Algebra Equations (Square, Cube, and LCD questions)

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szcoup

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Going over a review packet handed out in class and my brain is refusing to process two problems. Any help in pointing out where I have gone wrong, or what the next step should be would be appreciated.

3x (x - 1)^(1/2) + 2(x - 1)^(3/2).......................... those grouping symbols are very important. Without those - you have a different problem
3x square root (x - 1) + 2 cube root (x -1)^2
My question here is, I know I need to get rid of the roots. With only square roots, I would normally isolate and and square. But with a cube and a square, do I cube one and square the other, or do I cube both? Absolute brain meltdown! The given answer is 1, but I can't seem to figure it out. Can someone show me the next step to jog my memory please?

Here is the other one, which seems to be quite easy, but to which I haven't been able to come up with the given anser (-3 plus/minus square root 21 all over 6)

1/x - 1/(x+1) = 3 .......................... those grouping symbols are very important. Without those - you have a different problem
LCD is (x)(x+1) so,
(x+1) - x = 3x (x+1)
1 = 3x^2 + 3x
3x^2 + 3x - 1

If this is correct, what comes next?? Hint please. If this is wrong, please indicate where I went wrong.

Summer was obviously too long! Thanks -
 
Last edited by a moderator:
szcoup said:
Going over a review packet handed out in class and my brain is refusing to process two problems. Any help in pointing out where I have gone wrong, or what the next step should be would be appreciated.

3x (x - 1)^1/2 + 2(x - 1)^3/2
3x square root (x - 1) + 2 cube root (x -1)^2
My question here is, I know I need to get rid of the roots. With only square roots, I would normally isolate and and square. But with a cube and a square, do I cube one and square the other, or do I cube both? Absolute brain meltdown! The given answer is 1, but I can't seem to figure it out. Can someone show me the next step to jog my memory please?

Here is the other one, which seems to be quite easy, but to which I haven't been able to come up with the given anser (-3 plus/minus square root 21 all over 6)

1/X - 1/x+1 = 3
LCD is (x)(x+1) so,
(x+1) - x = 3x (x+1)
1 = 3x^2 + 3x
3x^2 + 3x - 1

If this is correct, what comes next?? Hint please. If this is wrong, please indicate where I went wrong.

Summer was obviously too long! Thanks -
Click to expand...

For the second problem, you did not write an equation in the final line. It should be

3x^2 + 3x - 1 = 0

Use the quadratic formula to solve. You will get the stated answer.

In the first problem, is there an equation in there somewhere that you are not sharing with us???
 
szcoup said:
3x (x - 1)^(1/2) + 2(x - 1)^(3/2)
Click to expand...

That is not an equation because there is no equals sign.

Whatever the instructions are, did you consider factoring out (x - 1)^(1/2) ?

Note the important grouping symbols that I placed around the exponents. Without them, your texting is not correct.



1/X - 1/x+1 = 3

...

3x^2 + 3x - 1
Click to expand...

Good up to this point, but what happened to your equation?

To find the roots of this quadratic polynomial, you may Complete the Square or use the Quadratic Formula.

We know that this polynomial does not factor nicely (to be able to apply the Zero-Product Property) because the Discriminant (which equals 21) is not a perfect square.

By the way, you did not place grouping symbols around the denominator in 1/(x - 1).

Without those grouping symbols, your texting means this:

\(\displaystyle \frac{1}{x} - \frac{1}{x} + 1 = 3\)

Cheers :cool:

PS: Don't interchange symbols x and X, either. In algebra, these two symbols are not equivalent.
 
Clarification

Thanks - The other questions involved factoring, and I totally blanked out the quadratic equation. So,

3x^2 + 3x -1 = 0
a = 3, b = 3, c = -1
Quadratic Equation:
x = -b plus/minus square root b^2 - 4ac over 2a
x = -3 plus/minus square root 9 - 4(3)(-1) over 2(3)
x = -3 plus/minus square root 21 over six

Thanks for the push in the right direction. A duh moment!

As for the other problem, maybe I didn't describe it correctly. I have attached a hand written equation. I guess I am not clear as to what comes next. I remember what to do with just squaring, but this is throwing me. Can you guide me in the right direction? Or is my line two incorrect? (I left off = 0 on the thumbnail, sorry).

Thanks
 

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szcoup said:
Thanks - The other questions involved factoring, and I totally blanked out the quadratic equation. So,

3x^2 + 3x -1 = 0
a = 3, b = 3, c = -1
Quadratic Equation:
x = -b plus/minus square root b^2 - 4ac over 2a
x = -3 plus/minus square root 9 - 4(3)(-1) over 2(3)
x = -3 plus/minus square root 21 over six

Thanks for the push in the right direction. A duh moment!

