Math Help

[Meness]

Hey guys,

I haven't posted in a while cause I was busy getting ready for school and all that. And well now I'm here in Canada all the way from Malaysia. Anyway, I need some help with a rationalizing question. I don't know if I'm doing anything wrong or this is all that I can do. I feel dumb as I didn't learn this in high school (or home school lol), and the first two lectures were supposedly just reviews uch:

Spoiler: Image

MANY THANKS GUISE

 

[coyotecraft]

On the first line you only need to multiply the conjugate of the square root; you don't need the 3+ in front.
no wait, you do.
Honestly, I don't remember doing this at all either in highschool so I looked up an example here:
http://www.wtamu.edu/academic/anns/mps/ ... nalize.htm

 

[coyotecraft]

I've followed your math up to the 3rd step. But I have no idea what you did with the denominator in the 4th step.
Can you combine X√(x+7) and -2√(x+7) as like terms? (x-2)√(x+7)

 

[dadevvtsvre]

I just checked it over and everything looks fine. I can't really see anywhere you might have made a mistake. Is that not the answer?

 

[Meness]

For the 4th step, I applied a factoring thing I learned from quadratic equations/expressions.
e.g in the equation x^2 – 3x – 4 = 0, after you find the correct factors and split 3x into two terms, it becomes:
x^2 - 4x + 1x - 4 = 0
Now we factor common factors:
x(x-4) + 1(x-4) = 0
I don't really remember the theory behind this, but whenever you have something like z(a±b) ± y(a±b), it can be further factored into (z±y)(a±b). So :
(x+1)(x-4) = 0

This can be proven to be correct by going backwards:
(x+1)(x-4) = 0
x^2 - 4x + x - 4 = 0
x^2 -3x - 4 = 0

sorry I read the wrong step lol.

Basically what I did was find the common factors so I could factor them:
where in 3x - 6 the common factor is 3 so I factored to 3(x-6).
You could actually factor it any way you want. For e.g
3x+x√x+7 the common factor is x, so x(3+√x+7)
-6-2√x+7 could be factored to -2(3+√x+7)
AND SO you end up with x(3+√x+7)-2(3+√x+7) which can be further made into:
(x-2)(3+√x+7)

I'm not really sure whether that answers your question, though, sorry THANKS FOR CHECKING THIS, BTW! :heart:

 

[Meness]

Hey dev,

This is actually an assignment, so I don't know the answers. I asked someone else in the lecture hall the other day whether I need to rerationalize the denominator again and they said it doesn't matter since it cancels out, so I got a bit skeptical about my answer :| Thanks for taking the time to check through it, btw!

 

[Meness]

Just wanted to post an update for whoever is reading this or will google upon this in the future.

I went to my prof's office today, and apparently you're supposed to invoke negativity into the equation so that you may cancel out some of the terms.

From
(2 - x)/[(3 + √x+7)(x-2)]
You add a negative sign to the front of the expression so it becomes
-{[2-x]/[(3 + √x+7)(x-2)]}
Then APPARENTLY you first open up the numerator terms to affect the signs
which is -(2-x) = -2+x = x-2
Then you can cancel x+2 from both numerator and denominator
AND ONLY THEN do you affect the denominator with negativity
and so the final answer is
-[1/(3 + √x+7)]

I say apparently, because, I'm not actually sure you can do this...but since the prof himself said it's okay...I'll give him the answer he wants.
1/(3

 

[Glitchfinder]

Just wanted to post an update for whoever is reading this or will google upon this in the future.

I went to my prof's office today, and apparently you're supposed to invoke negativity into the equation so that you may cancel out some of the terms.

From
(2 - x)/[(3 + √x+7)(x-2)]
You add a negative sign to the front of the expression so it becomes
-{[2-x]/[(3 + √x+7)(x-2)]}
Then APPARENTLY you first open up the numerator terms to affect the signs
which is -(2-x) = -2+x = x-2
Then you can cancel x+2 from both numerator and denominator
AND ONLY THEN do you affect the denominator with negativity
and so the final answer is
-[1/(3 + √x+7)]

I say apparently, because, I'm not actually sure you can do this...but since the prof himself said it's okay...I'll give him the answer he wants.
1/(3

That's actually correct. Did the professor explain why? It's a matter of multiplying the equation by -1/-1, which means you're really multiplying it by 1. Since multiplying a fraction by another fraction means multiplying the numerator and denominator separately, you're easily allowed to do that.