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Bunch of AS level math problems(quadratics,indices etc)

  1. Sep 4, 2009 #1
    Ok, so I have very little time to nail my AS level math papers and I got stuck at a couple places with a bunch of problems from various chapters. I've collected all my snags and I'm posting them together.
    Problem 1
    Use the completed square form to factorize 6x2 - 5x - 6

    The attempt at a solution
    (x + 5/2)2 - 25/4 - 6
    {(x + 5/2)2 - (5/2)2} - 6
    (by a2 - b2 = (a-b)(a +b))
    [((x + 5/2) - (5/2))((x + 5/2) + (5/2)) ]- 6
    [x(x + 10/2)] - 6 >>>>> Which is totally weird because the answer is supposed to be:
    (3x +2)(2x-3)

    Problem 2
    Find the point of intersection of the foll. curves:
    x + 2y = 3; x2 + xy = 2
    I know I need to substitute one equation in the other. I managed all the other problems in the section except this one.
    The expected answer is : (1,1) (-4,3.5)

    Problem 3
    Find values of k such that the straight line y = 2x + k meets the curve with equation x2 + 2xy + 2y2 = 5 exactly once.
    I dont really know what to do, I tried substituting like the previous problem and came up with a complicated mess of an equation with k.

    Problem 4
    Show that the equation 2x+1 + 2x-1=160 can be written in the form 2.5 X 2x = 160. Find the value of x which satisfies the equation.
    I'm a little embarrassed that i do now know how to really tackle this one!?:redface:

    Problem 5
    Solve the inequality : (x-1)(x-6)<0
    There something in my book about a table of signs of the factors but I can't make any sense of it!

    I'm studying math on my own and I'm a little absent minded sometimes and I miss the obvious when i can even understand complex things, so I'm sorry if any of the questions were silly.

  2. jcsd
  3. Sep 4, 2009 #2


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    Homework Helper

    1) First step to make the completed square form is this 6(x2-5x/6)-6.

    2) You'd need to show us your working to say if you did something wrong

    3) If you substitute for y, you'll get a quadratic in x. If

    [tex]x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}[/tex]

    because it meets the curve only once, means there is only one solution for 'x'. Looking at the formula above. How do you get it so that x is one value and not in the form x=(1±2)/2 or something like that?

    4) This will help you am*an=am+n. Work backwards with this one.

    5) What I'd do is see that the curve crosses the x-axis at 1 and 6. Draw the axis and see for what range of x, the curve lies for less than zero...or below the x-axis as it means.
  4. Sep 4, 2009 #3
    This is not correct.

    Start by factoring 6.


    Now use the fact that 169/144 = (c/d)2

    Next use a2-b2=(a-b)(a+b)

    Problem 2

    Hint: for the first equation

    Now substitute for y in the second equation and solve for x :smile:

    Problem 5

    Ok, if you have a*b<0 what "sign" should the values of a, b have?

    Hint: a<0 & b>0 or a>0 & b<0
    Last edited: Sep 4, 2009
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