real or imagined solutions x-1/x+2= 2x-3/x4
- Thread starter rn50nurse
- Start date
That's not 'imagined'. It's imaginary. :roll:
That's a good one.
Please, don't spell \(\displaystyle \pi\) as 'pie'. Please don't.
Is this what you mean?:
\(\displaystyle \L\\\frac{x-1}{x+2}=\frac{2x-3}{x^{4}}\)
Please use proper grouping symbols and use ^ when you mean a power, such as x^2 (x squared).
That's a good one.
Please, don't spell \(\displaystyle \pi\) as 'pie'. Please don't.
Is this what you mean?:
\(\displaystyle \L\\\frac{x-1}{x+2}=\frac{2x-3}{x^{4}}\)
Please use proper grouping symbols and use ^ when you mean a power, such as x^2 (x squared).
sorry!
First i apologize for not being more literate in the mathematical world nor being able to write out the problems here--i have attempted to impliment the TeX but seem to not have the hang of it. With that said, i do appreciate the help you all provide. If I were more of a mathematician, I wouldn't need this site, and I am very glad to have it available.
The problem is as you wrote except the denominator of the equation is x+4. thank you for your help.
First i apologize for not being more literate in the mathematical world nor being able to write out the problems here--i have attempted to impliment the TeX but seem to not have the hang of it. With that said, i do appreciate the help you all provide. If I were more of a mathematician, I wouldn't need this site, and I am very glad to have it available.
The problem is as you wrote except the denominator of the equation is x+4. thank you for your help.
OK, gotcha. Just wanted to be sure before I done the wrong problem.
\(\displaystyle \L\\\frac{x-1}{x+2}=\frac{2x-3}{x+4}\)
You can simply cross multiply:
\(\displaystyle \L\\(x+4)(x-1)=(x+2)(2x-3)\)
\(\displaystyle \L\\x^{2}+3x-4=2x^{2}+x-6\)
Can you finish now?.
I am sorry, but I had to rib you a little about the 'imagined'. :wink:
If you want to learn more about how to type in LaTex, just click on 'quote' at the upper right corner of my post and see the code I used. I just typed it; No special software or anything.
\(\displaystyle \L\\\frac{x-1}{x+2}=\frac{2x-3}{x+4}\)
You can simply cross multiply:
\(\displaystyle \L\\(x+4)(x-1)=(x+2)(2x-3)\)
\(\displaystyle \L\\x^{2}+3x-4=2x^{2}+x-6\)
Can you finish now?.
I am sorry, but I had to rib you a little about the 'imagined'. :wink:
If you want to learn more about how to type in LaTex, just click on 'quote' at the upper right corner of my post and see the code I used. I just typed it; No special software or anything.
reply
I assume now i factor- so :
(x+2)(x-2)=(2x+2)(x-2) Is this correct so far?
I don't blame you chuckling at my misuse of the language- i am a nurse and know when I speak "medicalese" others don't have a clue what I am saying--now I know how they feel!
I assume now i factor- so :
(x+2)(x-2)=(2x+2)(x-2) Is this correct so far?
I don't blame you chuckling at my misuse of the language- i am a nurse and know when I speak "medicalese" others don't have a clue what I am saying--now I know how they feel!
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- Apr 12, 2005
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Factor? Why would you do that? Didn't you just multiply the factors to get what is there? How about if you gather everything over to one side and see if it looks any better. Then you may wish to factor something.
Note: What were you factoring? The factors are right there in the previous step. You were making it up, I think.
Note: Great try on the LaTeX. Get all your text outside the 'tex' boxes and your spacing will return to life.
Note: What were you factoring? The factors are right there in the previous step. You were making it up, I think.
Note: Great try on the LaTeX. Get all your text outside the 'tex' boxes and your spacing will return to life.
Get all the terms to one side.
You should get \(\displaystyle \L\\2x^{2}-x^{2}+x-3x-6+4\)
Now add like terms and you have a quadratic to solve.
This doesn't have integer solutions, so perhaps try the quadratic formula or completing the square.
You should get \(\displaystyle \L\\2x^{2}-x^{2}+x-3x-6+4\)
Now add like terms and you have a quadratic to solve.
This doesn't have integer solutions, so perhaps try the quadratic formula or completing the square.
Re: imaginary numbers cont...
x-1/x+2 means 2 operations: x minus 1/x plus 2;
since 1 operation is the intent: x-1 divided by x+2,
then you MUST use brackets: (x-1) / (x+2);
SO: in a division, ALWAYS show the numerator in brackets,
then the denominator in brackets; clear nuff?
So your problem properly presented is:
(x-1) / (x+2) = (2x-3) / (x+4)
Your next step is correct, but should be identified as "cross multiplication":
(x+4)(x-1)=(x+2)(2x-3)
Then the actual multiplication:
2x^2 + x - 6 = x^2 + 3x - 4
You then combined terms and got: 3x^2 + 4x - 10 = 0 : NO NO!
You added the terms; that's a serious sin, punishable only by permanent
assignment to rectal thermometers for life :wink:
2x^2 + x - 6 = x^2 + 3x - 4
Get all terms on left (see galactus' post):
2x^2 - x^2 + x - 3x - 6 + 4 = 0
Do the math:
x^2 - 2x - 2 = 0
So now you're ready to solve for x.
Coming up with "real or imaginary solutions" SIMPLY means SOLVE FOR x !
That equation cannot be "factored", so use the quadratic:
a = 1, b = -2, c = -2
If you're unfamiliar with the quadratic formula: do a google search :idea:
Good luck.
Nursey, you have to first understand the importance of BRACKETS;rn50nurse said:
x-1/x+2 means 2 operations: x minus 1/x plus 2;
since 1 operation is the intent: x-1 divided by x+2,
then you MUST use brackets: (x-1) / (x+2);
SO: in a division, ALWAYS show the numerator in brackets,
then the denominator in brackets; clear nuff?
So your problem properly presented is:
(x-1) / (x+2) = (2x-3) / (x+4)
Your next step is correct, but should be identified as "cross multiplication":
(x+4)(x-1)=(x+2)(2x-3)
Then the actual multiplication:
2x^2 + x - 6 = x^2 + 3x - 4
You then combined terms and got: 3x^2 + 4x - 10 = 0 : NO NO!
You added the terms; that's a serious sin, punishable only by permanent
assignment to rectal thermometers for life :wink:
2x^2 + x - 6 = x^2 + 3x - 4
Get all terms on left (see galactus' post):
2x^2 - x^2 + x - 3x - 6 + 4 = 0
Do the math:
x^2 - 2x - 2 = 0
So now you're ready to solve for x.
Coming up with "real or imaginary solutions" SIMPLY means SOLVE FOR x !
That equation cannot be "factored", so use the quadratic:
a = 1, b = -2, c = -2
If you're unfamiliar with the quadratic formula: do a google search :idea:
Good luck.