Question

Solve 5 over x plus 8 equals 3 over x for x and determine if the solution is extraneous or not.

x = 12, non-extraneous

x = 12, extraneous

x = -12, non-extraneous

x = -12, non-extraneous

Answer 1

@phi

Answer 2

can you use the equation editor? It is hard to read.

Answer 3

sure

Answer 4

\[\frac{5}{x+8}=\frac{3}x\]Cross multiply as a first step:
\[5*x = 3*(x+8)\]Expand using the distributive property
\[5x = 3x + 24\]Can you do it from there?

Answer 5

yes its 3

Answer 6

Uh, no. That fact that none of your answers feature the number 3 should be a clue that perhaps you didn't do something right :-)

Answer 7

but thats not an answer choice and @phi she had put the equation on here

Answer 8

Show me your work...

Answer 9

i had combine 3x and 5x got 8x=24 divide final answer 3

Answer 10

Explain just how you combined them? \[5x=3x+24\]

Answer 11

omg my bad its suppose to be 2x=24 which equals to 12 so the answer is A

Answer 12

there you go :-) except you also need to test the solution to see if it is extraneous. If you plug x=12 into the original equation, do you get a number sentence that is true, or false?

Answer 13

its false so the correct answer is B @whpalmer4

Answer 14

Whoa, hang on, how did you decide it was false? Can you show me your work again?

Answer 15

\[\frac{5}{x+8}\]If we substitute \(x=12\), what does that fraction become?

Answer 16

i replace x with 12

Answer 17

what are the two fractions you get after you replace \(x\) with \(12\)?

Answer 18

i was looking at the wrong problem

Answer 19

do you think you could help me with one more

Answer 20

One of the most valuable things learned in math class is working carefully, with attention to detail...

Sure, what's the other question?

Answer 21

What are the possible number of positive, negative, and complex zeros of f(x) = -3x4 - 5x3 - x2 - 8x + 4?

Positive: 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

Positive: 1; Negative: 3 or 1; Complex: 2 or 0

Positive: 3 or 1; Negative: 1; Complex: 2 or 0

Positive: 4 or 2 or 0; Negative: 2 or 0; Complex: 4 or 2 or 0

Answer 22

@whpalmer4

Answer 23

i know there is only one sign change

Answer 24

I get a notification when you respond on a post where I've commented, it isn't necessary to tag me again and clutter up my life with additional notifications, thanks...

Okay, you've got the polynomial written in the right order (descending powers of the exponent), and I agree, there is one sign change in \(f(x)\). That means we have 1 positive real root, or 1 negative real root or 1 complex root (you tell me which)?

Answer 25

oh well my bad i didnt know because i dont get a notification when you respond back

Answer 26

1 positive real root

Answer 27

I think the system assumes that you'll be watching your own problem intently :-) Anyhow, if I haven't responded within a few minutes, then feel free to tag me, just don't do it immediately.

Yes, 1 positive real root is correct. What's the next step in the procedure?

Answer 28

ok i got you:) and next you should try to find the negative real root right

Answer 29

Try to find the count of the possible negative real root(s), that is. What do you do to do that?

Answer 30

look at a number that is negative right

Answer 31

No, you need to count the possible negative roots. You change \(f(x)\) to \(f(-x)\) and do the sign change count again. That gives you your possible negative real roots.

Answer 32

so would it be 3

Answer 33

\[f(x) =-3x^4 - 5x^3 - x^2 - 8x + 4\]
\[f(-x) = -3x^4+5x^3-x^2+8x+4\]
3 sign changes in \(f(-x)\) so we have either 3 negative real roots or 1 negative real root and 2 complex roots.

How many roots will we have in total?

Answer 34

4

Answer 35

Right. So what are the possible scenarios for the composition of the roots of this equation, if we don't know what they actually are?

Answer 36

B right

Answer 37

No, I want you to spell out the combinations in your own words...

Answer 38

ok so there is 3 or 1 negative roots and one positive root

Answer 39

Hey, OS, don't mess with what I type :-(

The whole story:

We can have:

1 positive real root, 3 negative real roots, and 0 complex roots, OR
1 positive real root, 1 negative real root, and 2 complex roots in a conjugate pair

conjugate pair means they can be written as \(a\pm bi\) where \(i = \sqrt{-1}\)

Answer 40

we know the complex roots are in a conjugate pair because we only have real numbers as the coefficients of this polynomial. To get the complex roots to not contribute any complex coefficients, they have to get combined as the difference of squares:

\[(a+bi)(a-bi) = a^2 -abi + abi -b^2i^2 = a^2-b^2i^2=a^2-b^2(-1) = a^2+b^2\]Look ma, no \(i\)!