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Math Help - What do I do wrong when solving square inequality?

  1. #1
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    What do I do wrong when solving square inequality?

    I have this enequality:

    sqr(x+5)-sqr(8-x)<=1

    1) I find the area of determination (sorry if I misuse math slang),
    because the expression under roots cannot be negative

    x+5>=0 x>=-5 and
    8-x>=0 x<=8

    which means -5<=x<=8

    2) Now I work with my inequation

    sqr(x+5)-sqr(8-x)<=1
    sqr(x+5)<=1+sqr(8-x)

    Now because both left and right sides of the inequality ARE positive
    (because squre roots are positive) I can make them both in 2nd power and the truthfulness of the statement won't change.

    (sqr(x+5))^2<=(1+sqr(8-x))^2
    x+5<=1+2*sqr(8-x)+8-x
    x+5-1-8+x<=2*sqr(8-x)
    2x-4<=2*sqr(8-x)
    x-2<=sqr(8-x)
    (x-2)^2<=(sqr(8-x))^2
    x^2-4x+4<=8-x
    x^2-3x-4<=0

    I solve this enequality and find area of x-s which lie into the determination area (however this is called
    -1<=x<=4

    which I thought was the answer two the exersice,but
    the correct answer (thank's God we have answers in the math book)
    is -5<=x<=4

    What do I do wrong?

    Thanks
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  2. #2
    Senior Member OReilly's Avatar
    Joined
    Mar 2006
    Posts
    340
    Quote Originally Posted by Val21 View Post
    I have this enequality:

    sqr(x+5)-sqr(8-x)<=1

    1) I find the area of determination (sorry if I misuse math slang),
    because the expression under roots cannot be negative

    x+5>=0 x>=-5 and
    8-x>=0 x<=8

    which means -5<=x<=8

    2) Now I work with my inequation

    sqr(x+5)-sqr(8-x)<=1
    sqr(x+5)<=1+sqr(8-x)

    Now because both left and right sides of the inequality ARE positive
    (because squre roots are positive) I can make them both in 2nd power and the truthfulness of the statement won't change.

    (sqr(x+5))^2<=(1+sqr(8-x))^2
    x+5<=1+2*sqr(8-x)+8-x
    x+5-1-8+x<=2*sqr(8-x)
    2x-4<=2*sqr(8-x)
    x-2<=sqr(8-x)
    (x-2)^2<=(sqr(8-x))^2
    x^2-4x+4<=8-x
    x^2-3x-4<=0

    I solve this enequality and find area of x-s which lie into the determination area (however this is called
    -1<=x<=4

    which I thought was the answer two the exersice,but
    the correct answer (thank's God we have answers in the math book)
    is -5<=x<=4

    What do I do wrong?

    Thanks
    When you did 2nd power of
    \sqrt {8 - x}  \ge x - 2 you forgot to do one other thing that is needed for all solutions of this irrational inequation.

    These types of irrational inequations are solved like this:
    \begin{array}{l}<br />
 \sqrt {a(x)}  \ge b(x) \\ <br />
 \left( {a(x) \ge b^2 (x) \wedge b(x) \ge 0} \right) \vee \left( {a(x) \ge 0 \wedge b(x) \le 0} \right) \\ <br />
 \end{array}<br />

    You have done one condition for solution:
    \sqrt {8 - x}  \ge x - 2 \Leftrightarrow 8 - x \ge (x - 2)^2  \wedge x - 2 \ge 0

    But there is also second:

    \begin{array}{l}<br />
 \sqrt {8 - x}  \ge x - 2 \Leftrightarrow 8 - x \ge 0 \wedge x - 2 \le 0 \\ <br />
 x \le 8 \wedge x \le   2 \\ <br />
 \end{array}<br />

    So, all x \le   2 are also solutions.

    Of course, you have condition from original inequation that x \ge  - 5 so solutions are indeed  - 5 \le x \le 4
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  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Thanks
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    Hello, Val21!

    Is a puzzlement . . .


    I have this enequality: . \sqrt{x+5} - \sqrt{8-x} \leq 1

    I solved this inequality: . -1 \leq x \leq 4

    But the correct answer is: . -5 \leq x \leq 4

    What do I do wrong?

    I sketched a graph of the situation. .The domain is [-5,\,8].

    We have the curve: . y_1 \:=\:\sqrt{x + 5}
    This is the upper half of a right-opening parabola with vertex (-5, 0).

    We have the curve: . y_2 \:=\:\sqrt{8 - x}
    This is the upper half of a left-opening parabola with vertex (8, 0).

    We want the x-values where the difference y_1 - y_2 \:\leq \:1
    Code:
                  |
          *..     |              *
          ::::::* |       .*
          ::::::::|::::*::::
          ::::::::*        *
          ::::*   |           *
          ::*     |             *
          ::      |
        --*-------+--------+-----*--
         -5       |        4     8

    And the solution is indeed: .  -5 \leq x \leq 4

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