- Saylilbaby

help pls im begging you all

- schrodinger

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- anonymous

with?

- pooja195

It helps i you actually asked your question.

- pooja195

*if

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- Saylilbaby

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- whpalmer4

@saylilbaby are you here now?

- Saylilbaby

yes @whpalmer4

- Saylilbaby

- whpalmer4

Okay, let's look at the first one. Are you able to factor the numerator and denominator?

- whpalmer4

\[\frac{n^4-11n^2+30}{n^4-7n^2+10}\]

- whpalmer4

looking at the numerator, \(n^4-11n^2+30\)
what are two numbers which when multiplied give you \(30\), but when added give you \(-11\)? Hopefully it is obvious that both must be negative numbers.

- Saylilbaby

is it C @whpalmer4 ?

- Saylilbaby

or A im confused @whpalmer4

- whpalmer4

Let's work through the problem together, and you should be certain of your answer by the time we are done.

- Saylilbaby

i already worked thru i just needto state the restriction @whpalmer4

- whpalmer4

Ok, didn't want to assume that...but it makes life easier :-)
So, let's look at the factored denominator of the original expression:
\[n^4-7n^2+10 = (n^2-5)(n^2-2)\]Agreed?

- whpalmer4

When we find the restrictions, we have to do so on the unsimplified version, because we still have a "hole" in the function at those spots where we simplified away a term in the denominator.
That means our restrictions on \(n\) here are located at all values of \(n\) where
\[n^2-5=0\]and\[n^2-2=0\]
Any value of \(n\) that makes either of those true will cause a division by \(0\) in the original expression.

- whpalmer4

We can see that A is not the right answer, because the restrictions it imposes are \(n\ne5\) and \(n\ne2\), but those values do not cause the original denominator to be equal to \(0\).
\[(5^2-5)(5^2-2) = 460\]\[(2^2-5)(2^2-2) = -2\]

- whpalmer4

D is also not a correct answer for the same reason.
\[(5^2-5)(5^2-2) = 460\]\[(-2)^2-5)((-2)^2-2) = -2\]

- whpalmer4

B has the wrong fraction, so our hopes are left with C. Let's check it!
\[n\ne \pm \sqrt{5}\]\[n\ne\pm\sqrt{2}\]
If we evaluate the original denominator at \(n=\sqrt{5}\)
\[(\sqrt{5}^2-5)(\sqrt{5}^2-2) = (5-5)(5-2)= 0 \]
If we evaluate the original denominator at \(n=-\sqrt{5}\)
\[((-\sqrt{5})^2-5)((-\sqrt{5})^2-2) = (5-5)(5-2)=0 \]
So far, so good.
Now trying \(n=\pm\sqrt{2}\)
\[(\sqrt{2}^2-5)(\sqrt{2}^2-2) = (2-5)(2-2) = 0\]
\[((-\sqrt{2})^2-5)((-\sqrt{2})^-2) = (2-5)(2-2) = 0\]
Those are all good, so C is our answer.

- whpalmer4

To find the restrictions, factor the initial denominator. Set each product term \(=0\) and find all the values which satisfy that. The list of all of such values for all of the terms is the set of restrictions on the expression.

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