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Math Help - Is solution ok?

  1. #1
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    Is solution ok?

    I have to solve this irrational inequality:
    \sqrt {25 - x^2 }  + \sqrt {x^2  + 7x}  > 3

    Is this solution?
     \sqrt {25 - x^2 }  > 0 \wedge \sqrt {x^2  + 7x}  > 3
     \sqrt {25 - x^2 }  > 1 \wedge \sqrt {x^2  + 7x}  > 2
     \sqrt {25 - x^2 }  > 2 \wedge \sqrt {x^2  + 7x}  > 1
     \sqrt {25 - x^2 }  > 3 \wedge \sqrt {x^2  + 7x}  > 0
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  2. #2
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    Quote Originally Posted by DenMac21
    I have to solve this irrational inequality:
    \sqrt {25 - x^2 }  + \sqrt {x^2  + 7x}  > 3

    Is this solution?
    Hello,

    the bad news first: The right solution is: x \in \{0 \leq x \leq 5\}

    The first root is defined for x \in \{-5 \leq x \leq 5\}

    the second root is defined for x \in \{x \leq -7 \vee x \geq 0\}

    that means the LHS of your inequality is defined for x \in \{0 \leq x \leq 5\}

    For all x from 0 to 4 the first root is greater than 3, that means the inequality is true for x \in \{0 \leq x \leq 4\}

    For all x greater than 4 the second root alone has a value greater than \sqrt{44} \approx 6.6

    Therefore the inequality is true for all x \in \{0 \leq x \leq 5\}

    Best wishes

    Bye
    Last edited by earboth; February 16th 2006 at 07:58 AM.
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