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Math Help - The quadratic Problem

  1. #1
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    The quadratic Problem

    I am not able to solve this problem

    if a^2+b^2=1 then find the range of a+b

    what I am doing is
    (a+b)^2>0
    a^2+b^2+2ab>0 where a^2+b^2=1
    1+2ab>0
    2ab>-1

    Case 2
    (a-b)^2>0
    a^2+b^2-2ab>0 where a^2+b^2=1
    1-2ab>0
    2ab<1

    -1<2ab<1

    a+b=\sqrt{a^2+b^2+2ab}
    a+b=\sqrt{1+2ab}
    a+b=\sqrt{1-1} -1<2ab<1 to find minimum
    a+b=0


    a+b=\sqrt{1+1} -1<2ab<1 to find maximum
    a+b=\sqrt{2}


    The range I am getting is (0,\sqrt{2}) but the answer is ( (-\sqrt{2},\sqrt{2})
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  2. #2
    MHF Contributor

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    Re: The quadratic Problem

    I would do this a completely different way. If we use x and y rather than a and b (it is still exactly the same problem, of course), we can think of "a^2+ b^2= 1" (x^2+ y^2= 1) as the equation of a circle with center at (0, 0) and radius 1. "a+ b= c" or x+ y= c, for c a fixed number, is a line sloping down to the left at a 45 degree angle. The smallest and largest possible values of c are when that line is tangent to the circle which happen at the endpoints of the diameter perpendicular to the line: that is the line y= x which crosses the circle x^2+ y^2= 1 when x^2+ x^2= 2x^2= 1 so that x= y= \pm\sqrt{2}/2. Taking x= a= b= \sqrt{2}/2, a+ b= \sqrt{2}/2+\sqrt{2}/2= \sqrt{2}. But taking x= a= b= -\sqrt{2}/2 we get a+ b= -\sqrt{2}.

    Perhaps you forgot the \pm on the square root?
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