Results 1 to 10 of 10

Math Help - Questions!

  1. #1
    Member princess_anna57's Avatar
    Joined
    Jul 2007
    Posts
    78

    Exclamation Questions!

    Solve for x:

    a) 4x^2 + 3x - 2 = 0
    b) 4x + cuberoot x - 2 = 0. (Hint: rename cuberoot x = z)
    c)[(x + 4) / (x - 2)] - [(x - 3) / (x - 4)] = 0.
    d) 3x = squareroot(3 + x)
    Last edited by princess_anna57; July 22nd 2007 at 01:38 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by princess_anna57 View Post
    a) 4x^2 + 3x - 2 = 0
    This is a quadratic. If you don't know what else to do with it, use the quadratic formula. As this doesn't factor, that's what I'm going to do.
    4x^2 + 3x - 2 = 0

    So....
    x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}

    x = \frac{-3 \pm \sqrt{9 + 32}}{8}

    x = \frac{-3 \pm \sqrt{41}}{8}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by princess_anna57 View Post
    b) 4x + cuberoot x - 2 = 0. (Hint: rename cuberoot x = z)
    4x + \sqrt[3]{x} - 2 = 0

    Why were you given this problem? It seems above the level that I thought you were at.

    Let z = \sqrt[3]{x}, then x = z^3.

    Thus
    4z^3 + z - 2 = 0

    This has no rational factors, so the only way to do this (neglecting Cardano's method, which I'll do if you really want me to) is to use a graphing calculator to estimate it, or numerically calculate it.

    I get that
    z \approx 0.689398, -0.344699 \pm 0.778751 \cdot i

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by princess_anna57 View Post
    c)[(x + 4) / (x - 2)] - [(x - 3) / (x - 4)] = 0.
    \frac{x + 4}{x - 2} - \frac{x - 3}{x - 4} = 0

    First note that the solution set cannot contain 2 or 4, since that is outside the domain of the original problem.

    Now add the fractions by getting a common denominator:
    \frac{x + 4}{x - 2} \cdot \frac{x - 4}{x - 4} - \frac{x - 3}{x - 4} \cdot \frac{x - 2}{x - 2} = 0

    \frac{(x + 4)(x - 4) - (x - 3)(x - 2)}{(x - 2)(x - 4)} = 0

    \frac{(x^2 - 16) - (x^2 - 5x + 6)}{(x - 2)(x - 4)} = 0

    \frac{5x - 22}{(x - 2)(x - 4)} = 0

    The only way the LHS can be 0 is for the numerator be 0. Thus set
    5x - 22 = 0

    x = \frac{22}{5}

    (Note that x is neither 2 or 4, so this is a valid solution.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by princess_anna57 View Post
    d) 3x = squareroot(3 + x)
    3x = \sqrt{3 + x}

    Square both sides:
    9x^2 = 3 + x

    9x^2 - x - 3 = 0

    Again this is a quadratic that doesn't factor. So use the quadratic formula. I got:
    x = \frac{1 \pm \sqrt{109}}{18}

    Now, we need to check each of these in the original equation because squaring both sides tends to add spurious solutions. In this case we may simply note that if we use the "-" sign for x we don't have a solution. (I used the decimal approximation of the two x values to check in the original equation.)

    So the solution is:
    x = \frac{1 + \sqrt{109}}{18}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member princess_anna57's Avatar
    Joined
    Jul 2007
    Posts
    78
    With b), can you show me how to do the equation if it wasn't a cuberoot ie 4x + 3 squarerootx - 2 = 0?
    Thank you so much for answering, I really appreciate it.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1
    Quote Originally Posted by princess_anna57 View Post
    With b), can you show me how to do the equation if it wasn't a cuberoot ie 4x + 3 squarerootx - 2 = 0?
    Thank you so much for answering, I really appreciate it.
    4x + 3 \sqrt{x} - 2 = 0

    Let z = \sqrt{x}, that is to say let x = z^2.

    Then
    4z^2 + 3z - 2 = 0

    This is, in fact, the equation from part a), which I already solved. So we know that
    z = \frac{-3 \pm \sqrt{41}}{8}

    So
    x = z^2 = \left ( \frac{-3 \pm \sqrt{41}}{8} \right ) ^2

    x = \frac{9 \mp 6\sqrt{41} + 41}{64} = \frac{50 \pm 6\sqrt{41}}{64} <-- We don't really care which order the + or - comes in.

    x = \frac{25 \pm 3 \sqrt{41}}{32}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member princess_anna57's Avatar
    Joined
    Jul 2007
    Posts
    78
    For d), how did you get 109 when you used the quadratic formula?
    Many thanks.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by princess_anna57 View Post
    For d), how did you get 109 when you used the quadratic formula?
    Many thanks.
    Recall the quadratic formula.

    For a quadratic equation of the form ax^2 + bx + c = 0

    the roots of the equation are given by x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}

    For 9x^2 - x - 3 = 0, a = 9, b = -1 and c = -3

    so under the square root we get: b^2 - 4ac = (-1)^2 - (4)(9)(-3) = 109
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member princess_anna57's Avatar
    Joined
    Jul 2007
    Posts
    78

    Thumbs up Thanks Jhevon

    Thanks Jhevon, I knew the formula, I just put the wrong numbers in the wrong place. Thanks again.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum