Tuesday, June 9, 2009

1.5 Solving Nonfactorable Quadratic Equations

pg 126-128 #22,24,32,36,38,40,44,50,96,108,110,116

NS

5 comments:

  1. 16) I have -3x + 12.25=the square root on 15.25. I think that I subtract 12.25 to the 15.25 making it the square root of 3 so then I would divide that by -3x. Hopefully you understand this enough to help

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  2. 116) I got to 16t squared - 120t + 212=0, and I'm stumped after that.

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  3. 96)
    2X2 – 5X -3 = 0  First divide everything by 2
    X2 – (5/2)X -3/2 = 0 Divide the “b” coefficient (5/2) by 2 and then square it
    ((5/2)/2)2 =(5/4)2= (25/16)  This is going to be what you want for the c, so convert the c to the same denominator
    X2 – (5/2)X – 24/16 = 0  Just changed (3/2) to (24/16). So now, how do we get 25/16?
    X2 – (5/2)X – 24/16 + 49/16 = 49/16  We added49/16 to both sides so that we could get 25/16 for the c coefficient
    X2 – (5/2)X + 25/16 = 49/16  Simplified by adding – 24/16 + 49/16 to get 25/16.
    (X – (5/4))*( X – (5/4)) = 49/16  Factored!
    (X-(5/4))2 = 49/16  Simplified
    X-5/4 = sqrt (49/16)  Square root of both sides
    X – 5/4 = +- 7/4  Simplified the sqrt (49/16)
    X = 5/4 +- 7/4  Solved
    X = 3 OR -1/2  Solutions are both rational solutions!

    Hope this helps
    NS

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  4. In this problem (116) they are giving you the equation h= -16t2 +120t -106. This means the initial height is -106 (cavern is 106 feet deep) and the initial velocity is 120 ft/sec. Since the height is measured down from the top of the canyon wall, they are telling you that the top of the canyon wall is 0 feet. So h = 0!
    This gives us the following equation:
    0 = -16t2 +120t -106; where h = 0.
    So we solve like we would any non-factorable equation, either with the method used in 96, or with the quadratic formula. Let’s use the quadratic formula this time:
    (-b +/- sqrt (b2 – 4ac))/2a where
    a =-16; b= 120; c=-106
    Putting those into the equation, we get that
    t= 1.02 seconds, I leave the exact form for you!

    NS

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