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chelle21689's avatar

Can someone please help me with my final exam review?

Asked by chelle21689 (5128 points ) December 5th, 2010

I’m having trouble with a few questions. Answer whichever you can and please help explain so that I can undrestand.

#19. When solving equations algebraically, we have seen two types of equations that can lead to extraneous solutions. What are these two types of equations? How do you check for extraneous solutions in the two cases?

#22. a^4 – 7a^2 + 12 = 0

Solve: (3x-12)(X=5)(2x-3)>0

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9 Answers

saraaaaaa's avatar

Isn’t that cheating?

chelle21689's avatar

Why would it be cheating? It’s a study guide to help me prepare for the final exam. I’ve done all the questions but there are a few questions that I don’t understand…so to answer your question…no it’s not cheating.

Vortico's avatar

For 19, use Google to find a definition and example for extraneous solutions. #22 is made simpler when setting x = a^2.

chelle21689's avatar

How about the last one?

CyanoticWasp's avatar

For #19:
Just guessing here… but it seems that one of the ‘extraneous solutions’ that can occur is something like: x^2 + 4 = 0. It’s an impossible solution, since a square of a number will never be negative. But I don’t really know how to answer the question about “these two types of equations”. Equations that have impossible roots are unresolvable.

For #22:
a^4 – 7a^2 + 12 = 0 resolves to:
(a^2 – 4) * (a^2 – 3) = 0

So the two solutions are a = 2 or a = sqrt of 3

I can’t make sense of the second equation listed under that… it seems to be two equations mixed together somehow.

LuckyGuy's avatar

I’m guessing the (3x-12)(X=5)(2x-3)>0 Should be (3x-12)(X+5)(2x-3)>0 . Right?

chelle21689's avatar

@worriedguy. yeah you’re right..oops

roundsquare's avatar

For the last one, here is a hint.

If you multiple 3 numbers and you want the product to be positive, you need either
a) all should be positive
b) two should be negative and one positive

Also, none of the numbers should be zero.

Make a number line for x and see where that takes you…

ben's avatar

For #22, you just factor the term. Your final solution will look something like:
(a^2 + or – n) * (a^2 + or – m)

You know it’s a^2 in front because it’s often just the square root of the first term. The tricky part is figuring out n and m. The way to think about it english is “what two numbers add up to -7 and multiply to 12?” When you multiply the two terms, you can see that’s the same mathematically at that sentence. Solving that problem can be tricky, and I’ve found the easiest way is to intuit it… there’s usually only a few ways it can go. (In your head, think of which factors of twelve could add to 7): 3 and 4! And clearly they’ll both have to be negative.

so your solution is

(a^2 – 4 ) * (a^2 + – 3)

And you can verify by multiplying it out.

Makes sense? Read more on wikipedia

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