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Finding the minimum

  1. Jan 15, 2013 #1
    Right so, I actually know the math, my problem is that I think, I am getting confused with wording. I will explain better where I show you where my problem lies.

    Question: Show that [itex]2x^2-36x+175[/itex] mat be written in [itex]a(x-b)+c[/itex] where the values a,b,c are to be found.

    state, with reason, the least possible value of [itex]2x^2-36x+175[/itex]

    so basically I complete the square, I get [itex]2(x-9)+26[/itex] the book ans [itex]1(x-9)+13[/itex] so all they have done is not multiplied back through by 2.

    The when it comes to find the least possible value x=9 and y= 13. I did check this with differentiation and get x=9 which don't get me wrong I can see that, but what I don't understand, my books when ever completing the square always multiply through with the number factored out. This has cause some confusion on my part.
     
  2. jcsd
  3. Jan 15, 2013 #2

    Mark44

    Staff: Mentor

    That would have to be a(x - b)2 + c, which is not what you wrote. You are consistently omitting this exponent, so it doesn't seem to be merely a typo.
    Both answers are incorrect, which can be seen if you expand each expression. Neither one expands to 2x2 - 36x + 175.

     
  4. Jan 15, 2013 #3
    Nope it was a typo, I was doing quickly on a break, did not have a lot of time, so rushed through, still no excuses; apologise on my behalf . Here is how I worked it out: [itex] 2x^2-36x+175 \rightarrow 2(x^2-18x+\frac{175}{2}) \rightarrow 2((x-9)^2+\frac{175}{2}-\frac{162}{2}) \rightarrow 2((x-9)^2+\frac{13}{2}) \rightarrow 2(x-9)^2+13[/itex]

    which I believe is correct??

    I did multiply back through and got the correct ans.

    I do actual think I have ans my own question in doing this, but wouldn't mind the input.

    Once again I do apologise for my pervious mistake, I do not wish to think I was wasting your time.
     
  5. Jan 15, 2013 #4

    Mark44

    Staff: Mentor

    Yes, this is correct. The only change I would make is to use = between each pair of expressions instead of an arrow.
     
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