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Homework Help: Three quizzes from our exam(junior middle school grade 3)

  1. Aug 23, 2004 #1
    #1, x , y are real number, x>=y>=1, 2*x^2-xy-5x+y+4=0,then x+y=?
    #2,look at the Attach image,ABCD is a square,and EFGH is a quadrangle,
    angle BEG and angle CFH are both sharp angles. EG=3(units),FH=4(units),
    the acreage of EFGH is 5, question: what is the acreage of ABCD?
    #3 the quadratic equation about x is (k^2-6k+8)*x^2+(2k^x-6k-4)*x+k^2=4
    and the two roots of x are integer,question: find out all the value of k
    (k is a real number).

    Attached Files:

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  2. jcsd
  3. Aug 23, 2004 #2
    Nobody is going to solve the questions for you, how about you post how far you got and where you a re stuck.
  4. Aug 23, 2004 #3
    What is your grade 3 equivalent to in the US system, or rather, how old are you? I WISH I was doing this stuff in US 3rd grade...
  5. Aug 23, 2004 #4
    Is there any thing unique about sharp angles?
  6. Aug 23, 2004 #5


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    Hint Hint, quadratics tend to have a highest power of _

    Otherwise it's a geometric... and then some.
  7. Aug 23, 2004 #6
    I would wish as well. These questions are a lot more challenging than the ones at my highschool.

    I hope somebody can post a better hint to these problems, because i am interested in how to solve them.
    Last edited: Aug 23, 2004
  8. Aug 23, 2004 #7
    Hint for 1: Graph the equations you have:
    (1) x>=y
    (2) y>=1
    (3) 2x2-xy-5x+y+4=0

    Now do it algebraically...
  9. Aug 23, 2004 #8

    Are you sure the second term of question #3 is [tex](2k^x-6k-4)x[/tex] ?
  10. Aug 23, 2004 #9


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    #1 ) There is an infinite set of solutions, given by x+y = 2*x^2-xy-6x+2y+4. That seems a funny question to ask.

    #3) k^x seems unlikely.
  11. Aug 24, 2004 #10
    For the above posts , the #3) k^x is k^2,i am sorry i typed wrong.
    i am 17 now,but the three questions were in one exam when i was 16 years old.
    i'd solved the 3 questions in my exam when i was 16,and i guess if you can do it.
    just quizzes of chinese stype,i hope you like it.just tell me your answer!
  12. Aug 24, 2004 #11
    if you want the answers,you can send me a message to get.
  13. Aug 24, 2004 #12


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    #1 ) I didn't notice the condition x>=y before...So the only real solution is (2,2) so x+y=4.
  14. Aug 24, 2004 #13
    Using [tex]b^2-4ac \geq 0 [/tex] because the roots is a real number,i found that k equals to any positive real number includes zero.

    since nicky has done these questions before, why don't you post the answers here so anyone can give it a try ?
    Last edited: Aug 24, 2004
  15. Aug 24, 2004 #14


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    Leong, the question says that the roots are both integers. B^2 > 4AC is a necessary condition, but not sufficient. And clearly, k=0 does not give 2 integer roots (x = 1, and x = -1/2).
  16. Aug 24, 2004 #15


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    #2) I get Area(ABCD) = 44/5 = 8.8 sq. units. I hope I'm not making some stupid mistake...
  17. Aug 24, 2004 #16
  18. Aug 25, 2004 #17
    i gave up. i guess i am not as good as you guys.
  19. Aug 25, 2004 #18


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    Hmmm...I never did problems like these till I got to high school. If you do them in junior-middle school, you folks have a really high standard in math.

    Wait, when I was 16, I was in high school - I guess the nomenclature is different.
    Last edited: Aug 25, 2004
  20. Aug 25, 2004 #19
    Solution to question #1

    Rearrange :
    [tex] y=\frac{2x^2-5x+4}{x-1}[/tex]
    Plot graph y(x).
    Consider the inequality conditions given.
    1. [tex] y\geq1[/tex]
    2. [tex] x\geq1[/tex]
    3. [tex]x\geq y[/tex]

    These conditions give point (2,2) as the only solution. the answer = x+y=2+2=4.
  21. Aug 25, 2004 #20
    Solution to question #2

    Let x=length of the side of the square


    Formula for area of a quadrangle = [tex]\frac{1}{2}d_{1}d_{2}sin\theta[/tex] with
    [tex]d_{1}[/tex]=length of the first diagonal
    [tex]d_{2}[/tex]= length of the other diagonal
    [tex] \theta[/tex]=angle between the diagonals

    [tex]sin \theta = \frac{5}{6}[/tex]
    [tex]cos \theta = -\frac{\sqrt{11}}{6}[/tex]
    [tex]\theta[/tex] is an obtuse angle

    [tex]\alpha + \beta + \theta = \pi[/tex]
    [tex] cos \beta =sin \theta sin\alpha - cos \theta cos\alpha[/tex]
    [tex] cos\beta=\frac{1}{6}[5sin\alpha+\sqrt{11}cos\alpha][/tex].........(2)

    Substitute (2) into (1); get

    Area of the square = [tex](4sin\alpha)^2[/tex]
    = [tex]\frac{44}{8}[/tex]
    = 8.8
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