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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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How do i solve for the variable 7w equals 112?
7w=122 122/7= 17.43 w=17.43
How do you rearrange v divided by 7x equals w divided y to make x the subject?
To rearrange the equation v/(7x) = w/y to solve for x, you can start by multiplying both sides by 7x to eliminate the denominator on the left side. This gives you v = 7wx/y. Next, isolate x by dividing both sides by 7w/y, resulting in x = v/(7w/y), which simplifies to x = vy/(7w).
How do you factor trinomials?
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).
