Results 1 to 6 of 6
Like Tree5Thanks
  • 2 Post By romsek
  • 3 Post By skeeter

Thread: Solutions to quadratic equation

  1. #1
    Newbie
    Joined
    Mar 2017
    From
    New York
    Posts
    18

    Solutions to quadratic equation

    Hi All,

    What are the solutions of the equation, (5x-1)2 = (2x+7)2 ?


    My answer:8/3. Are there any other solutions?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,768
    Thanks
    2417

    Re: Solutions to quadratic equation

    $(5x-1)^2 = (2x+7)^2$

    $(5x-1)^2 -(2x+7)^2 = 0$

    $21 x^2-38 x-48=0$

    $(3x-8)(7x+6) = 0$

    $x = \dfrac 8 3,~-\dfrac 6 7$
    Thanks from allo95 and topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,106
    Thanks
    3647

    Re: Solutions to quadratic equation

    Quote Originally Posted by allo95 View Post

    What are the solutions of the equation, (5x-1)2 = (2x+7)2 ?
    $(5x-1)^2 - (2x+7)^2 = 0$

    $[(5x-1)-(2x+7)] \cdot [(5x-1)+(2x+7)] = 0$

    $(3x-8)(7x+6) = 0$
    Thanks from deesuwalka, romsek and topsquark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Feb 2015
    From
    Ottawa Ontario
    Posts
    1,671
    Thanks
    313

    Re: Solutions to quadratic equation

    Yes, there is also a negative answer.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2016
    From
    India
    Posts
    84
    Thanks
    18

    Re: Solutions to quadratic equation

    There are two answers of quadratic.

    Factors of the quadratic
    $$(3x-8)(7x+6)=0$$
    Therefore,$$\begin{equation}\begin{split}3x-8&=0 \\3x&=8 \\x&=\dfrac{8}{3} \end{split} \end{equation}$$
    and$$\begin{equation}\begin{split}7x+6&=0\\7x&=-6 \\ x&=\dfrac{-6}{7} \end{split} \end{equation}$$
    Hence, $x=\dfrac{8}{3},\;\dfrac{-6}{7}$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2016
    From
    Earth
    Posts
    134
    Thanks
    55

    Re: Solutions to quadratic equation

    It good to know multiple ways of solving this type of equation.

    There is relatively more work involved in the method in post # 2, because 1) it requires expanding the square of two binomials,
    2) collecting like terms, and 3) factoring a higher degree polynomial, a quadratic, whose first and last coefficients have a
    relatively large total number of ways of breaking them down into usable factors.


    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


    (5x - 1)^2 \ = \ (2x + 7)^2

    5x - 1 \ = \ \pm\sqrt{2x + 7}

    5x - 1 \ = \ -2x - 7 \ \ \  \ or \ \ \  \ 5x - 1 \ = \ 2x + 7

    7x \  =  \ -6 \ \ \ \ or \ \ \  \ 3x \ = \ 8


    x \ = \ \dfrac{-6}{7} \ \ \  \ or \ \ \  \ x \ = \ \dfrac{8}{3}
    Last edited by greg1313; Mar 21st 2017 at 11:05 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic Solutions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jun 3rd 2011, 07:42 PM
  2. Extra Solutions of a Quadratic Diophantine Equation
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: May 21st 2011, 05:20 AM
  3. Quadratic solutions - why usually two?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Dec 17th 2009, 06:59 AM
  4. Solutions of quadratic equation
    Posted in the Algebra Forum
    Replies: 15
    Last Post: May 14th 2009, 06:11 AM
  5. Quadratic Equation Solutions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 19th 2008, 09:40 AM

/mathhelpforum @mathhelpforum