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Math Help - Three tough problems.

  1. #1
    Senior Member sakonpure6's Avatar
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    Three tough problems.

    Sorry for such vague title.
    Anyway, I am doing some math evaluation test when I ran across these bad boys.

    For question #1, I have no idea where to begin.
    For question #2, I forgot how to deal with absolute values.
    For question #3:
    y= x+1 So, y^2 = x^2 + 2x +1 Sub that into the other equation and we get \frac{x^2}{4} + \frac{x^2 +2x+1}{9} = 1
    9x^2 + 4x^2 + 8x + 4 = 36
    13x^2 + 8x - 32 = 0 but we do not get any nice roots, and all the options for the questions are rational.

    If some one can show me how to solve these, I would appreciate it
    Three tough problems.-question.pngThree tough problems.-question-2.pngThree tough problems.-question-3.png
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  2. #2
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    Re: Three tough problems.

    Hello, sakonpure61

    \text{The two curves }y \:=\:x+1\text{ and }\frac{x^2}{4} + \frac{y^2}{9} \:=\:1

    \text{ intersect in exactly two points }(a,b)\text{ and }(c,d).

    \text{Find the value of }a+c.
    y\:=\: x+1 \quad\Rightarrow\quad y^2 \:=\: x^2 + 2x +1

    Sub that into the other equation and we get . \frac{x^2}{4} + \frac{x^2 +2x+1}{9} \:=\: 1

    9x^2 + 4x^2 + 8x + 4 \:=\: 36 \quad\Rightarrow\quad 13x^2 + 8x - 32 \:=\: 0

    But we do not get any nice roots, and all the options for the question are rational.

    I bet your work is correct!
    Re-read the question.

    Quadratic Formula: . x \:=\:\frac{\text{-}8 \pm\sqrt{1728}}{26} \:=\:\frac{\text{-}4\pm12\sqrt{3}}{13}

    The points are:
    . . (a,b) \:=\:\left(\frac{\text{-}4+12\sqrt{3}}{13},\:\frac{9+12\sqrt{3}}{13} \right)\;\text{ and }\;(c,d) \:=\:\left(\frac{\text{-}4-12\sqrt{3}}{13},\;\frac{9-12\sqrt{3}}{13}\right)

    \text{Now add }a\text{ and }c\!: \;\frac{\text{-}4+12\sqrt{3}}{13} + \frac{\text{-}4 - 12\sqrt{3}}{13} \;=\;\cdots
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  3. #3
    Senior Member sakonpure6's Avatar
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    Re: Three tough problems.

    Oh! I gave up too soon, thank you for the reply.
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  4. #4
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    Re: Three tough problems.

    Hey Sakon, hope that's a HOCKEY jersey in your avatar!!
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  5. #5
    Senior Member sakonpure6's Avatar
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    Re: Three tough problems.

    hehe, kind of an embarrassing situation, just look the other way ;D .
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  6. #6
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    Re: Three tough problems.

    The answer to 10 is B, and here I will teach you how to cheat lol.
    If |b| > a > 0, that requires b>a or b<-a. Hence, it is impossible for b to stand between -a and a.
    That's why B is incorrect. 6-x^2 cannot stay between -3 and 3 if |6-x^2| >=3.
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  7. #7
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    Re: Three tough problems.

    For the first question, I am gonna find out the equation for line BC first, and you shall see why.
    slope: (0-1)/(1- -3) = -1/4, the line should be y=(-1/4)x+b. Plugging C(1, 0) in, you should get b= 1/4 so the line BC is y=(-1/4)x+(1/4)
    Here's the thing: since HA and BC are perpendicular, THE PRODUCT OF THEIR SLOPES will be -1.
    Hence, you know that the slope for HA is 4.
    Then, y=4x+b for HA. Plugging A(-1, 2) in should get b= 6. Then line HA is y=4x+6.
    Since H is the intersection point of BC and HA, that means y=4x+6=(-1/4)x+(1/4).
    To solve that, I am gonna multiply both sides by 4: 16x+24 = -x+1 => 17x = -23, x=-23/17, y=4(-23/17) + 6 = 10/17.
    A?
    Last edited by dennydengler; July 25th 2014 at 03:09 AM.
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