As for the other problem, maybe I didn't describe it correctly. I have attached a hand written equation. I guess I am not clear as to what comes next. I remember what to do with just squaring, but this is throwing me. Can you guide me in the right direction? Or is my line two incorrect? (I left off = 0 on the thumbnail, sorry).

Thanks
Click to expand...

It is incorrect - because

(x-1)3/2 = (x-1)1+1/2 = (x-1) * (x-1)1/2
 
Thanks

Thank you for showing me both ways. By the way, sorry about leaving off the "=0" from the equation - I edited and made a note of that on my earlier response. Bad habit, I know! I followed the logic and process in the responses, but I was surprised at how to come to the answer of one - not coming to a straight answer numerically. I will be discussing this one with my teacher in class during review to make sure I fully understand. I was orginally confused over whether to cube or square and I now see the logic of cubing. I had not done any peviously the way mmm suggested. I am going to work it through both ways to try and "get" it. Thank you for the many explanations!
 
One More Time

Okay - I've reviewed the other posts, and I am going to take another shot at this. I really want to understand this. I've come up with the right answer, but I want to check that I have in fact gotten it correctly. If not, I will get some help in class.

OOPS that should be 2(SQRT 13) / 13 on the left - the one on my thirteen blended in on the step before. Mea culpa.
 

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szcoup said:
I edited and made a note of that on my earlier response. Bad habit, I know!
Click to expand...

OIC. Yes -- I was in the composition field, when you made that edit (excuse'm whaa).

I have a bad habit, too. I begin drafting a post, then go do something for a few minutes. Type some more, take care of something else. A lot can happen, during 20 minutes.
 
szcoup said:
OOPS that should be 2(SQRT 13) / 13 on the left - the one on my thirteen blended in on the step before.
Click to expand...

Huh? Did you intend to type "right" ?

Are you saying that your solution is 2√(13)/13 ?

That number is not a correct solution. There are two solutions.

We may see by inspection that one of the solutions is x=1 simply because the expression x-1 equals zero when x=1; hence, everything gets multiplied by zero, when x=1.

Looking at your work, I'm concerned about how you're squaring expressions.

In going from your line3 to line4, it looks like you skipped writing down the separation of radicals to each side of the equation before squaring both sides of the equation.

More importantly, however, is how you squared the binomial (2x-2). We may not simply square each term inside the parentheses, to arrive at 4x^2-4.

Are you familiar with something called the FOIL algorithm?

(a + b)^2 = a*a + a*b + b*a + b*b

Try applying this to (2x - 2)^2.

I did not look at your work, beyond line4.
 
szcoup said:
Going over a review packet handed out in class and my brain is refusing to process two problems. Any help in pointing out where I have gone wrong, or what the next step should be would be appreciated.

3x (x - 1)^(1/2) + 2(x - 1)^(3/2).......................... those grouping symbols are very important. Without those - you have a different problem
3x square root (x - 1) + 2 cube root (x -1)^2
Click to expand...
\(\displaystyle x^{3/2}\) is not "cube root x^2". That would be \(\displaystyle x^{2/3}\)

My question here is, I know I need to get rid of the roots. With only square roots, I would normally isolate and and square. But with a cube and a square, do I cube one and square the other, or do I cube both? Absolute brain meltdown! The given answer is 1, but I can't seem to figure it out. Can someone show me the next step to jog my memory please?
Click to expand...
Repeat after me:
There is no cube root! There is no cube root! There is no cube root!

Here is the other one, which seems to be quite easy, but to which I haven't been able to come up with the given anser (-3 plus/minus square root 21 all over 6)

1/x - 1/(x+1) = 3 .......................... those grouping symbols are very important. Without those - you have a different problem
LCD is (x)(x+1) so,
(x+1) - x = 3x (x+1)
1 = 3x^2 + 3x
3x^2 + 3x - 1

If this is correct, what comes next?? Hint please. If this is wrong, please indicate where I went wrong.

Summer was obviously too long! Thanks -
Click to expand...
 
I know :(

I was going over my work the next morning and saw my error on the squaring . . . thanks for catching it though. I hate it when one question just seems out to get you :D! Thanks for all the help.
 
szcoup said:
I hate it when one question just seems out to get you
Click to expand...

Tell me about it! Some exercises seem to be selectively "cursed". :lol:

Did you check your new solutions, by substituting them into the original equation and ensuring that each of them leads to a true statement?

Cheers
 
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