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Math Help - Irrational euqations!!!

  1. #1
    map
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    Irrational euqations!!!

    I tried to solve those euqations but made errors. Could you please solve those irrational euqations!

    ____ __
    √4x+9 - 2√x = 1
    ____ ____
    √3x+4+ √x-3 = 5


    I tried to write that 4x+9 all together is under √ sign not only 4x and same for 3x+4. I hope you understand what I mean!
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  2. #2
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    Quote Originally Posted by map View Post
    I tried to solve those euqations but made errors. Could you please solve those irrational euqations!

    _____ __
    √4x+9 - 2√x = 1
    ____ ___
    √3x+4+ √x-3 = 5
    I have no clue what that is. I will assume it is as follows:

    sqrt(4x + 9) - 2*sqrt(x) = 1 and

    sqrt(3x + 4) + sqrt(x-3) = 5 ?

    Or is it all under the sqrt? Write it the way I have written it, or use LaTeX.
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  3. #3
    map
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    4x + 9 are under sqrt, not only 4x

    3x + 4 are under sqrt, not only 3x
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  4. #4
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    Quote Originally Posted by map View Post
    ____ __
    √4x+9 - 2√x = 1
    \sqrt{4x + 9} - 2\sqrt{x} = 1

    \sqrt{4x + 9} = 2\sqrt{x} + 1 <-- Square both sides:

    4x + 9 = (2\sqrt{x} + 1)^2

    4x + 9 = 4x + 4\sqrt{x} + 1

    8 = 4\sqrt{x}

    \sqrt{x} = 2 <-- Square both sides again:

    x = 4

    Now, whenever we square both sides of the equation we should ALWAYS check the solution in the original equation. (Squaring both sides of an equation can add extra solutions.) As it turns out, x = 4 does work.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by map View Post
    ____ ____
    √3x+4+ √x-3 = 5
    \sqrt{3x + 4} + \sqrt{x - 3} = 5

    \sqrt{3x + 4} = 5 - \sqrt{x - 3} <-- Square both sides:

    3x + 4 = (5 - \sqrt{x - 3} )^2

    3x + 4 = 25 - 10\sqrt{x - 3} + (x - 3)

    2x - 18 = -10\sqrt{x - 3} <-- Square both sides again:

    (2x - 18)^2 = 100(x - 3)

    4x^2 - 72x + 324 = 100x - 300

    4x^2 - 172x + 624 = 0

    4(x^2 - 43x + 156) = 0

    4(x - 4)(x - 39) = 0

    Thus x = 4 and x = 39.

    Again, we must check these solutions in the original equation. It turns out that only x = 4 will work.

    Note the process that was followed in both of these: isolate one of the square roots on one side of the equation, then square both sides. Then isolate the remaining square root and square both sides again. If you have more than two radical expressions you can repeat this process as many times as necessary.

    -Dan
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  6. #6
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    Quote Originally Posted by map View Post
    4x + 9 are under sqrt, not only 4x

    3x + 4 are under sqrt, not only 3x
    Okay;

    The first was already answered;

    The second:

    sqrt(3x + 4) + sqrt(x - 3) = 5

    5 - sqrt(x - 3) = sqrt(3x + 4)

    Square each side;

    (5 - sqrt(x - 3))^2 = 3x + 4

    (5 - sqrt(x - 3))*(5 - sqrt(x - 3)) = 3x + 4

    25 - 5sqrt(x - 3) - 5sqrt(x - 3) + sqrt(x - 3)^2 = 3x + 4

    25 - 10sqrt(x - 3) + x - 3 = 3x + 4

    -10sqrt(x - 3) = 2x - 18

    Square both sides;

    100*(x - 3) = (2x - 18)^2

    100x - 300 = 4x^2 -72x +324

    4x^2 -172x + 624 = 0

    4(x - 39)*(x - 4) = 0

    x = 39 or x = 4

    x = 39 does not work, therefore throw that out; check to see that x = 4 works.

    It does.

    Final answer: x = 4
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  7. #7
    map
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    Thanks for answers but I don't wan't to bother you but I rechecked answers and in second example I couldn't get asnwer 4 only 39 and 2... I used this formula D=b*b-4*a*c and then x=-bą√D/2*a

    Where is my mistake?
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  8. #8
    map
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    sorry my mistake you're right!
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  9. #9
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    Quote Originally Posted by map View Post
    Thanks for answers but I don't wan't to bother you but I rechecked answers and in second example I couldn't get asnwer 4 only 39 and 2... I used this formula D=b*b-4*a*c and then x=-bą√D/2*a

    Where is my mistake?
    Quote Originally Posted by map View Post
    sorry my mistake you're right!
    Either way, thanks for checking. Someone has to keep us honest!

