Question

(5/(x^3+5x^2))=(4/(x+5))+(1/x^2)
I found the extraneous solution for 0, and I need to solve for x. I want to isolate x out of the denominator, what can I multiply this with?

Answer 1

I don't know what I can multiply all of these with so that the denominator disappears

Answer 2

first of all, can x^3+5x^2 be factored?

Answer 3

yes into x^2(x+5)

Answer 4

\[\frac{5}{x^2(x+5)}=\frac{4}{x+5}+\frac{1}{x^2}\]

Second of all, we definitely no x cannot be what two numbers?
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third of all, what is the least common multiply of (x+5) and x^2?
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fourth of all, the lcm is what we will choose to multiply both sides by.

Answer 5

So second question,
we know x cannot be what two numbers?

Answer 6

0 and -5 right?

Answer 7

Right on! :)

Answer 8

Ok so if we get a solution when we solve this that is one of those numbers we will know that is a contradiction and it definitely is not a solution

Answer 9

Anyways, the third question:
what is the least common multiply (lcm) of x^2 and (x+5)

Answer 10

if you don't know, sometimes it is just best to find a common multiple
and you can do that pretty quick by multiplying the x^2 and the (x+5)
notice these were factors in the bottom in the fractions

Answer 11

the x^2 and x+5 both are x I think?

Answer 12

lcm of x^2 and (x+5) is just x^2(x+5)

Answer 13

and that is what we will multiply both sides by

Answer 14

oh ok so do I multiply that out of all the denominators?

Answer 15

\[\frac{x^2(x+5)}{1} \cdot \frac{5}{x^2(x+5)}=\frac{x^2(x+5)}{1} \cdot \frac{4}{x+5}+\frac{x^2(x+5)}{1} \cdot \frac{1}{x^2}\]
cancel away

Answer 16

i got 5=4x^2+x^2

Answer 17

the only one I'm worried about is your last term there

Answer 18

sorry I replied before looking at your image

Answer 19

its k
it looks almost good
just the last term needs to be corrected

Answer 20

sorry last one is x+5

Answer 21

messed that up

Answer 22

yep yep :)

Answer 23

so you need to solve 5=4x^2+x+5

Answer 24

Thanks ! I got 4x^2+x=0
sorry I'm kinda stuck right here.... how should I simplify this?
I know it's a ax^2+bx+c form...

Answer 25

4x^2+x=0
is not too scary
you can actually factor the 4x^2+x

Answer 26

both 4x^2 and x have what as a common factor?

Answer 27

x

Answer 28

so 4x^2+x=x times (what? )

Answer 29

4x

Answer 30

4x(x)=4x^2 right?

Answer 31

\[4x^2+x=x(\frac{4x^2}{x}+\frac{x}{x})=x(4x+1)\]

Answer 32

how come there is a +1 at the end?

Answer 33

factor an x from each term you need to divide each term by x

Answer 34

wait I figured it out sorry.

Answer 35

x times (4x+1)=4x^2+x

Answer 36

ok

Answer 37

so you know then we have 4x^2+x=0
which we rewrote as x(4x+1)=0

Answer 38

x(4x+1)=0
means we have either x=0 or 4x+1=0 or both=0

Answer 39

so what is our solution

Answer 40

so I might need a little more help on this: I know in order to find the solution I need the x isolated.

Answer 41

ya

Answer 42

x=0 (we already know this isn't a solution because remember we said x could not be 0 at the beginning)
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so we only need to solve 4x+1=0

Answer 43

oh i see
then it's x=-1/4

Answer 44

yes

Answer 45

that is correct
good job nick

Answer 46

you were able to answer a lot of questions on this problem that I gave
you did really well

Answer 47

thank you... so to sum it up, the extraneous solutions are only 0, -5 and the x is -1/4?
Thank you so so so so so much!

Answer 48

well the extraneous solutions was only x=0 (I did not include x=-5 in the extraneous solution set because we did not get -5 in the end)
but the only solutions (non-extraneous) is x=-1/4

Answer 49

but technically -5 is an extraneous solution, correct?

Answer 50

http://www.mathwords.com/e/extraneous_solution.htm
"A solution of a simplified version of an equation that does not satisfy the original equation.-extraneous solution "
This is the definition I'm using.
4x^2+x=0 was a simplified version of the original equation (the equation we started with)
this simplified equation 4x^2+x=0 didn't give us x=-5 so I do not include in the extraneous solution set
I only include the x=0 in the extraneous solution set
but the actual solution to the problem is x=-1/4

Answer 51

Oh I see what you mean thanks so much again! I wouldn't have been able to do this without you! :P

Answer 52

Np. http://hotmath.com/hotmath_help/topics/extraneous-solutions.html
Here is another example. Notice in that first problem there they do not call x=-2 an extraneous solution but they only call x=2 an extraneous solution because it works for the simplified version but not the original version.

Answer 53

Ill take a look at it. Thanks!

Answer 54

I will give you a medal for your awesome work into this problem. You did super.