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Math Help - Hi, some help with year 11 algebra please

  1. #1
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    Hi, some help with year 11 algebra please

    Ok so the question is as follows: solve for y

    y^2 + 6y + 9 = 0

    I don't know if I'm correct but so far this is what I've got
    y^2 + 6y = -9
    y^2 + y = -9
    6
    y^2 + y= -1.5

    I don't know if there's a way to type to the second on here?? so ^2 will do for now
    Last edited by dylandriscoll79; April 17th 2013 at 12:35 PM.
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  2. #2
    Member dokrbb's Avatar
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    Re: Hi, some help with year 11 algebra please

    working with a second degree equation of the type ay^2+ by + c = 0, you normally use

    \y = \frac {(-b) \pm \sqrt{b^{2}-4ac}}{2a}

    the sign  \pm shows that you may have two(2) roots for this equation and this depends on the values of the discriminant \sqrt{b^{2}-4ac}}

    now just plug in the values and do the math,

    dokrbb
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  3. #3
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    Re: Hi, some help with year 11 algebra please

    This particular question you can also do by splitting the middle term. the algorithm is
    for ax^2+ bx + c = 0; find the product ac which in this case is 9. next find such factors of this product such that their sum is b, the coefficient of x. in this case 9 = 3 x 3 and 3 + 3 = 6
    rewrite the equation with middle term as the sum.
    i.e., y^2+ 3y + 3y = 9=0
    Now take common and proceed to factorize.
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  4. #4
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    Re: Hi, some help with year 11 algebra please

    dokrbb is correct that the quadratic equation above will always work, sometimes you can factor it for a much less messy approach. Which if this was a HW problem is the approach wanted here.

     y^{2}+6y+9  = (y+3)(y+3) = 0

    So for that to equal zero, either y+3 = 0 , or y+3 = 0 (Most of the time the factors will be different.. but in this case they are the same so we'll get only 1 answer:
     y=-3
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