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Why is z^2+2z-15=0 {z=3,z=5}? How are those numbers derived?

Guest Mar 26, 2017
 #1
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 z+ 2z - 15 = 0

 

You can use the quadratic formula, complete the square, or just factor it by looking at it and telling what the factors are. The last one takes the most practice, but it is the fastest once you get the hang of it.

 

On this one I would just factor it, but since I don't know how to explain my thought process very well, I will just show you how to use the quadratic formula on it.

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

x = z

a, b, and c are the constants of each term.

a = 1, b = 2, c = -15

Just plug these into the formula.

\(z = {-2 \pm \sqrt{2^2-4(1)(-15)} \over 2(1)} \\~\\ z = {-2 \pm \sqrt{64} \over 2} \\~\\ z = {-2 \pm 8 \over 2} \\~\\ z = -1 \pm 4\)

 

z = -5  or  z = 3

hectictar  Mar 26, 2017
edited by hectictar  Mar 26, 2017

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