    -Dan
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  10. #10
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    OK still trying to improve my skills everything is going quiet nice although not too easy any way I tried to do few other easy looking exercises and... errors OK I won't give up anyway but could you please help me on few exercises?
    1st one:

    sqrt(3x + 4) + sqrt(x + 2) = 8

    Everything was OK but then I got error went back to recheck but didn't find anything, then rechecked next day still nothing...


    P.s. How it's possible write it like topsquark is it that LaTeX?
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    Quote Originally Posted by map View Post
    OK still trying to improve my skills everything is going quiet nice although not too easy any way I tried to do few other easy looking exercises and... errors OK I won't give up anyway but could you please help me on few exercises?
    1st one:

    sqrt(3x + 4) + sqrt(x + 2) = 8

    Everything was OK but then I got error went back to recheck but didn't find anything, then rechecked next day still nothing...


    P.s. How it's possible write it like topsquark is it that LaTeX?
    sqrt(3x + 4) + sqrt(x + 2) = 8

    sqrt(x + 2) = 8 - sqrt(3x + 4)

    Square both sides;

    x + 2 = (8 - sqrt(3x + 4))^2

    x + 2 = (8 - sqrt(3x + 4))*(8 - sqrt(3x + 4))

    x + 2 = 68 - 8sqrt(3x + 4) - 8sqrt(3x + 4) + 3x

    x + 2 = 68 - 2*8*sqrt(3x + 4) + 3x

    x + 2 = 68 - 16sqrt(3x + 4) + 3x

    x - 66 = -16sqrt(3x + 4) + 3x

    (-2x - 66)/(-16) = sqrt(3x + 4)

    Square both sides;

    (-2x - 66)^2/256 = 3x + 4

    (-2x - 66)^2 = 768x + 1024

    4x^2 + 264x + 4356 = 768x + 1024

    4x^2 - 504x + 3332 = 0

    4(x - 119)*(x - 7) = 0

    x = 7 V x = 119

    Go back to the original and check:

    sqrt(3x + 4) + sqrt(x + 2) = 8

    sqrt(3(7) + 4) + sqrt(7 + 2) = 8 ? Yes.

    sqrt(3(119) + 4) + sqrt(119 + 2) ? No.

    Thus, the final answer is x = 7.
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  12. #12
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    Hello I tried to solve those euqations and I tried to found out more easy way how to resolve them faster and without big figures. so one guy showed this methode he us. He said that many teachers won't do it in this way but it works. So I tried ...and it works .... What do you think? Is this correct way how to solve those euqations.


    sqrt(x + 2) + sqrt(x - 2)=sqrt(3x - 2)

    sqrt(x + 2) = sqrt(3x - 2) - sqrt(x - 2)

    sqrt(x + 2) = sqrt(3x - 2 - x + 2)

    sqrt(x + 2) = sqrt(2x) then square both sides.

    x + 2 = 2x

    -x = -2

    x = 2



    Check:


    sqrt(2 + 2) + sqrt(2 - 2) ? sqrt(6 - 2)

    sqrt(2 + 2 + 2 - 2) ? sqrt(4)

    sqrt(4) = sqrt(4)

    2=2

    I did it also in usual way and I got same answer,so...? What do you think?
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  13. #13
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    Quote Originally Posted by map View Post
    Hello I tried to solve those euqations and I tried to found out more easy way how to resolve them faster and without big figures. so one guy showed this methode he us. He said that many teachers won't do it in this way but it works. So I tried ...and it works .... What do you think? Is this correct way how to solve those euqations.


    sqrt(x + 2) + sqrt(x - 2)=sqrt(3x - 2)

    sqrt(x + 2) = sqrt(3x - 2) - sqrt(x - 2)

    sqrt(x + 2) = sqrt(3x - 2 - x + 2)
    OUCH!!! I think this is either a trick problem or that your friend needs to pay more attention to the details. Look at the RHS for the last two lines. For arbitrary a and b (such that a, b, and a - b are positive):
    \sqrt{a} - \sqrt{b} \neq \sqrt{a - b}

    Just pick a couple of a and b values to prove it to yourself. So either the problem was designed to give a correct result despite the error in the Mathematics, or your friend was very lucky indeed to get the correct answer using this method.

    -Dan
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    Quote Originally Posted by topsquark View Post
    OUCH!!! I think this is either a trick problem or that your friend needs to pay more attention to the details. Look at the RHS for the last two lines. For arbitrary a and b (such that a, b, and a - b are positive):
    \sqrt{a} - \sqrt{b} \neq \sqrt{a - b}

    Just pick a couple of a and b values to prove it to yourself. So either the problem was designed to give a correct result despite the error in the Mathematics, or your friend was very lucky indeed to get the correct answer using this method.

    -Dan
    I find it quite funny many students think that,
    (a+b)^n=a^n+b^n.
    The funny thing about this is that it works in a finite field of prime charachteristic.
    (a+b)^p=a^p+b^p.
    Thus, what you always do wrong is sometimes right.
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