Results 1 to 7 of 7

Thread: Difference of the square - Tricky question for me...

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    2

    Difference of the square - Tricky question for me...

    Hi ! First post here !

    I'm having some difficulty with a math problem in my math book and I'm starting to think the author made a mistake.

    I'm trying to factorize $\displaystyle (a-b)^2 -4$ using the "difference of the squares" technique. According to my book, it should result in the following factors :

    $\displaystyle (a-b-2) (a-b+2)$

    Now, if I multiply those two factors together, it gives me this :

    $\displaystyle a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

    then

    $\displaystyle a^2-2ab+b^2-4$

    so, I'm tempted to conclude that $\displaystyle (a-b-2) (a-b+2)$ does not equal $\displaystyle (a-b)^2 -4$, and therefore that my book is wrong about it.

    Am I right ? Did I make a mistake ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by mindraked View Post
    Hi ! First post here !

    I'm having some difficulty with a math problem in my math book and I'm starting to think the author made a mistake.

    I'm trying to factorize $\displaystyle (a-b)^2 -4$ using the "difference of the squares" technique. According to my book, it should result in the following factors :

    $\displaystyle (a-b-2) (a-b+2)$

    Now, if I multiply those two factors together, it gives me this :

    $\displaystyle a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

    then

    $\displaystyle a^2-2ab+b^2-4$

    so, I'm tempted to conclude that $\displaystyle (a-b-2) (a-b+2)$ does not equal $\displaystyle (a-b)^2 -4$, and therefore that my book is wrong about it.

    Am I right ? Did I make a mistake ?

    $\displaystyle a^2-2ab+b^2-4 = (a-b)^2 -4$

    because

    $\displaystyle (a-b)^2 = a^2-2ab+b^2$

    Hopefully you didn't make the mistake of thinking

    $\displaystyle (a-b)^2 = a^2-b^2\ \ \ \text{(Wrong)}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Your book is correct.

    $\displaystyle a^2-2ab+b^2-4= (a-b)(a-b)-4= (a-b)^2-4$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member grgrsanjay's Avatar
    Joined
    May 2010
    From
    chennai,tamil nadu
    Posts
    143
    Thanks
    1
    Quote Originally Posted by mindraked View Post
    Hi ! First post here ! welcome!!



    $\displaystyle (a-b-2) (a-b+2)$

    Now, if I multiply those two factors together, it gives me this :

    $\displaystyle a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

    then

    $\displaystyle a^2-2ab+b^2-4$

    so, I'm tempted to conclude that $\displaystyle (a-b-2) (a-b+2)$ does not equal $\displaystyle (a-b)^2 -4$, and therefore that my book is wrong about it. they both are equal and book is right

    Am I right ? Did I make a mistake ? may be wrong in factorizing

    it comes out to be true
    (a-b-2)(a-b+2) =(a-b)^2 - 2^2 =$\displaystyle (a-b)^2 - 4$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,779
    Thanks
    3028
    Quote Originally Posted by mindraked View Post
    Hi ! First post here !

    I'm having some difficulty with a math problem in my math book and I'm starting to think the author made a mistake.

    I'm trying to factorize $\displaystyle (a-b)^2 -4$ using the "difference of the squares" technique. According to my book, it should result in the following factors :

    $\displaystyle (a-b-2) (a-b+2)$

    Now, if I multiply those two factors together, it gives me this :

    $\displaystyle a^2 - ab + 2a - ab + b^2 - 2b - 2a + 2b - 4$

    then

    $\displaystyle a^2-2ab+b^2-4$
    And, of course, $\displaystyle a^2- 2ab+ b^2= (a- b)^2$ so that this is exactly $\displaystyle (a- b)^2- 4$.

    so, I'm tempted to conclude that $\displaystyle (a-b-2) (a-b+2)$ does not equal $\displaystyle (a-b)^2 -4$, and therefore that my book is wrong about it.

    Am I right ? Did I make a mistake ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Sep 2010
    Posts
    2
    Quote Originally Posted by grgrsanjay View Post
    it comes out to be true
    (a-b-2)(a-b+2) =(a-b)^2 - 2^2 =$\displaystyle (a-b)^2 - 4$
    Thank you for this quick answer (and all the others), but, may I ask, how do you choose not to multiply each variable of the fist factor by each variable of the second factor and instead decide to couple the two $\displaystyle a-b$ into a $\displaystyle (a-b)^2$ ?

    My understanding of all this is a bit fragile for the moment. Why would I chose to use two variables together or independently ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by mindraked View Post
    Thank you for this quick answer (and all the others), but, may I ask, how do you choose not to multiply each variable of the fist factor by each variable of the second factor and instead decide to couple the two $\displaystyle a-b$ into a $\displaystyle (a-b)^2$ ?

    My understanding of all this is a bit fragile for the moment. Why would I chose to use two variables together or independently ?
    Factoring, remember that x^2 - y^2 = (x-y)(x+y). Here you can let x = a-b and y = 2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. tricky square root problem
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: Mar 28th 2011, 04:29 AM
  2. help with difference of square?
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Feb 13th 2011, 01:22 PM
  3. Difference of Chi Square Random Variables
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: Mar 15th 2010, 12:21 PM
  4. A tricky question on %'s....
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Aug 9th 2009, 12:03 AM
  5. square difference
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 13th 2007, 12:45 PM

Search Tags


/mathhelpforum @mathhelpforum