College Algebra | MathHelp.com - Free Educational videos for Students in K-12 | Lumos Learning

College Algebra | MathHelp.com - Free Educational videos for Students in k-12


College Algebra | MathHelp.com - By MathHelp.com



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00:00 think of the one in this problem as 1/1 .
00:07 The common denominator for 41 and X is four X
00:13 . So multiply both sides of the equation by four
00:17 x . To get rid of the fractions . When
00:26 multiplying four X times X minus 2/4 before is cancel
00:32 . And we have X times parentheses X minus two
00:39 . four x times positive one is positive for x
00:44 . And when multiplying four X times 12 over X
00:48 the X's cancel . And we have 12 times four
00:52 or 48 distributing through the parentheses . We have X
01:00 squared minus two , X plus four X equals 48
01:08 . And simplifying further on the left side , we
01:11 have X squared plus two , X equals 48 .
01:19 Notice that our equation has an X squared term in
01:23 it . In this situation remember we must first set
01:28 the equation equal to zero , then factor . So
01:33 our next step is to move the 48 to the
01:36 left side of the equation By subtracting 48 from both
01:40 sides . To get x squared plus two , X
01:47 minus 48 equals zero . Now we can factor to
01:54 get X plus eight times , X minus six equals
02:01 zero . So either X plus eight equals zero ,
02:06 or x minus six equals zero , which means that
02:10 our solutions set is negative eight six . Finally make
02:17 sure that neither of our solutions will make a denominator
02:22 in the original equation equals zero , which they don't
02:27 . So we can keep both answers .
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