Results 1 to 4 of 4

Math Help - How do you solve a quadratic equation where b = the sum of square roots?

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    2

    How do you solve a quadratic equation where b = the sum of square roots?

    I understand the quadratic formula. I can derive the quadratic formula. I do not understand what to do when b= the sum of two square roots, as in the following problem:

    Use the quadratic formula to solve:

    r squared + [square root (3) - square root (2)]*r = square root (6)

    r^2+(√(3)-√(2))r = √(6)

    I know the answer is r = -(√(3)) or √(2). I don't know how to get to that answer. I have tried a number of ways. Plugging a, b, and c straight into the quadratic formula, I get:

    r = 1/2[√(2)-√(3)+√(5+(2√(6)))] or 1/2[√(2)-√(3)-√(5+(2√(6)))]

    This is right, but I don't understand the following steps:

    1/2[√(2)-√(3)+√(5+(2√(6)))] --> √(2)
    1/2[√(2)-√(3)-√(5+(2√(6)))] --> -(√(3))

    Thanks for your help! (I've been working on this problem for WAY too long.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,319
    Thanks
    1238
    r^2 + (\sqrt{3} + \sqrt{2})r = \sqrt{6}

    r^2 + (\sqrt{3} + \sqrt{2})r - \sqrt{3}\sqrt{2} = 0


    \displaystyle{r = \frac{-(\sqrt{3} + \sqrt{2}) \pm \sqrt{(\sqrt{3} + \sqrt{2})^2 - 4(1)(\sqrt{3}\sqrt{2})}}{2(1)}}

    \displaystyle{= \frac{-\sqrt{3} - \sqrt{2} \pm \sqrt{(\sqrt{3})^2 + 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2 - 4\sqrt{3}\sqrt{2}}}{2}}

    \displaystyle{= \frac{-\sqrt{3} - \sqrt{2} \pm \sqrt{(\sqrt{3})^2 - 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2}}{2}}

    \displaystyle{= \frac{-\sqrt{3}-\sqrt{2} \pm \sqrt{(\sqrt{3} - \sqrt{2})^2}}{2}}

    \displaystyle{= \frac{-\sqrt{3} - \sqrt{2} \pm (\sqrt{3} - \sqrt{2})}{2}}

    \displaystyle{r = \frac{-2\sqrt{2}}{2}} or \displaystyle{r = \frac{-2\sqrt{3}}{2}}

    \displaystyle{r = -\sqrt{2}} or \displaystyle{r = -\sqrt{3}}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    2
    This answer helped me solve the problem! Thank you. The question answered here is a little different from mine. This question answers r^2+(√(3)+√(2))r = √(6) instead of r^2+(√(3)-√(2))r = √(6), but it was really helpful as an example format to use to solve the latter question. The key that I was missing was factoring √((√3)^2 + 2√(3)√(2) + (√(2))^2) ---> √((√(3) + √(2))^2). Thank you again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,080
    Thanks
    66
    Quote Originally Posted by Prove It View Post
    r^2 + (\sqrt{3} + \sqrt{2})r = \sqrt{6}
    \displaystyle{r = -\sqrt{2}} or \displaystyle{r = -\sqrt{3}}.
    Should be [SQRT(3) - SQRT(2)]r
    So answer is r = SQRT(2) or r = -SQRT(3)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: January 25th 2011, 02:33 PM
  2. Equation with square roots and logs
    Posted in the Algebra Forum
    Replies: 6
    Last Post: August 28th 2010, 04:58 PM
  3. Roots of quadratic equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 12th 2008, 11:05 PM
  4. Quadratic equation from given roots.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 15th 2007, 09:11 AM
  5. Equation with square roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 7th 2006, 10:56 PM

Search Tags


/mathhelpforum @mathhelpforum