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Math Help - Solve for X

  1. #1
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    Solve for X

    How do I solve for x here? I know I need to move things over to one side but I don't know how to handle the x:

    a-.5bx-.5bx = c -.5dx -.5dx

    TIA.
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  2. #2
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    Yep, move all to the same side then where

    0=ax^2+bx+c \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}
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  3. #3
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    Thanks, pickslides. Unfortunately I need more information. This is for work and I unfortunately haven't had algebra in decades. I recognize the reference to using the quadratic equation and am guessing:

    on left side:
    a= -.5b
    b= -.5b
    c= 1

    on the right side:
    a= -.5d
    b= -.5d
    c= 1
    Would I then move one quadratic equation over to the other side by subtracting it from each so that everything equals '0'?

    or do I move everything over first and then plug them into a quadratic equation, ie:

    a-.5bx-.5bx = c -.5dx -.5dx
    a-c-.5bx-.5bx +.5dx +.5dx = 0 (/edit: +.5dx +.5dx)

    If this is the case, I'm confused what a,b,c would be with so many variables.

    As you can see, it's been a long time.

    (Btw, what program do you use to create your equations and to display as an image?)
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  4. #4
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    Quote Originally Posted by quartz View Post
    How do I solve for x here? I know I need to move things over to one side but I don't know how to handle the x:

    a-.5bx-.5bx = c -.5dx -.5dx

    TIA.
    a-0.5bx^2-.5bx = c-0.5dx^2-.5dx<br />

    a-c-0.5bx^2+0.5dx^2-.5bx+0.5dx = 0<br />

    -0.5bx^2+0.5dx^2-.5bx+0.5dx+a-c = 0<br />

    (-0.5b+0.5d)x^2+(0.5d-0.5b)x+(a-c) = 0<br />

    Now substitute the bracketed parts into the eqn in post #2.

    Quote Originally Posted by quartz View Post
    (Btw, what program do you use to create your equations and to display as an image?)
    No program needed, go to the sub-forum titled "Latex Help" read the first couple of posts. All will be clear.
    Last edited by pickslides; July 22nd 2010 at 03:47 PM. Reason: formatting
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  5. #5
    Senior Member eumyang's Avatar
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    a - 0.5bx^2 - 0.5bx = c - 0.5dx^2 - 0.5dx

    Yes, try to move everything onto one side.
    a - c - 0.5bx^2 + 0.5dx^2 - 0.5bx + 0.5dx = 0

    It's helpful to rearrange the terms so that the x's are in decreasing order -- first the x2 terms, then the x terms, then the constant terms.
    0.5dx^2 - 0.5bx^2 + 0.5dx - 0.5bx + a - c = 0
    You'll notice that I also put the terms with "d" in front of the terms with "b". I did that because I like putting positive terms first.

    Before we go on I am going to multiply both sides by 2 to get rid of the decimals:
    dx^2 - bx^2 + dx - bx + 2a - 2c = 0

    Now, factor out the x2 and the x from the terms that have them:
    (d - b)x^2 + (d - b)x + (2a - 2c) = 0

    This is where it gets ugly. If you want to use the quadratic formula
    x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}
    (I'll use capital letters here)

    ...plug in (d - b) for A, (d - b) for B, and (2a - 2c) for C:
    x = \frac{-(d - b) \pm \sqrt{(d - b)^2 - 4(2a - 2c)(d - b)}}{2(2a - 2c)}

    EDIT: Too slow!
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  6. #6
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    Quote Originally Posted by quartz View Post
    How do I solve for x here? I know I need to move things over to one side but I don't know how to handle the x:

    a-.5bx-.5bx = c -.5dx -.5dx

    TIA.
    both sides are equal, so "twice both sides" are equal

    2a-bx^2-bx=2c-dx^2-dx

    both sides are equal, subtracting them gives zero...

    2a-2c-bx^2--dx^2-bx--dx=0

    2(a-c)+(d-b)x^2+(d-b)x=0

    divide both sides by (d-b) if "d" is not equal to "b"

    x^2+x+2\frac{a-c}{d-b}=0

    x=\frac{-1\pm\sqrt{1-8\frac{a-c}{d-b}}}{2}
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  7. #7
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    Thanks so much everyone. It's really helpful seeing the steps taken. Each of them are very understandable (and actually pretty cool) altho definitely not something I could have done at this rusty stage without your help.

    Quote Originally Posted by eumyang View Post

    Now, factor out the x2 and the x from the terms that have them:
    (d - b)x^2 + (d - b)x + (2a - 2c) = 0

    This is where it gets ugly. If you want to use the quadratic formula...
    I'm curious, what would another solution from here look like without using the quadratic formula? Does it matter that X will always be >0?
    Last edited by quartz; July 22nd 2010 at 06:20 PM.
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  8. #8
    Senior Member eumyang's Avatar
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    I guess if you take everything but the last line from Archie Meade's post you could complete the square, like this:
    \begin{aligned}<br />
x^2 + x + 2\frac{a-c}{d-b} &= 0 \\<br />
x^2 + x &= -2\frac{a-c}{d-b} \\<br />
x^2 + x + \frac{1}{4} &= -2\frac{a-c}{d-b} + \frac{1}{4} \\<br />
\left( x + \frac{1}{2} \right)^2 &= \frac{-2\frac{a-c}{d-b}}{1} + \frac{1}{4} \\<br />
\left( x + \frac{1}{2} \right)^2 &= \frac{-8\frac{a-c}{d-b} +1}{4} \\<br />
x + \frac{1}{2} &= \pm \sqrt{\frac{1 - 8\frac{a-c}{d-b}}{4}} \\<br />
x &= -\frac{1}{2} \pm \frac {\sqrt{1 - 8\frac{a-c}{d-b}}}{2} \\<br />
x &= \frac {-1 \pm \sqrt{1 - 8\frac{a-c}{d-b}}}{2} \\<br />
               \end{aligned}
    I skipped a step here and there, but hopefully you'll get the gist.
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  9. #9
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    I see. It will take me some time to chew on that, lol. Thanks for taking the time to go thru all those steps.
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  10. #10
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    n/m
    Last edited by quartz; July 23rd 2010 at 08:01 AM.
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