# Three quizzes from our exam(junior middle school grade 3)

1. Aug 23, 2004

### NickyuTse

#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?
and the two roots of x are integer,question: find out all the value of k
(k is a real number).

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2. Aug 23, 2004

Nobody is going to solve the questions for you, how about you post how far you got and where you a re stuck.

3. Aug 23, 2004

### K.J.Healey

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...

4. Aug 23, 2004

### needhelpperson

Is there any thing unique about sharp angles?

5. Aug 23, 2004

### Alkatran

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

Otherwise it's a geometric... and then some.

6. Aug 23, 2004

### needhelpperson

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
7. Aug 23, 2004

### zefram_c

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...

8. Aug 23, 2004

### Leong

Are you sure the second term of question #3 is $$(2k^x-6k-4)x$$ ?

9. Aug 23, 2004

### Gokul43201

Staff Emeritus
#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.

10. Aug 24, 2004

### NickyuTse

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!

11. Aug 24, 2004

### NickyuTse

if you want the answers,you can send me a message to get.

12. Aug 24, 2004

### Gokul43201

Staff Emeritus
#1 ) I didn't notice the condition x>=y before...So the only real solution is (2,2) so x+y=4.

13. Aug 24, 2004

### Leong

Using $$b^2-4ac \geq 0$$ 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
14. Aug 24, 2004

### Gokul43201

Staff Emeritus
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).

15. Aug 24, 2004

### Gokul43201

Staff Emeritus
#2) I get Area(ABCD) = 44/5 = 8.8 sq. units. I hope I'm not making some stupid mistake...

16. Aug 24, 2004

### NickyuTse

17. Aug 25, 2004

### Leong

i gave up. i guess i am not as good as you guys.

18. Aug 25, 2004

### Gokul43201

Staff Emeritus
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
19. Aug 25, 2004

### Leong

Solution to question #1

$$2x^2-xy-5x+y+4=0$$
Rearrange :
$$y=\frac{2x^2-5x+4}{x-1}$$
Plot graph y(x).
Consider the inequality conditions given.
1. $$y\geq1$$
2. $$x\geq1$$
3. $$x\geq y$$

These conditions give point (2,2) as the only solution. the answer = x+y=2+2=4.

20. Aug 25, 2004

### Leong

Solution to question #2

Let x=length of the side of the square

$$x=4sin\alpha$$
$$x=3cos\beta$$
$$4sin\alpha=3cos\beta$$......(1)

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

$$\frac{1}{2}(3)(4)sin\theta=5$$
$$sin \theta = \frac{5}{6}$$
$$cos \theta = -\frac{\sqrt{11}}{6}$$
$$\theta$$ is an obtuse angle

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

Substitute (2) into (1); get
$$tan\alpha=\frac{\sqrt{11}}{3}$$
$$sin\alpha=\frac{\sqrt{11}}{2\sqrt{5}}$$

Area of the square = $$(4sin\alpha)^2$$
= $$\frac{44}{8}$$
= 8.8