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Math Help - Radicals Problem

  1. #1
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    Radicals Problem

    x + root(x+3x+5) = 7 - 3x


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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Trentt View Post
    x + root(x+3x+5) = 7 - 3x


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    Ok, so this problem gets really messy--too messy to type some parts of it. If anyone out there has a shorter way of doing, be my guest and post it, cause this is the only way I see how. Here goes:

    x^2 + sqrt(x^2 + 3x + 5) = 7 - 3x
    => sqrt(x^2 + 3x + 5) = -x^2 - 3x + 7
    => x^2 + 3x + 5 = (-x^2 - 3x + 7)^2.......................squared both sides
    Here comes the monstrosity:
    => x^2 + 3x + 5 = x^4 + 3x - 7x^2 + 9x^2 + 3x^3 - 21x - 7x^2 - 21x + 49
    Whew! Let's simplify a bit
    x^4 + 6x^3 - 6x^2 - 45x + 44 = 0

    We see x-1 is a factor (do you know how to spot the factors?), so we do long division or synthetic division (whichever one you're better at) and divide x^4 + 6x^3 - 6x^2 - 45x + 44 by x-1 to see that

    x^4 + 6x^3 - 6x^2 - 45x + 44 = (x-1)(x^3 + 7x^2 + x - 44) = 0

    Now for x^3 + 7x^2 + x - 44 we see that x+4 is a factor. Doing long division again we see that x^3 + 7x^2 + x - 44 = (x+4)(x^2 + 3x - 11)

    So finally we have:

    (x-1)(x+4)(x^2 + 3x - 11)=0
    so x-1 = 0 => x=1
    or x+4 = 0 => x= -4
    or x^2 + 3x -11 = 0
    the latter does not foil nicely, so we can compute by the quadratic formula or completing the square to see that in this case
    x = [-3 +/- sqrt(53)]/2 .......Note the plus or minus sign, this is two answers in one. There are four values for x, as should be expected since we have an x^4 function
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    Ok, so this problem gets really messy--too messy to type some parts of it. If anyone out there has a shorter way of doing, be my guest and post it, cause this is the only way I see how. Here goes:

    x^2 + sqrt(x^2 + 3x + 5) = 7 - 3x
    => sqrt(x^2 + 3x + 5) = -x^2 - 3x + 7
    => x^2 + 3x + 5 = (-x^2 - 3x + 7)^2.......................squared both sides
    Here comes the monstrosity:
    => x^2 + 3x + 5 = x^4 + 3x - 7x^2 + 9x^2 + 3x^3 - 21x - 7x^2 - 21x + 49
    Whew! Let's simplify a bit
    x^4 + 6x^3 - 6x^2 - 45x + 44 = 0

    We see x-1 is a factor (do you know how to spot the factors?), so we do long division or synthetic division (whichever one you're better at) and divide x^4 + 6x^3 - 6x^2 - 45x + 44 by x-1 to see that

    x^4 + 6x^3 - 6x^2 - 45x + 44 = (x-1)(x^3 + 7x^2 + x - 44) = 0

    Now for x^3 + 7x^2 + x - 44 we see that x+4 is a factor. Doing long division again we see that x^3 + 7x^2 + x - 44 = (x+4)(x^2 + 3x - 11)

    So finally we have:

    (x-1)(x+4)(x^2 + 3x - 11)=0
    so x-1 = 0 => x=1
    or x+4 = 0 => x= -4
    or x^2 + 3x -11 = 0
    the latter does not foil nicely, so we can compute by the quadratic formula or completing the square to see that in this case
    x = [-3 +/- sqrt(53)]/2 .......Note the plus or minus sign, this is two answers in one. There are four values for x, as should be expected since we have an x^4 function
    The solution method is correct, but it's not quite that simple. Whenever you square both sides of an equation you might be introducing extra solutions, so you should make a habit of checking all solutions in the original equation.

    As it happens, the solution for x is only 1 and -4.

    -Dan
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  4. #4
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    Hello, Trentt!

    This problem is pretty awful . .


    . . . . .__________
    x + √x + 3x + 5 .= .7 - 3x
    . . . . . . . . .__________
    We have: .√x + 3x + 5 .= .7 - 3x - x
    . . . . . . . . . . . . . . . . __________
    Square both sides: . (√x + 3x + 5) .= .(7 - 3x - x)

    . . . . x + 3x + 5 .= .49 - 42x - 5x + 6x + x^4


    We have: .x^4 + 6x - 6x - 45x + 44 .= .0


    Using the Factor Theorem and Remainder Theorem, I found x = 1 is a root.

    . . So we have: .(x - 1)(x + 7x + x - 44) .= .0


    I found that x = -4 is a root of the cubic.

    . . Then we have: .(x - 1)(x + 4)(x + 3x - 11) .= .0

    - - - . . . . . . . . . . . . . . . . . . . . . . . . . . .__
    And the quadratic has roots: .x .= .(-3 √53)/2


    The answer are: .x .= .1, -4

    Neither of the irrational roots satisfies the equation.
    . . They are both extraneous.

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