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Homework Help: Need help with lots of different types of problems

  1. Sep 7, 2006 #1
    I already posted a thread about a "physics" (although I am actually in basic college algebra) but I have a few more that I need help with.

    1) A jumbo chocolate bar with a rectangular shape measures 12 cm in length, 7cm in width and 3 cm in thickness. Due to escalating costs of cocoa, management decides to reduce the volume of the bar by 20%. To accomplish this reduction, management decides that the new bar should have the same 3 cm thickness but the length and width of each should be reduced an equal number of centimeters. What should the dimensions of the new candy bar be?

    ~Now I know that the formula for volume is lwh. So I took the original measurements 12 x 7 x 3 and got 252. I reduced that by 20% which would be 252-50.4= 201.6 to get the new volume of the reduced bar. Now I know that the thickness is going to stay the same so I think the new equation would be 201.6=lw x 3. I need to make sure that the length and width get reduced the same amount so I have to make sure that they stay 5 cm apart (if original l=12 and w=7 12-7=5cm) so I made the equation say 201.6= w+5 x w x 3 (x's are multiplication symbols by the way) but now what do I do..do I subtract 5 from both sides and make the w's w^2 and then divide both sides by 3 and square root the answer to get w? And then once I know that I can find l? I don't know what to do from here!

    2)Find k such that the equation x^2 - kx +4 =0

    Ok so I subtacted 4 from each side to get x^2 - kx= -4 Now I am horrible at these so I am probably way off but then I divded both sides by x....so that I had x-k = -4 over x (I don't know how to type fractions, sorry!) but I had no idea what to do from there...I am supposed to only keep a constant on the right correct? So moving x over seemed wrong to me, but with only one given number in an equation I don't know what to do.

    I guess that is it for now, I have more but maybe I can do them once I know how to do these ones. Thanks guys!
  2. jcsd
  3. Sep 7, 2006 #2
    #1 has me a bit stumped at the moment :s
    but #2 you did good so far getting [tex]x-k=\frac{-4}{x}[/tex] and now you just to have to continue solving for k by having k by itself on one side of the equation
  4. Sep 7, 2006 #3
    I just realized that I didn't type the entire problem for number 2....it should say "find k such that the equation x^2-kx+4=0 has a repeated real solution." So I should move x over in which case Iwould have -k=-4-x....which would make k=4+x and that is as far as I can go right? But what about the repeated real solution part that I am supposed to find??? Kinda confusing to me
  5. Sep 7, 2006 #4
    uh repeated real solution? never heard of that sorry. Does your textbook have any mention of repeated real solution?

    but in the solution [tex]k=4+x[/tex] you droped the /x on the 4
  6. Sep 7, 2006 #5
    In my text book all it says is "When the left side factors into two linear equations with the same solution, the quadratic equation is said to have a repeated solution." It doesn't say the word real in the definition but I am assuming they are referring to a repeated solution with real numbers....still a bit confusing to me. ohh so the solution should be k= 4/x + x , can you add that together and make it k=4/2x which would just = 2x right? so is k=2x the answer?
  7. Sep 7, 2006 #6
    ok finally i solved #1 :s so i can help you out with it:

    so you know from the question that 12 and 7 will be lowered by the same amount right? so one way to solve it is to create 3 equations to solve this problem:

    the original


    so ill let you have a go at it and see if you can solve it
  8. Sep 7, 2006 #7
    ok can you give me the next step to get started on because I am stumped...
  9. Sep 7, 2006 #8
    wait! ok so 201.6 = (12-x)(7-x)(3) is that the right direction?
  10. Sep 7, 2006 #9
    ooo i get that now, what they mean is like in the quadratic equation a (real) repeated solution would be when the determinant [tex]b^2-4ac=0[/tex] which would give a repeated solution, what they also want is this to be a real solution, no negative square roots.

    so using the original equation you just plug in for b, a, c and solve for k

    just in case you forgot the quadratic equation is in form [tex]ax^2+bx+c=0[/tex]
  11. Sep 7, 2006 #10
    so should I foil the first 2 and then multiply by 3? ugh but then how do I separate what is l and what is w?
  12. Sep 7, 2006 #11
    yes that's the right direction, but i recomend before plugging in the value's for l and w to move the 3 to the right side to make things a bit easier, and then do foil and solve for x
    Last edited: Sep 7, 2006
  13. Sep 7, 2006 #12
    so after foiling 201.6= 252-57x+3x^2

    that doesn't seem right to me... ok I am confused again
  14. Sep 7, 2006 #13
    try dividing both sides by 3, and then foiling, then you'll have a quadratic equation which you can solve using the quadratic formula

    or you could do that now by moving the 201.6 over to the right side which would give you
    Last edited: Sep 7, 2006
  15. Sep 7, 2006 #14
    so if I divide both sides by 3 I have 67.2 = (12-x)(7-x)/3 is that right so far?
  16. Sep 7, 2006 #15
    but how do I foil it if it is all over 3?? or when I divide by 3 does it just cancel out the (3) and not affect the others. Should it be 67.2=(12-x)(7-x)
  17. Sep 7, 2006 #16
    no, if you divided by 3 you'd just have 67.2=(12-x)(7-x)
  18. Sep 7, 2006 #17
    ok I thought you had to divide every single thing by 3...my bad(jeepers it has been so long since I have done this!), ok so now I have 67.2=84-19x+x^2 so now I can plug it into the equation you showed me?
  19. Sep 7, 2006 #18
    the quadratic formula I mean
  20. Sep 7, 2006 #19
    well if you had it lw3 then it would be 3l*3w, so dividing by 3 would just leave lw or (12-x)(7-x)

    edit: sorry didnt see your last posts

    kz you got everything right so far, now just move the 67.2 over to the right side so its all =0 and plug a/b/c intot he quadratic formula to solve for x
    Last edited: Sep 7, 2006
  21. Sep 7, 2006 #20
    ok I got x=12 or x=7 but those were the original 2 numbers. lol...how did i get that? I will try to show what I did. I foiled and got 67.2= 84-19x+x^2 I plugged it in and got x= -(-19) +- square root of 19^2-4(1)(84) over 2(1) (you might want to write that out to visualize it) so then I got it down to x=19+- square root of 361-336 over 2. Which = x=19 =- square root of 25 over 2, which is then x=19 +- 5 over 2. So then I did x=19-5 over 2 which is x=14 over 2 which is x=7. Then 19+5 over 2 which is x=24 over 2 which is 12....but those can't be right becuse then it would be the SAME EXACT as before I reduced it by 20%....how did that happen?
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