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What is the answer here? Maths! watch

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    (Original post by CoolDude)
    Therefore, x=5 or x=0....it's a quadratic equation that touches the x-axis at two points.......so it HAS TWO ANSWERS!!!!!!!
    no because if you plug 0 back in to the equation you'll see that it's incorrect. by squaring both sides you create an artificial answer.
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    (Original post by CoolDude)
    Therefore, x=5 or x=0....it's a quadratic equation that touches the x-axis at two points.......so it HAS TWO ANSWERS!!!!!!!
    You got to put it back into the original equation.

    When X = 0
    \sqrt {x+4}+2 = x
    \sqrt {4}+2 \not= x

    When X = 5
    \sqrt {x+4}+2 = x
    \sqrt9 +2 = 5
    x = 5
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    (Original post by Procrastinate)
    You got to put it back into the original squation.

    When X = 0
    \sqrt {x+4}+2 = x
    \sqrt {4}+2 \not= x

    When X = 5
    \sqrt {x+4}+2 = x
    \sqrt9 +2 = 5
    x = 5
    ah DRAT!!!! ok I accept my mistake sorry!
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    (Original post by mf2004)
    no because if you plug 0 back in to the equation you'll see that it's incorrect. by squaring both sides you create an artificial answer.
    ah DRAT!!!! ok I accept my mistake sorry! didn't think of that!
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    (Original post by SuperiorHuman)
    Looks like you can't speak english either.
    You make it too obvious.
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    (Original post by Procrastinate)
    You got to put it back into the original equation.

    When X = 0
    \sqrt {x+4}+2 = x
    \sqrt {4}+2 \not= x

    When X = 5
    \sqrt {x+4}+2 = x
    \sqrt9 +2 = 5
    x = 5
    Only if you accept the arbitrary rule that \sqrt{4}=2 rather than \sqrt{4}=-2.

    Break free!
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    (Original post by Mbob)
    Only if you accept the arbitrary rule that \sqrt{4}=2 rather than \sqrt{4}=-2.

    Break free!
    Oh my god, novice mistake :mad: +rep lol
 
 
 
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