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An "extraneous solution" is not a characteristic of an equation, but has to do with the methods used to solve it. Typically, if you square both sides of the equation, and solve the resulting equation, you might get additional solutions that are not part of the original equation. Just do this, and check each of the solutions, whether it satisfies the original equation. If one of them doesn't, it is an "extraneous" solution introduced by the squaring.

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Q: What is the extraneous solution to w equals sqrt 7w?

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7w=122 122/7= 17.43 w=17.43

v/7x = w/y Multiply both sides of the equation by x, to get: v/7 = xw/y Multiply both sides of this equation by y/w, to get: vy/7w = x .

you do 2 sets of parenthesis and check it. for example: w2(w squared)-7w-8 (w+1) (w-8) *if you add 1w and -8w you will get -7w, which is what they want you to get. and w & w multiply to get w2(w squared), which is also what the factoring wants. another example: 3w2 (3w squared)+2-8 (3w-4) (w+2) *same thing applies with 3w x w = 3w2, and -4 +2=2, which is the answer. use this theory in all of them, unless there is a greatest common factor (GCF).

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127w = 847w/7 = 84/7w = 12

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7w=122 122/7= 17.43 w=17.43

224

t

-63=7w

16

-4

25y2 - 49w2 = (5y)2 - (7w)2 = (5y - 7w)(5y + 7w)

4w - 2 = -7w11w - 2 = 011w = 2w = 2/11

(5y + 7w)(5y - 7w)

56w2 + 17w - 3 = 56w2 + 24w - 7w - 3 = 8w(7w + 3) - 1(7w + 3) = (7w + 3)(8w - 1)

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