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Math Help - How Is This Solved

  1. #1
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    How Is This Solved

    -4\sqrt{x}+4>12
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  2. #2
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    I hope I get this right:

    -4\sqrt{x}+4>12

    -4\sqrt{x}>8

     \sqrt{x}<-2

     x < (-2)^2

     x< 4
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  3. #3
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    No, that isn't correct.

    #1, when dividing by -4 the inequality has to flip.

    #2 the answer has to be a range, not just one number.

    The answer is x>4, but I don't know how to arrive at the answer after:

    \sqrt{x}<-2
    x<4 (how do we get to x>4?)
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  4. #4
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    Quote Originally Posted by jgv115 View Post

    -4\sqrt{x}>8

     \sqrt{x}>-2
    dividing by a negative number will change the inequality.

    therefore

    -4\sqrt{x}>8

     \sqrt{x}<-2
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  5. #5
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    Quote Originally Posted by jgv115 View Post
    I hope I get this right:

    -4\sqrt{x}+4>12

    -4\sqrt{x}>8

     \sqrt{x}>-2

     x > (-2)^2

     x= 4
    Wrong. If you multiply or divide by a negative number, it changes the direction of the inequality.

    First, it is important to note that x \geq 0, since you can not have the square root of a negative number.


    -4\sqrt{x} + 4 > 12

    -4\sqrt{x} > 8

    \sqrt{x} < -2.

    x < (-2)^2

    x < 4.


    Putting it all together, we have our solution set

    0 \leq x < 4.


    Notice that if you use any number less than 4, you need to use the NEGATIVE square root to make the sentence make sense...
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  6. #6
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    That isn't a correct solution set.

    Substitute some values back into the inequality. For example if x=1, the inequality isn't true.

    -4(-1)+4=8 which isn't greater than 12.

    -4(1)+4=0 which isn't greater than 12.

    The answer is x>4, but how do we arrive at that?
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  7. #7
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    Yea, i know..

    sorry about the mistake. I did it right on paper but just didn't type it right.
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  8. #8
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    -4\sqrt{x}+4>12

    -4\sqrt{x}>8

    \sqrt{x}<-2

    -1 \sqrt{x}> 2

    (-1 \sqrt{x})^2> (2)^2

    (-1)^2 \sqrt{x}^2> 2^2

    1 x > 4

    x > 4

    Okay ?
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  9. #9
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    Much better
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    Thanks.
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  11. #11
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    Quote Originally Posted by fcabanski View Post
    -4\sqrt{x}+4>12
    There is no solution since the inequality reduces to \sqrt{x} < -2 and the left hand side is always greater than or equal to zero.

    Quote Originally Posted by Bacterius View Post
    -4\sqrt{x}+4>12

    -4\sqrt{x}>8

    \sqrt{x}<-2

    -1 \sqrt{x}> 2

    (-1 \sqrt{x})^2> (2)^2

    (-1)^2 \sqrt{x}^2> 2^2

    1 x > 4

    x > 4

    Okay ?
    Not okay. For example, x = 25 lies in this 'solution'. Does it satisfy the original equation ....?
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    There is no solution since the inequality reduces to \sqrt{x} < -2 and the left hand side is always greater than or equal to zero.
    I was wondering when someone was gonna point that out.
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  13. #13
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    Quote Originally Posted by VonNemo19 View Post
    I was wondering when someone was gonna point that out.
    I was thinking that, but then I thought "every square root has two solutions, one positive and one negative".

    E.g. \sqrt{9} = 3 or \sqrt{9} = -3.

    And -3 < -2...
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  14. #14
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    Quote Originally Posted by Prove It View Post
    I was thinking that, but then I thought "every square root has two solutions, one positive and one negative".

    E.g. \sqrt{9} = 3 or \sqrt{9} = -3.

    And -3 < -2...
    No. \sqrt{9} = 3. -\sqrt{9} = -3.

    If x^2 = 9 then it's true to say that x = \pm \sqrt{9} = \pm 3. But the notation \sqrt{9} means the positive square root of 9.
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  15. #15
    Senior Member Stroodle's Avatar
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    Quote Originally Posted by fcabanski View Post
    -4\sqrt{x}+4>12
    It looks to me like there is no solution as:

    -4\sqrt{x}+4>12

    4\sqrt{x}+8<0

    There are no real values of x that satisfy this inequality.
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