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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

    l*w*3=201.6
    [tex]l=12-x[/tex]
    [tex]w=7-x[/tex]

    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
    [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

    or you could do that now by moving the 201.6 over to the right side which would give you
    [tex]3x^2-57x-50.4=0[/tex]
     
    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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