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Quadratic equations problem?

  1. Apr 6, 2005 #1

    I have a test today and cant figure out how to solve certain quadratic equations.

    We are doing solving the equations by completing the square using these steps:

    1. divide each side by coefficient of x squared
    2. rewrite the equation with the constant on right side
    3. complete the square: add the square of one half of the coefficient of x to both sides
    4. write the left side as a square and simplify the right side
    5. equate the square root of the left side to the principle square root of the right side and its negative.
    6. solve the 2 resulting equations.

    Heres my problem:

    2s squared+5s=3

    heres what im doing

    s squared+5/2s=2/3

    but when you hald 5/2 and then square it you get a fraction...how are you supposed to factor a number with a fraction in it?

    help please!
  2. jcsd
  3. Apr 6, 2005 #2
    You don't need to even think about the factorization. The thing added to [itex]x[/itex] in the factorization is always just half the coefficient of [itex]x[/itex] (including a negative sign, if the coefficient of [itex]x[/itex] has one!!!). You made a little mistake here anyways though, so:

    [tex]2s^2 + 5s = 3 \Longrightarrow s^2 + \frac{5}{2}s = \frac{3}{2}[/tex]

    [tex] \Longrightarrow s^2 + \frac{5}{2}s + \frac{25}{16} = \frac{3}{2}+\frac{25}{16}[/tex]

    [tex] \Longrightarrow \left(s+\frac{5}{4}\right)^2 = \frac{49}{16}[/tex]

    you can finish...
  4. Apr 6, 2005 #3
    where did you get the 25/16 from?
  5. Apr 6, 2005 #4
    by the way the answer is x=-3,1/2
  6. Apr 6, 2005 #5
    [itex]25/16[/itex] is half of [itex]5/2[/itex] squared. If you solve from the point I left off, you'll get those answers~
  7. Apr 6, 2005 #6
    5/2 squared is 25/4 how can 25/16 be half of it???

    im not trying to say you dont know what your doing lol i just am still totally lost lol
  8. Apr 6, 2005 #7
    Other way. You half it, then square it :smile:
  9. Apr 6, 2005 #8
    for final answer all i get is x=5/4+ square root of 49/16
    x=5/4- square root of 49/16
  10. Apr 6, 2005 #9
    nevermind i got it :surprised :surprised
  11. Apr 6, 2005 #10
    Here's a little explanation as to why, since you seem confused:

    When you square a binomial, say, [itex]a+b[/itex], you get the following:

    [tex](a+b)^2 = (a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2[/tex]

    now, what happens if [itex]a = x[/itex]? Well, then this is

    [tex]x^2 + 2bx + b^2.[/tex]

    since we put no restrictions on [itex]b[/itex], we find that a polynomial

    [tex]cx^2 + dx + z^2[/tex]

    can be factored into a perfect square if and only if

    [tex] z = \left(\frac{d}{2}\right) \Longrightarrow z^2 = \left(\frac{d}{2}\right)^2[/tex]

    and that in this case,

    [tex]cx^2 + dx + z^2 = (x+z)^2.[/tex]

    so for your example, we want to do something to make the left side of

    [tex]s^2 + \frac{5}{2}s = \frac{3}{2}[/tex]

    into a perfect square. Well, here, on the left side, using the letters from above, [itex]c=1, \ d = 5/ 2, \ z = 0[/itex]. We need [itex] z = (d/2) = 5/4 \Longrightarrow z^2 = (d/2)^2 = (5/4)^2[/itex], so we add

    [tex] \left( \frac{5/2}{2} \right)^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16}[/tex]

    to each side, and thus it factors to

    [tex](s+z)^2 = \left(s + \frac{5}{4}\right)^2.[/tex]
    Last edited: Apr 6, 2005
  12. Apr 6, 2005 #11
    thanks...ok i tried to do the same thing for another of the same type question but got the wrong answer again lol.

    3y squared=3y+2

    so i did

    3y squared-3y=2
    y squared-y=2/3
    y squared-y-2/4=2/3-2/4
    y=2/4 +or- square root of 1/6

    but the answer is x= 1/6 (3 +or- square root of 33)
  13. Apr 6, 2005 #12


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    [tex] (a-b)^{2}=a^{2}-2ab+b^{2} [/tex]

  14. Apr 6, 2005 #13
    i tried a different way to do it still no luck

    3y squared-3y=2
    y squared-y=2/3
    y squared-y+1/4=2/3+1/4
    y+1/2=square root of 11/12

    i still get the wrong answer :grumpy:
  15. Apr 6, 2005 #14
    anyone? lol
  16. Apr 6, 2005 #15


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    Of course

    [tex] x^{2}-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^{2} [/tex]

    according to the formula i wrote above...

  17. Apr 6, 2005 #16
    ok i changed it to that but i still get the wrong answer

    now i get x=1/2 +or- square root of 11/12

    the answer is supposed to be 1/6(3+or- square root of 33)
  18. Apr 6, 2005 #17


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    I little algebra may show you that the two (apparently different) real numbers are one & the same...

  19. Apr 6, 2005 #18
    Your answer is correct. It's the same as theirs in a different form.

    [tex]\frac{1}{2} \pm \sqrt{\frac{11}{12}} = \frac{1}{2} \pm \frac{\sqrt{11}}{2\sqrt{3}} [/tex]

    [tex]= \frac{1}{3}\left(\frac{3}{2} \pm \frac{3}{2}\frac{\sqrt{11}}{\sqrt{3}}\right) = \frac{1}{6}\left(3 \pm \frac{3\sqrt{11}}{\sqrt{3}}\right)[/tex]

    [tex] = \frac{1}{6}\left( 3 \pm \sqrt{3}\sqrt{11}\right) = \frac{1}{6}\left(3 \pm \sqrt{33}\right)[/tex]
  20. Apr 6, 2005 #19
    i see it now...thanks data you were a huge help
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