Question

Solve for s
(Question will be posted below)

Answer 1

\[\large \frac{ 3 }{ s-1 } + 1= \frac{ 12 }{ s^2 - 1 }\]

Answer 2

@mathstudent55

Answer 3

would the common denominator be
\((s-1)(s^2-1)\)

Answer 4

your common multiple of s-1 and s^2-1 is not the least common multiple but it could work

factor s^2-1 is a good start if you want to see least common multiple
and make note of the restricted v alues

Answer 5

restricted values*

Answer 6

okay
\(s^2-1s+1s-1\)
\(s(s-1) + 1(s-1)\)
\((s+1)(s-1)\)

Answer 7

right so your least common multiply of s^2-1 and s-1 is...

Answer 8

\((s+1)(s-1)\)

Answer 9

\[\frac{3}{s-1}+1=\frac{12}{s^2-1} \\ \frac{3}{s-1}+1=\frac{12}{(s-1)(s+1)} \\ \text{ \right so multiply both sides by } (s-1)(s+1) \\ \text{ keep in mind the final solution can definitely not be } 1 \text{ or } -1 \]

Answer 10

ok

Answer 11

if it is 1 or -1 then it will be an extraneous solution right?

Answer 12

right

Answer 13

ok. can you give me a moment as i solve the problem?

Answer 14

k

Answer 15

would it be \(s=5 \text{ }\text{and}\text{ } s=2\)?

Answer 16

half of your answer is right
the other half is close

Answer 17

can I see the quadratic you ended up solving (before factoring if you factored )

Answer 18

\(s^2+3s-10\)

Answer 19

and how did you factor that quadratic expression

Answer 20

\(s^2 -2s + 5s - 10\)
\(s(s-2) + 5(s-2)\)

ok. i realize my mistake
\(s = -5\)
\(s = 2\)

Answer 21

that is right

Answer 22

thank you!

i just wanted to get a tip: if i notice a quadratic expression on any of the denominators then i should factor it and go from there right?

Answer 23

yep or even a cubic or greater degree
just so you can choose the least common multiple

Answer 24

ok. i need help on another one. can you help me?

Answer 25

it's basically on factoring a quadratic.
\(6b^2 - 43b - 40\)

Answer 26

we can try to find factors of a*c that multiply to be a*c and have sum b
that is:
we can try to find factors of 6(-40) that multiply to be -240 and have sum -43

Answer 27

now I realize these numbers are a bit juicy so it might be a little intimidating

Answer 28

factors of -240
1, -240 (vice versa)
2, -120 (vice versa)
3, -80 (vice versa)
4, -60 (vice versa)
5, -48 (vice versa)

okay so i think we can use -48 and 5
\(6b^2 + 5b - 48b - 40\)
\(b(6b + 5) + -8(6b +5)\)
\((b-8)(6b+5)\)

Answer 29

wow you did good

Answer 30

thank you!

Answer 31

you thought about this before asking I guess

Answer 32

i was processing what to do while typing it out.

Answer 33

but thank you for guiding me through these two questions! 😀

Answer 34

you do have problems finding two such factors you could cheat and acquire assistance from the quadratic formula
\[6b^2-43b-40=0 \\ b=\frac{43 \pm \sqrt{43^2-4(6)(-40)}}{2(6)} =\frac{43 \pm 53}{12} =\frac{-10}{12},\frac{96}{12}=\frac{-5}{6},8 \\ b=\frac{-5}{6} , b=8 \\ 6b=-5, b=8 \\ 6b+5=0 , b-8=0 \\ \text{ the two factors of } 6b^2-43b-40 \text{ are } (6b+5) \text{ and } b-8\]
does require calculator though since no one wants to square 43 :p

Answer 35

when i do know when it will be much more convenient to use the quadratic formula over the AC method or vice-versa?

Answer 36

If the numbers get much more bigger I said use the quadratic formula
because you will certainly have a lot of numbers to go through if the numbers get bigger

Answer 37

okay thanks for the advice!

Answer 38

i do prefer AC method though when the numbers are easy to work with or small
because I don't have to go through many possibilities to find what works