In the equation c2 – 4 = 0, c is equal to
A. 8.
B. ±2.
C. 16.
D. ±4.
In the equation c2 – 4 = 0, c is equal to
A. 8.
B. ±2.
C. 16.
D. ±4.
1. a^{2} - b^{2} = (a - b)(a + b)
2. If x^{2} = a (a ≥ 0) then x = ± √a
~~~~~~~~
c^{2} - 4 = 0
1. c^{2} – 2^{2} = 0
(c - 2)(c + 2) = 0
Product equal "0" , if, at least, one factor equal "0"
c - 2 = 0 -----> c_{1} = 2
c + 2 = 0 -----> c_{2} = - 2
c = ± 2
Or
2. c^{2} - 4 = 0
c^{2} = 4
c = ± √4
c = ± 2
So, the answer is (B)
Amber, you have typed quite a few multiple choice problems up here. I am going to help you with a few of them but I do hope you are taking the time to understand the math so that you can do well on any tests you may have to take at school.
c2-4 = 0
Algebra is basically "undoing" an equation to find a single component (in this case the variable C). Since we are "undoing" we should be working backwards. If our order of operations is PEMDAS, then in algebra we are going to use SADMEP.
That said, let's start by trying to isolate that letter C.
c2-4 = 0
c2 - 4 + 4 = 0 + 4
c2 = 4
Notice I added 4 to each side of the equation. I can do ANYTHING I WANT mathematically speaking to one side of the equation as long as I also do it to the other side. Think of the equal sign as the pivot point of a balance and we always want to keep that balance even.
Next, in order to isolate the C we will have to remove that number 2 somehow. when there is a number next to a variable it means to multiply. Well the opposite of multiply is to divide, so if we divide by 2 on both sides of the equation let's see what we get:
C2 = 4
C2 / 2 = 4/2
C = 2
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Natalia really goes in depth with the "perfect square" aspect (and the +/- bit). My way is the shortcut way (you'll still find the correct answer), but in all honesty her way is the "mathematician" way. If you really want to understand it more in depth study how she solved it, if you are just trying to get through the course because you have to and, well, most people don't really need to do this sort of thing outside of school.
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