Course:MATH110/Archive/2010-2011/003/Math Forum/Webwork A1

WebWorks Questions

There were two WebWorks questions that I didn't know how to solve and am hoping for some answers.

The first one is : $P(x)=x^{4/3}-7x^{2/3}+6$

The second is : $x^{2}/x+13=14$ ; solve for x.

Thanks!

Re: WebWorks Questions

Hey, I am not that sure for the first question and I would appreciate anyone to give a comment on that cuz im also interested if the answer is correct!

first question

i) Solve for $x$: $x^{4/3}-x^{2/3}+6=0$

ii) Take a factor from $x^{2/3}$

iii) $x^{2/3}(x^{2}-7)+6=0$

iv) Take $x^{2/3}$ to the other side, which gives us: $x^{2}-7+6=0$

v) $x^{2}-1=0$

vi) $x=\pm 1$

By the way, this isn't a correct solution, check out these answers, they won't work. Where's the mistake? What's the good answer?
--DavidKohler

It isn't that hard to see that both $x=0$ and $x=1$ are not solutions. For $x=-1$ it seems equally wrong after all, so none of these numbers are solutions of the equation, which means something more important must be wrong.

DavidKohler

Sorry about that I thought David wrote out the steps so I assumed. Also I noticed a little mistake in step 1. Instead of $x^{4/3}-x^{2/3}+6=0$, the question really is $x^{4/3}-7x^{2/3}+6=0$. (I made a little modification to your fractions, they didn't look right, but otherwise good job! DavidKohler)

References:

second question

i) Solve for x: $x^{2}/x+13=14$

ii) Do the following: $x^{2}*x^{(}-1)+13=14$

iii) $x^{1}=-1$

iv) x = -1

Lets c if this is right!

Albert

---

Three comments:

• Please have a look at the math formatting that I edited. Mainly, remember to put brackets and not parentheses for exponent notations. You also need to add an empty line for wiki to know you mean you want to get a new line, otherwise it does weird stuff.
• Can we get the problem numbers?
• Now as far as the math goes, the only thing I'll say for now is: did you check your solutions against the original problem?

--DavidKohler

Question 17

Hi everyone! how do i submit the answer to question 17n if it has a square root on it????

Put sqrt right before the rest...For instance, sqrt(3V/h) for ${\sqrt {3V \over h}}$.

Try to use exponent notation instead of the square root. So write $x^{1/2}$ instead of ${\sqrt {x}}$.

-DavidKohler 05:58, 20 September 2010 (UTC)

Another note.....make sure you put the 1/2 in brackets so that it reads (x)^(1/2). Otherwise it will divide ${\frac {x^{1}}{2}}$. Took me a few tries to figure this one out.

TrevorShumka 06:18, 20 September 2010 (UTC)

Question 26

Hello there, I get my question in the following form and it needs be solved and expressed in interval notation form:

${6 \over x-1}-{6 \over x}\geqq 1$

I still believe the correct answer is $(-\infty ,-2]\cup [3,\infty )$ but that and a myriad of other solutions I tried were apparently all wrong. I'd be grateful for suggestions on possible ways to solve this.

I dont understand how to solve this question or answer it on the website, could anyone please help me by telling me how you go about solving this question? Thanks! JustineVallieres

Question 15

Hi guys,

I have issues with this question: Find an equation y = m x + b of the perpendicular bisector of the line segment joining the points A(8,7) and B(14,1).

I worked out that the slope of the line segment joining points A(8,7) and B(14,1)is 1 but I am unsure of how to find b. I mean I should still be able to use formula y-b=m(x-0), right? Sooo confused, grrr! Could someone help? Thanks kindly.

ArabellaCynthiaOlomide

Re: Question 15

Hi Arabella,

This is how I solved for b in question 15. There may be a simpler way but this way just makes sense to me. I think everyone is given different values for the questions though because the points I have for question 15 are A (7,6) and B (13,0).

I didn't use y-b = m(x-0). Instead I used this formula to calculate the midpoint. Note that y2+y1 is the same as y1+y0 if that is what you're used to: ((y2+y1)/(2)) , ((x2+x1)/(2))

Just plug in your x and y-values from the 2 points, A and B, given and you get the following (Although you have different coordinates so your answer will be different):

((0+6))/(2)) , ((13+7)/(2)) = (y,x) Therefore midpoint (x,y) = (10,3)

Since I now know the midpoint (x,y) = (10,3) and the slope (m) = 1 all that is left to do is plug these numbers into the equation y=mx+b and I can use this to solve for b:

3=1(10) + b Now I just isolate b and solve: b=7

Now I know the equation of the perpendicular bisector in the form y=mx+b to be y=x+7

I hope this explanation helps you. Like I said there is probably a simpler way to solve for b but I tend to use the way that makes the most sense to me. Let me know if you need me to clarify anything I wrote here! :):):) --SteffanyChwedoruk 22:03, 18 September 2010 (UTC)

Hi Steffany,

Your explanation makes totally sense! I was omitting the midpoint part!

Thank you. Arabella.

Question 7

Hello, I'm struggling on question 7 on our first webwork assignment. Here is the question:

Find the point (0,b) on the y-axis that is equidistant from the points (2,2) and (6,-5).

So what I did was that I first used the distance formula for point (0,b) with point (2,2):

${\sqrt {(2-0)^{2}+(2-b)^{2}}}$

and with point (6,-5):

${\sqrt {(6-0)^{2}+(-5-b)^{2}}}$

and then I made them equal to each other to solve for b.

This simplifies to

$2+(2-b)=6+(-5-b)$

b is then found to be

$-4.25$

But when I plugged in $-4.25$ into the formulas to double check the answer, it is obviously not correct. I do not know where I went wrong. Any ideas?

--AdamsNguyen 02:26, 18 September 2010 (UTC)

The PDF Version that I printed out says (2,2) and (6,-5) :( --AdamsNguyen 07:41, 18 September 2010 (UTC)

Hello guys,

For problem #7, I have different data: Find the point (0, b) on the y-axis that is equidistant from the points (3, 3) and (4, -3).

I tried Adam's method but my answer is not a solution!

I initially, I worked as follows though:

-I found the slope m= -3-3/ 4-3= -6/1

-Then I equated using the formula y-b=m(x-0)

for point (3, 3)

3-b=-6(3-0)

b=21

for point (4, -3)

-3-b=-6(4-0)

b=21

However, like Adam, the program is not accepting my answer. I'm not sure what I'm doing wrong. Thank you.

Arabella.

Re: Question 7

Hi everyone,

We must all be given different values for the same question.

Adam - Your math loses me during simplification but I used the distance formula to determine the distance between the point (0,b) and each given coordinate, in my case (2,2) and (4,-3) and got the correct answer. I also tried my method with your values for practice and got the correct answer so you must just be simplifying something incorrectly or plugging a value into the distance formula wrong. I also did not need to determine slope or anything else to solve it, just the distance formula (since the question asks for the point equidistant from the given coordinates). Therfore, the distance of one coodinate to (0,b) equals the distance of the other coordinate to (0,b). You will get the right answer I think if you go back through your work because I was able to solve it using this method with your values. Here is another hint: I think your mistake may have had something to do with squaring and roots at the beginning of the simplifying.

Good Luck! --SteffanyChwedoruk 03:26, 19 September 2010 (UTC)

Steffany,

I am following what you're saying about using the distance formula but when I went to solve for 'b' by putting them equal to one another, I get confused. Can you step me through your simplifying?

Thanks! LaBri Krahn --

Hi,

When solving for b, try to get rid of the square roots by squaring both sides of the equation and see if that helps you. This should make things much easier to follow.

Steffany