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{10 \pm sqrt{256}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{10 \pm 16}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = 5 \pm 8]


I sincerely hope that you were not surprised that all three methods produced the same result.


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