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Math Help - Quadratic Problem

  1. #1
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    Quadratic Problem

    I am unable to find the answer of this question. Any help will be appreciated

    If A^2 + B^2 =1 then find the range of (a+b) where a and b is real.
    Last edited by varunkanpur; July 30th 2013 at 03:03 AM.
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  2. #2
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    Re: Quadratic Problem

    Are you aware that the graph of x^2+ y^2= 1 is a circle with center at (0, 0) and radius 1? So that the circle extends from x= -1 to x= 1 horizontally, and y= -1 to 1 vertically.
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  3. #3
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    Re: Quadratic Problem

    Quote Originally Posted by varunkanpur View Post
    I am unable to find the answer of this question. Any help will be appreciated

    If A^2 + B^2 =1 then find the range of (a+b) where a and b is real.
    Technical point: a and A are not the same variable.

    -Dan
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  4. #4
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    Re: Quadratic Problem

    Assuming you haven't done geometry of the circle you can think of it this way.

    A^2+B^2=1

    A^2=1-B^2

    A=\sqrt{1-B^2}

    Think of the maximum and minimum vales of B so that A is real. Then do the same thing for A instead of B.
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  5. #5
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    Re: Quadratic Problem

    hmm if you've done trigonometry,try this. Let A=sin x and B=cos x[Parametric equation of circle]. Now range of A+B is Range of sin x + cos x i.e [-1/sqrt(2),1/sqrt(2)].
    Last edited by smatik; August 2nd 2013 at 09:51 PM.
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  6. #6
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    Re: Quadratic Problem

    Hi,
    smatak has a good idea, but unfortunately he gave the wrong answer.

    sin(x)+cos(x)=\sqrt2sin(x+\pi/4)

    So the range of sin(x)+cos(x) is [-\sqrt2,\sqrt2]
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  7. #7
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    Re: Quadratic Problem

    oh yea.i calculated wrong.... btw its smatik
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