Question

Algebra 2 corrections. Fan and medal for help

See attachment

Answer 1

I think you are right.

Answer 2

It said incorrect :/

Answer 3

Darn. :/ Well... There goes that....

Answer 4

So I have no idea what it is

Answer 5

I suck at this stuff... I'm out of here. I'm an assassin any way...

Answer 6

Thx anyway

Answer 7

@phi @ganeshie8

Answer 8

let \(t = n^2\)

Answer 9

the rational expression becomes :
\[\large \dfrac{t^2-10t+24}{t^2-9t+18}\]
yes ?

Answer 10

can you factor numerator and denominator ?

Answer 11

ganesh showed you the "trick"
if you let t= n^2 (and so t^2 = n^2 * n^2 = n^4) you can rewrite the expression, so it looks like a quadratic.
You (hopefully) know how to factor a quadratic (assuming it factors nicely):
the top
t^2 -10 t + 24= (t + a) (t+b)

the +24 means a*b= 24 and "a" and "b" have the same sign.
the -10 means a+b= -10 and the largest of "a" and "b" is negative.
(so the two "rules" together mean both a and b are negative)
now list the factors that give 24
1,24
2,12
3, 8
4, 6
add these factors together (and make both negative because of the rules up above)
only 1 pair: -4 + -6 = -10 so -4 and -6 are what we want
(t - 4) (t-6) is how we factor t^2 -10 + 24

you can do the same for the bottom
you should get
(t-3)(t-6)

Answer 12

the simplified expression (and replace t with n^2) is
\[ \frac{n^2-4}{n^2-3} \]
but notice that the original expression would have a "divide by zero" if n^2 = 6
so we would not allow n= \( \pm \sqrt{6} \)
so to get the *exact* equivalent expression we should continue to not allow that value.
and we will still want to exclude n^2 = \( \pm \sqrt{3} \)

All of your choices show the correct "fraction"
\[ \frac{n^2-4}{n^2-3} \]
except choice B.

it appears that choice D is the correct answer *except* it has a typo!
I would ask your teacher about it, because the excluded values should look like in choice B. (and not how they show them)

Answer 13

thank you both for explaining this to me :) i appreciate it

Answer 14

thanks @ganeshie8 and @phi