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Thread: Is solution ok?

  1. #1
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    Is solution ok?

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

    Is this solution?
    $\displaystyle \sqrt {25 - x^2 } > 0 \wedge \sqrt {x^2 + 7x} > 3 $
    $\displaystyle \sqrt {25 - x^2 } > 1 \wedge \sqrt {x^2 + 7x} > 2 $
    $\displaystyle \sqrt {25 - x^2 } > 2 \wedge \sqrt {x^2 + 7x} > 1 $
    $\displaystyle \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:
    $\displaystyle \sqrt {25 - x^2 } + \sqrt {x^2 + 7x} > 3 $

    Is this solution?
    Hello,

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

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

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

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

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

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

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

    Best wishes

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