Results 1 to 10 of 10

Math Help - Solving a Two Radical Equation for Roots

  1. #1
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    435
    Thanks
    1

    Solving a Two Radical Equation for Roots

    How do I get the roots from this equation?

     \sqrt {x + 4} + \sqrt{1 - x} = 3
    Last edited by Jason76; October 19th 2012 at 04:18 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Solving a Two Radical Equation for Roots

    Quote Originally Posted by Jason76 View Post
    How do I get the roots from this equation?

     \sqrt {x + 4} + \sqrt{1 - x} = 3
    1. Isolate a radical

    2. Square both sides

    3. If the equation contains a radical go back to step1, if not move on to step 4

    4. Solve the equation

    5. check your solutions. extraneous solutions may have been introduced when the equation was squared.

    \sqrt{x+4}=3-\sqrt{1-x}

    (\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)

    -6 = -6\sqrt{1-x} \iff 1=\sqrt{1-x}

    1=1-x \iff x=0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    435
    Thanks
    1

    Re: Solving a Two Radical Equation for Roots

    I basically get the jist of it but I don't understand the following line:

    (\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)

    So

     \sqrt{x+4}

    was squared (to get rid of the square root), so now the same has to be done to the other side. However,

    3 (to the right of the equation)

    was not under square root. So why was it squared?

    Obviously,

    \sqrt{1-x}

    was squared to get rid of the radical sign.
    Last edited by Jason76; October 19th 2012 at 04:43 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Solving a Two Radical Equation for Roots

    Quote Originally Posted by Jason76 View Post
    I basically get the jist of it but I don't understand the following line:

    (\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)

    So

     x + 4

    which was square rooted was squared (to get rid of the square root), so now the same has to be done to the other side. However,

     - 3

    was not under square root. So why was it squared?

    If you multiply two binomials together you do not get

    (a+b)^2 \ne a^2+b^2

    You must use the distributive property to get

    (a+b)^2=(a+b)(a+b)=a^2+2ab+b^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    435
    Thanks
    1

    Re: Solving a Two Radical Equation for Roots

    Quote Originally Posted by TheEmptySet View Post
    If you multiply two binomials together you do not get

    (a+b)^2 \ne a^2+b^2

    You must use the distributive property to get

    (a+b)^2=(a+b)(a+b)=a^2+2ab+b^2
    Considering again:

    (\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)


    Right, but

     \sqrt{x+4}


    is a binomial, yet, after squaring, no distributive property is used on it. On the other hand,

    - \sqrt{1-x}


    (after squaring) gets a distributive property operation done to it. Plus, I don't understand where the

    9 - 6

    really came from.

    Again:

    (\sqrt{x+4}^2=(3-\sqrt{1-x})^2 \implies x+4=9-6\sqrt{1-x}+(1-x)


    Anyhow, what I'm trying to say is "Two binomials exist on both sides of the equation. Both of them should be squared (to get rid of the radical signs). Therefore, the binomial on the left is squared, as expected. However, the one on the right is different. It is squared and then distributed. Yet, the radical sign remains for one of the duplicate terms.". Plus, I still have no idea where the

    9 - 6

    came from. I'm assuming the  9 came from squaring.
    Last edited by Jason76; October 19th 2012 at 05:34 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    435
    Thanks
    1

    Re: Solving a Two Radical Equation for Roots

    Sorry to bump thread, but I couldn't edit the last post.

    Here is what I'm saying more clearly:

    How do I solve this for roots?

    \sqrt{x+4} + \sqrt{1-x} = 3

    Put one radical on the right side of the equation.

    \sqrt{x + 4} = 3 - \sqrt{1 - x}

    Square both sides to get rid of radicals.

    \sqrt{x + 4}^2 = (3 - \sqrt{1 -x})^2

    Squaring both sides (to get rid of radicals) should lead to this (by my logic):

    x + 4 = 9 - 1 - x

    But this is the correct answer (according to the book):

    x + 4 = 9 - 6 \sqrt{1-x} + (1 - x)

    That's main part I have trouble with. The rest is easy to understand.

    Of course, this only half of the problem. I haven't solved for roots yet.
    Last edited by Jason76; October 19th 2012 at 07:01 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Solving a Two Radical Equation for Roots

    Look at the rule TheEmptySet gave in post #4...this will tell you why the squaring of the side in question becomes what it does.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    435
    Thanks
    1

    Re: Solving a Two Radical Equation for Roots

    I think I got it now:

    Recognize it (right hand side of equation) as a "Square of a Binomial".

    (3 - \sqrt{1-x})^2

    (3 - \sqrt{1-x}) (3 - \sqrt{1-x})

    FOIL

    (3)(3) + (3) ( - \sqrt{1-x}) + (3) ( - \sqrt{1-x}) + ( - \sqrt{1-x}) ( - \sqrt{1-x})

    9 + (3) ( - \sqrt{1-x}) + (3) ( - \sqrt{1-x}) + ( 1-x )

    Next, add the two middle terms. In this case, it's adding negatives, so it's -b - b. The same as -b (+) -b

    9 - 3 \sqrt{1-x} - 3 \sqrt{1-x} + ( 1-x )

    9 - 6 \sqrt{1-x} + ( 1-x)

    Full equation:

    x + 4 = 9 - 6 \sqrt{1-x} + ( 1-x)

    Next, more algebra until the roots are found.
    Last edited by Jason76; October 20th 2012 at 06:28 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    435
    Thanks
    1

    Re: Solving a Two Radical Equation for Roots

    Ok. Here is the 2nd part of the equation:

    x + 4 = 9 - 6 \sqrt{1-x} + ( 1-x)

    Isolate the radical again by moving it to the left side.

    (6 \sqrt{1-x})^2 = (6 - 2x)^2

    36(1-x) = 66 - 24x + 4x^2

    Turn this into a quadratic equation by creating a 0 on one side.

    4x^2 + 12x = 0

    4x(x + 3) = 0

    (one linear factor): (x + 3)

    Roots:

    x = 0

    or

    x = -3

    Plugin those values into the original equation to see if each one is true.

    One question. Where did the 0 come from?
    Last edited by Jason76; October 20th 2012 at 08:00 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Solving a Two Radical Equation for Roots

    Quote Originally Posted by Jason76 View Post
    Ok. Here is the 2nd part of the equation:

    ...

    4x(x + 3) = 0

    (one linear factor): (x + 3)

    Roots:

    x = 0

    or

    x = -3

    Plugin those values into the original equation to see if each one is true.

    One question. Where did the 0 come from?
    \underbrace{4x}_{\text{1st factor}}\overbrace{(x + 3)}^{\text{2nd factor}} = 0

    A product equals zero if one of the factors equals zero. Therefore:

    4x = 0 ~\vee~x+3=0~\implies~\boxed{x=0~\vee~x=-3}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving a radical equation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 7th 2011, 12:02 AM
  2. Replies: 2
    Last Post: February 2nd 2011, 07:12 PM
  3. Solving a radical equation.......
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 24th 2011, 01:48 PM
  4. Solving this radical equation
    Posted in the Algebra Forum
    Replies: 7
    Last Post: June 1st 2010, 12:43 PM
  5. Solving an Equation w/ Radical
    Posted in the Algebra Forum
    Replies: 3
    Last Post: September 16th 2009, 01:14 PM

Search Tags


/mathhelpforum @mathhelpforum