Results 1 to 5 of 5

Math Help - Please help with solving for x when a 4th degree equation is involved.

  1. #1
    Junior Member
    Joined
    Feb 2007
    Posts
    53

    Please help with solving for x when a 4th degree equation is involved.

    I have mixed ideas as to how to solve this equation:

    2x^4-8x^2+12=0

    I found something in my book that says: to solve a 4th degree equation use substitution to rewrite the equation as a quadratic equation. So you'd let u = x^2. However there is a coefficient here, so I wasn't sure how I should attempt to factor it.

    I thought about:

    (2x^2 - 6) (x^2-2) = 0 but my answers didn't check. I had (2, -2).

    So, I think I'm going to need help (once again). Please provide any formulas or rules that are applicable to equations of this type. Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,899
    Thanks
    329
    Awards
    1
    Quote Originally Posted by lilrhino View Post
    I have mixed ideas as to how to solve this equation:

    2x^4-8x^2+12=0

    I found something in my book that says: to solve a 4th degree equation use substitution to rewrite the equation as a quadratic equation. So you'd let u = x^2. However there is a coefficient here, so I wasn't sure how I should attempt to factor it.

    I thought about:

    (2x^2 - 6) (x^2-2) = 0 but my answers didn't check. I had (2, -2).

    So, I think I'm going to need help (once again). Please provide any formulas or rules that are applicable to equations of this type. Thanks in advance!
    Only even powers of x appear in the equation, so I agree with making the substitution u = x^2.
    2u^2 - 8u + 12 = 0

    The reason your factoring doesn't give you the right answers is because you factored wrong. Try this first: each term has a common factor of 2, so:
    2u^2 - 8u + 12 = 0

    2(u^2 - 4u + 6) = 0

    u^2 - 4u + 6 = 0

    It should be relatively easy to see that this doesn't factor. So use the quadratic formula:

    u = [4 (+/-) sqrt{16 - 4*1*6}]/(2*1)

    u = [4 (+/-) sqrt{-8}]/2

    u = 2 (+/-) I*sqrt{2}, where I^2 = -1

    Now, u = x^2 so

    x^2 = 2 (+/-) I*sqrt{2}

    So
    x = (+/-)sqrt{2 (+/-) I*sqrt{2}} <-- Where the (+/-) signs are not related.
    (Note that there are 4 solutions and all of them are complex.)

    This will do for an answer, but it is not simplified.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by topsquark View Post
    Only even powers of x appear in the equation, so I agree with making the substitution u = x^2.
    2u^2 - 8u + 12 = 0

    The reason your factoring doesn't give you the right answers is because you factored wrong. Try this first: each term has a common factor of 2, so:
    2u^2 - 8u + 12 = 0

    2(u^2 - 4u + 6) = 0

    u^2 - 4u + 6 = 0

    It should be relatively easy to see that this doesn't factor. So use the quadratic formula:

    u = [4 (+/-) sqrt{16 - 4*1*6}]/(2*1)

    u = [4 (+/-) sqrt{-8}]/2

    u = 2 (+/-) I*sqrt{2}, where I^2 = -1

    Now, u = x^2 so

    x^2 = 2 (+/-) I*sqrt{2}

    So
    x = (+/-)sqrt{2 (+/-) I*sqrt{2}} <-- Where the (+/-) signs are not related.
    (Note that there are 4 solutions and all of them are complex.)

    This will do for an answer, but it is not simplified.

    -Dan
    Thanks for your response Dan. Gosh, this is complicated. I think I get stuck when something doesn't factor. So, I could've used the quadratic formula since it didn't factor .

    What do you mean exactly when you say that the answer is not simplified? Would an answer such as this be acceptable on a test?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,899
    Thanks
    329
    Awards
    1
    Quote Originally Posted by lilrhino View Post
    Thanks for your response Dan. Gosh, this is complicated. I think I get stuck when something doesn't factor. So, I could've used the quadratic formula since it didn't factor .

