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Math Help - Question about radical equations

  1. #1
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    Question about radical equations

    In the radical equation √-4x-4 = x, you get x=-2. (-4x-4 is the complete radicand.)

    Now, since radical equations only use principal roots, this comes out as no solution, correct? Although this confuses me, since technically √x = y


    Also, one other thing:

    16+36=c^2, so c=2√13, right? (It seemed really simple, so this answer seems a bit off.)
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  2. #2
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    The given equation does not have any real solutions.
    -2 cannot be a solution since every radical is positive and Vx= +y only.

    It it is highly important to see what values the variable can take before even attempting to solve the equation.
    In this case, for instance, x >= 0 because it is on the right side of a radical equation and -4x-4 >= 0 because it is under the sign of the radical. The second one gives x <= -1, so the two multitudes do not intersect, hence there is no real solution.

    -2 gives -2 = 2

    Now, 16+36=c^2 does equal c=2√13 and c=-2√13, unless it is known for c that it is greater than zero, like when regarding a geometry problem.
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  3. #3
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    c=+ and - 2 sqrt13

    Last edited by learnmath; April 14th 2009 at 08:48 PM.
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  4. #4
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    Talking

    Quote Originally Posted by tyler. View Post
    In the radical equation √-4x-4 = x, you get x=-2. (-4x-4 is the complete radicand.)

    Now, since radical equations only use principal roots, this comes out as no solution, correct?
    I'm not sure what you mean by this...?

    Yes, when dealing with "the" square root, you consider only the "principal" square root, which is the positive square root. But you're not dealing with taking square roots to solve; to solve the radical equation, you'll need to square.

    . . . . . \sqrt{-4x\, -\, 4}\, =\, x

    . . . . . -4x\, -\, 4\, =\, x^2

    . . . . . 0\, =\, x^2\, +\, 4x\, +\, 4

    . . . . . 0\, =\, (x\, +\, 2)^2

    So the only possible solution is x = -2 but, as you have pointed out, a positive square root cannot equal a negative value.

    Quote Originally Posted by tyler. View Post
    16+36=c^2, so c=2√13, right? (It seemed really simple, so this answer seems a bit off.)
    The solution to any "solving" problem can be checked by plugging it back into the original exercise. In your case, you would need to check that 2\sqrt{13}, when squared, returns a value of 16 + 36 = 42. If it does, then your answer is correct.

    . . . . . \left(2\sqrt{13}\right)^2\, =\, 4(13)\, =\, 42

    Note: Since, in this case, you did take a square root to solve, you then must include the possibility that the value squared was negative. So you'll need a "plus-minus" on your answer.
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