Results 1 to 6 of 6

Math Help - Solving equations containing square roots.

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    142

    Red face Solving equations containing square roots.

    a)  \sqrt{2x+1} - \sqrt{x} = 1
    True or false?

    b)  \sqrt{x+2} + \sqrt{7x+2} = 6
    True or false?

    Can you please help me understand this.
    Last edited by mr fantastic; February 1st 2010 at 12:13 AM. Reason: Changed post title
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Jan 2010
    Posts
    9
    Quote Originally Posted by Anemori View Post
    a)  \sqrt{2x+1} - \sqrt{x} = 1
    True or false?

    b)  \sqrt{x+2} + \sqrt{7x+2} = 6
    True or false?

    Can you please help me understand this.
    a)  \sqrt{2x+1} - \sqrt{x} = 1
    True or false?

    \sqrt{2x+1}=(1+\sqrt{x})

    Square both sides...

    2x+1=1+2\sqrt{x}+x

    Subtract 1 and x from the right...

    x=2\sqrt{x}

    We see that x=0

    Plugging it in shows that \sqrt{0+1}-\sqrt{0}=1

    So this is a true statement.

    You can try B...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    9
    Here's B anyway since it's more involved....

     \sqrt{x+2} + \sqrt{7x+2} = 6

    (\sqrt{x+2})=(6-\sqrt{7x+2})

    Square both sides...

    x+2=36-12\sqrt{7x+2}+7x+2

    Subtract 36, 7x, and 2 from the right.

    -6x-36=-12\sqrt{7x+2}

    Look, we can divide the equation by -6.

    (x+6)=(2\sqrt{7x+2})

    Square both sides again...

    x^2+12x+36=4(7x+2)

    Distribute the 4.

    x^2+12x+36=28x+8

    Subtract 28x and 8 from the right.

    x^2-16x+28=0

    This factors to...

    (x-14)(x-2)

    So our solutions are 14 and 2. Check them! You will find that 2 is the only solution and thus the statement is true.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2010
    Posts
    142
    Thank you that was awsome....
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    142
    Quote Originally Posted by halpo View Post
    a)  \sqrt{2x+1} - \sqrt{x} = 1
    True or false?

    \sqrt{2x+1}=(1+\sqrt{x})

    Square both sides...

    2x+1=1+2\sqrt{x}+x

    Subtract 1 and x from the right...

    x=2\sqrt{x}

    We see that x=0

    Plugging it in shows that \sqrt{0+1}-\sqrt{0}=1

    So this is a true statement.

    You can try B...
    i was confused with this one.. how come x = 0 on this x=2\sqrt{x} im sorry dumb question.. i just dont get it..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,368
    Thanks
    1313
    Were those really "true or false" questions? I would interpret "true or false" for an equation involving, say, x, to be asking if it was true for all x- and that's clearly false for both questions.

    To solve equations involving \sqrt{x}, you square to get rid of the square roots. After getting to x= 2\sqrt{x}, square again to get x^2= 4x which is the same as x^2- 4x= x(x- 4)= 0. Either x= 0 or x- 4= 0 so x= 4.

    Since squaring both sides like that can introduce "extraneous roots", you should always check by putting the numbers into the original equation. The original equation was [tex]\sqrt{2x+1}- \sqrt{x}= 1[tex]. With x= 0 that becomes \sqrt{2(0)+1}- \sqrt{0}= \sqrt{1}= 1 so x= 0 is a solution. With x= 4, it becomes \sqrt{2(4)+ 1}- \sqrt{4}= \sqrt{9}- \sqrt{4}= 3- 2= 1 so x= 4 is also a solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving equations by completing the square.
    Posted in the Algebra Forum
    Replies: 11
    Last Post: November 9th 2011, 11:01 AM
  2. Replies: 2
    Last Post: February 2nd 2011, 07:12 PM
  3. problem solving - square roots
    Posted in the Calculus Forum
    Replies: 5
    Last Post: December 4th 2009, 09:22 AM
  4. Solving for the square roots of binomials
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 26th 2009, 09:13 AM
  5. Solving equations with roots
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 5th 2008, 05:09 AM

Search Tags


/mathhelpforum @mathhelpforum