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Math Help - Solving this radical equation

  1. #1
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    Solving this radical equation

    How do I solve this radical equation? (the sqrt stands for square root)
    sqrt(x-3)+1=x


    sqrt(x-3)=x-1
    (sqrt(x-3))^2=(x-1)^2
    x-3=(x-1)(x-1)
    x-3=x^2-x-x-1
    x-3=x^2-2x+1
    0=x^2-3x+4

    I need an actual number so I can check the solution.
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  2. #2
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    Quote Originally Posted by thecalifornialife View Post
    How do I solve this radical equation? (the sqrt stands for square root)
    sqrt(x-3)+1=x


    sqrt(x-3)=x-1
    (sqrt(x-3))^2=(x-1)^2
    x-3=(x-1)(x-1)
    x-3=x^2-x-x-1
    x-3=x^2-2x+1
    0=x^2-3x+4

    I need an actual number so I can check the solution.
    b^2-4ac < 0 ... what's that tell you?
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  3. #3
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    Um.... i'm not sure how to connect the two equations. If I use the equation with what is left over, I get 7>0.
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  4. #4
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    Quote Originally Posted by thecalifornialife View Post
    How do I solve this radical equation? (the sqrt stands for square root)
    sqrt(x-3)+1=x


    sqrt(x-3)=x-1
    (sqrt(x-3))^2=(x-1)^2
    x-3=(x-1)(x-1)
    x-3=x^2-x-x-1
    x-3=x^2-2x+1
    \textcolor{red}{0=x^2-3x+4}
    you did everything correctly, and the last equation has no real solution.

    what does that tell you about any real solutions for the original equation?
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  5. #5
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    However, if I plug the equation to the quadratic formula, I get 1 /2(3+sqrt(7i))

    The solver on quickmath.com says the I is outside the square root. I dont know how they got that. But.... is that the solution?
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  6. #6
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    Quote Originally Posted by thecalifornialife View Post
    However, if I plug the equation to the quadratic formula, I get 1 /2(3+sqrt(7i))

    The solver on quickmath.com says the I is outside the square root. I dont know how they got that. But.... is that the solution?
    The equation has no real solutions, only imaginary (complex) ones. If you're looking for complex solutions, please specify.
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  7. #7
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    Is the imaginary solution I get incorrect? I would hate to teach my girlfriend this for her test and she gets it wrong.
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  8. #8
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    Quote Originally Posted by thecalifornialife View Post
    Is the imaginary solution I get incorrect? I would hate to teach my girlfriend this for her test and she gets it wrong.
    Okay, so

    x^2-3x+4=0

    x=\frac{3\pm\sqrt{9-16}}{2}

    x = \frac{3\pm\sqrt{-7}}{2}

    x = \frac{3\pm i\sqrt{7}}{2}

    The i should indeed be outside the radical.
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