    What do you mean exactly when you say that the answer is not simplified? Would an answer such as this be acceptable on a test?
    Probably. Let me show you how to simplify it, then you can decide for yourself if you would shoot your professor for making you simplify it.

    Each of the four solutions is a point on the Argand plane, that is to say of the form x + I*y = r*e^{I*t} = r*cos(t) + I*r*sin(t) (where I'm using t as shorthand for "theta.") We want to express 2 (+/-) I*sqrt{2} in terms of r and t, then we may easily take the square root.

    So. Let's take the "+" solution first.
    2 + I*sqrt{2} = r*cos(t) + I*r*sin(t)

    So
    2 = r*cos(t)
    sqrt{2} = r*sin(t)

    Dividing the two gives:
    [r*sin(t)]/[r*cos(t)] = sqrt{2}/2

    tan(t) = sqrt{2}/2 ==> t = 0.61548 rad <-- There is no exact solution for this.

    Thus:
    r = 2/cos(t) = 2.44949

    Thus
    2 + I*sqrt{2} = 2.44949*cos(0.61548) + I*2.44949*sin(0.61548) = 2.44949*e^{0.61548I}

    So
    x = (+/-)sqrt{2 + I*sqrt{2}} = (+/-)sqrt{2.44949*e^{0.61548I}}

    x = (+/-)sqrt{2.44949}*e^{0.61548I/2} = 1.56508*e^{0.30774I}

    x = (+/-)1.56508*e^{0.30774I} = (+/-)[1.49156 + 0.474073I] <-- Either form should be suitable.

    I'll let you do the two x solutions with the 2 - I*sqrt{2}.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by topsquark View Post
    Probably. Let me show you how to simplify it, then you can decide for yourself if you would shoot your professor for making you simplify it.

    Each of the four solutions is a point on the Argand plane, that is to say of the form x + I*y = r*e^{I*t} = r*cos(t) + I*r*sin(t) (where I'm using t as shorthand for "theta.") We want to express 2 (+/-) I*sqrt{2} in terms of r and t, then we may easily take the square root.

    So. Let's take the "+" solution first.
    2 + I*sqrt{2} = r*cos(t) + I*r*sin(t)

    So
    2 = r*cos(t)
    sqrt{2} = r*sin(t)

    Dividing the two gives:
    [r*sin(t)]/[r*cos(t)] = sqrt{2}/2

    tan(t) = sqrt{2}/2 ==> t = 0.61548 rad <-- There is no exact solution for this.

    Thus:
    r = 2/cos(t) = 2.44949

    Thus
    2 + I*sqrt{2} = 2.44949*cos(0.61548) + I*2.44949*sin(0.61548) = 2.44949*e^{0.61548I}

    So
    x = (+/-)sqrt{2 + I*sqrt{2}} = (+/-)sqrt{2.44949*e^{0.61548I}}

    x = (+/-)sqrt{2.44949}*e^{0.61548I/2} = 1.56508*e^{0.30774I}

    x = (+/-)1.56508*e^{0.30774I} = (+/-)[1.49156 + 0.474073I] <-- Either form should be suitable.

    I'll let you do the two x solutions with the 2 - I*sqrt{2}.

    -Dan
    I'll give it a shot Dan...phew this one makes me want to get out another can of Mountain Dew.

    It'll take me a while, my brain doesn't work as fast as I'd like it to. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 7th 2011, 10:08 AM
  2. Replies: 1
    Last Post: February 26th 2010, 05:30 AM
  3. solving an involved equation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 24th 2009, 11:53 PM
  4. Solving Exponential Equations (logs involved)
    Posted in the Algebra Forum
    Replies: 8
    Last Post: January 7th 2009, 08:14 AM
  5. Solving for unknown - natural log involved
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 10th 2008, 02:18 PM

Search Tags


/mathhelpforum @mathhelpforum