Question

Simplify the rational expression. State any restrictions on the variable.

Answer 1

\[\frac{ n^4-10n^2+24 }{ n^4-9^2+18 }\]

Answer 2

First lets find the restrictions. For that we solve for n : n^4 - 9^2 + 18 = 0

Answer 3

Than what? @math&ing001

Answer 4

@Hero

Answer 5

@satellite73

Answer 6

Let \(n^2 = y\) then try simplifying the expression.

Answer 7

Please help me I know its \[\frac{ n^2-4 }{ n^2-3}\] but i dont know the other half for my answer, so it's between A-C for the answer

Answer 8

(n+2)(n−2)(n2−6)
(n2−3)(n2−6)

Answer 9

How do you know it's \(\dfrac{n^2 - 4}{n^2 - 3}\)?

Answer 10

(n+2)(n−2)(n2−6)
(n2−3)(n2−6)

Answer 11

Multiply it back out to see if what you get matches the original expression.

Answer 12

Also the original expression you posted likely has a typo in it somewhere.

Answer 13

How do i do that?

Answer 14

@timo86m

Answer 15

@Astrophysics

Answer 16

@Michele_Laino can you please help me finish my math problem?

Answer 17

@mathmate

Answer 18

I just need help with the end of the problem. I think the answer is B though

Answer 19

@EllenJaz17
Please confirm that the problem is indeed:
\(\Large \frac{ n^4-10n^2+24 }{ n^4-9\color{red}{n}^2+18 } \)

Answer 20

yes

Answer 21

Yes it is @mathmate

Answer 22

hint:
we can factorize the denominator, using the same procedure for numerator, so
we can write this:
\[{n^4} - 9{n^2} + 18 = \left( {{n^2} - 6} \right)\left( {{n^2} - 3} \right)\]

Answer 23

(n+2)(n−2)(n2−6)
(n2−3)(n2−6)

Answer 24

\[\frac{ n^2-4 }{ n^2-3 }\]

Answer 25

yes!

Answer 26

I know thats the answer to the first half but i need to find out the second half

Answer 27

the denominator is equal to zero when these two conditions hold:
\[\begin{gathered}
{n^2} - 6 = 0 \hfill \\
\hfill \\
{n^2} - 3 = 0 \hfill \\
\end{gathered} \]

Answer 28

please remember, that we can not divide by zero

Answer 29

correct

Answer 30

please solve those quadratic equations for n, what do you get?

Answer 31

6 and 3?

Answer 32

are you sure?
we have 6^2= 36, and 3^2=9

Answer 33

No im not sure im just guessing

Answer 34

hint:
the solution to this equation:
\[{n^2} - k = 0\]
are:
\[n = \sqrt k ,\quad n = - \sqrt k \]

Answer 35

solutions*

Answer 36

since:
\[{\left( {\sqrt k } \right)^2} = {\left( { - \sqrt k } \right)^2} = {n^2}\]

Answer 37

being k a positive number or k=0

Answer 38

So is it answer B?

Answer 39

no, I don't think so, sorry!

Answer 40

hint:
what are the solution of this equation:
\[{n^2} - 3 = 0\] ?

Answer 41

So that only leaves A & C correct? because D isn't \[\frac{ n^2-4 }{ n^2-3 }\]

Answer 42

solutions*

Answer 43

\[n^{2}=3?\]

Answer 44

yes! and what is n=...?

Answer 45

\[n \neq \pm \sqrt{6}\pm \sqrt{3}?\]

Answer 46

from this equation:
\[{n^2} - 3 = 0\]
I get
\[n = \pm \sqrt 3 \]
am I right?

Answer 47

yes

Answer 48

ok! so we have to exclude those 2 values, since when:
\[n = \pm \sqrt 3 \]
the denominator is zero
Now do the same with the equation:
\[{n^2} - 6 = 0\]

Answer 49

\[n=\pm \sqrt{6}\]

Answer 50

perfect, we have to exclude those values too, since when:
\[n = \pm \sqrt 6 \]
the denominator is zero.
So what is the right option?

Answer 51

A

Answer 52

that's right! :)

Answer 53

Thank you so much for teaching me how to get to the answer!

Answer 54

:)