2012 Year 9 &10 Mathematics Marathon (1 Viewer)

kazemagic · Sep 11, 2012

Fawun said:
You shouldn't.

Nope. I can't even do Spirals question let alone yours LOL

Yeah I know circle geo

soon your going to realise that u dont

kazemagic · Sep 11, 2012

btw enoilgam for question 6b where it talks about the ratio to the unshaded area. does that include the unshaded area in both the circle and the square?

Fawun · Sep 11, 2012

kazemagic said:
soon your going to realise that u dont

No I do. Fack off

SpiralFlex said:
Fine. What question would you like next?

Anything in this list would be fine:

- Polynomials
- Logs
- Quadratics
- Trig
- Surds
- Graphs

anything besides

- coordinate geometry
- probability
- congruence and similarity
- statistics

and i'm fine. Although I only know how to do the substitution method for simultaneous equations.

enoilgam · Sep 11, 2012

kazemagic said:
btw enoilgam for question 6b where it talks about the ratio to the unshaded area. does that include the unshaded area in both the circle and the square?

It only includes the square.

kazemagic · Sep 12, 2012

enoilgam said:
It only includes the square.

wow mind fk question

D94 · Sep 12, 2012

kazemagic said:
wow mind fk question

If you draw separate diagrams with shaded areas and think of it as adding or subtracting areas then it becomes rather straightforward. Think of it as if you have those see through overhead sheets, and you're putting them on top of each other. One has a circle, one has a square, one has that diamond-ish shape etc., and you're just adding/subtracting areas.

Sy123 · Sep 12, 2012

$$i) Solve the equation using completing the square method$ \\ \\ ax^2+bx+c=0$

Next question

$$Given that the equation$ \\ (x-\alpha)(x-\beta)=x^2+bx+c \\ \\ $Find$ \\ $ii)$ \ \alpha \beta \\ $iii) \alpha + \beta \\ \\ \text{Find both in terms of b and c}$

Fawun · Sep 12, 2012

Sy123 said:
$$i) Solve the equation using completing the square method$ \\ \\ ax^2+bx+c=0$

Next question

$$Given that the equation$ \\ \\ \\ $Find$ \\ $ii)$ \ \alpha \beta \\ $iii) \alpha + \beta \\ \\ \text{Find both in terms of b and c}$

1. $ax^2+bx+c=0$

$x^2+\frac{bx}{a}+\frac{c}{a}=0$

$x^2+\frac{bx}{a}=-\frac{c}{a}$

$x^2+\frac{b}{a}x=-\frac{c}{a}$

$x^2+\frac{b}{a}x+ (\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2$

$(x+\frac{b}{2a})^2=\frac{4ac}{4a^2}+\frac{b^2}{4a^2}$

$(x+\frac{b}{2a})^2=\frac{b^2+4ac}{4a^2}$

$(x+\frac{b}{2a})=\sqrt{\frac{b^2+4ac}{4a^2}}$

$(x+\frac{b}{2a})=\frac{\sqrt{b^2+4ac}}{2a}$

$(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2+4ac}}{2a}$

$x=\frac{-b\pm \sqrt{b^2+4ac}}{2a}$

Turns out to be the quadratic formula

2. Okay this is a bit more tricky so i'm going to guess this one lol (i'm going to get it wrong but i'm trying!)

$(x-\alpha)(x-\beta)=x^2+bx+c$

$x^2-x\beta -\alpha x+\alpha \beta =x^2+bx+c$

Am I on the right track?

SpiralFlex · Sep 12, 2012

Fawun said:
1. $ax^2+bx+c=0$

$x^2+\frac{bx}{a}+\frac{c}{a}=0$

$x^2+\frac{bx}{a}=-\frac{c}{a}$

$x^2+\frac{b}{a}x=-\frac{c}{a}$

$x^2+\frac{b}{a}x+ (\frac{b}{2a})^2=-\frac{c}{a}+(\frac{b}{2a})^2$

$(x+\frac{b}{2a})^2=\frac{4ac}{4a^2}+\frac{b^2}{4a^2}$

$(x+\frac{b}{2a})^2=\frac{b^2+4ac}{4a^2}$

$(x+\frac{b}{2a})=\sqrt{\frac{b^2+4ac}{4a^2}}$

$(x+\frac{b}{2a})=\frac{\sqrt{b^2+4ac}}{2a}$

$(x+\frac{b}{2a})=\frac{\pm \sqrt{b^2+4ac}}{2a}$

$x=\frac{-b\pm \sqrt{b^2+4ac}}{2a}$

Turns out to be the quadratic formula

2. Okay this is a bit more tricky so i'm going to guess this one lol (i'm going to get it wrong but i'm trying!)

$(x-\alpha)(x-\beta)=x^2+bx+c$

$x^2-x\beta -\alpha x+\alpha \beta =x^2+bx+c$

Am I on the right track?

Keep going young fawn sy you shouldn't have made it monic for fawns intuition

deswa1 · Sep 12, 2012

Lol I thought Spiral answer the first one and I was like WTF- this is a year 10 thread. Well done Fawun on the first one. And yeah- you're on the right track for the second

kazemagic · Sep 12, 2012

deswa1 said:
Lol I thought Spiral answer the first one and I was like WTF- this is a year 10 thread. Well done Fawun on the first one. And yeah- you're on the right track for the second

ye same

SpiralFlex · Sep 12, 2012

Impressed how quickly you picked up LaTeX. Faster than I when I was your age young Fawn.

kazemagic · Sep 12, 2012

heres a good one

Fawun · Sep 12, 2012

SpiralFlex said:
Keep going young fawn sy you shouldn't have made it monic for fawns intuition

I don't know what to do next

Can you give me a HINT? Because I don't want the answer since I want to do it myself without any help.

Sy123 · Sep 12, 2012

HINT: (concept)

ak+bl=ck+dl
Therefore
a=c

b=d

EDIT: For the quadratic question it should be b^2 - 4ac lol,
silly mistake (and alot of people missed it too heh)

Fawun · Sep 29, 2012

Re: Years 9 & 10 2012 Chit Chat Thread

HSC2014 said:
What's everyone doing at this timeeee. Someone chuck me a challenging math question, so bored

enoilgam said:
View attachment 26350

I thought this was a pretty tough question - its from a Year 9 5.3 mid-semester exam.

⁮

enoilgam · Sep 29, 2012

Re: Years 9 & 10 2012 Chit Chat Thread

Fawun said:
⁮

Seriously?

kazemagic · Sep 29, 2012

Re: Years 9 & 10 2012 Chit Chat Thread

Fawun said:
⁮

u still havent solved that yet have u

Fawun · Sep 29, 2012

Re: Years 9 & 10 2012 Chit Chat Thread

enoilgam said:
Seriously?

?

kazemagic said:
u still havent solved that yet have u

No i haven't and I still have no clue as to how

and wot of it

HSC2014 · Sep 29, 2012

Re: Years 9 & 10 2012 Chit Chat Thread

Yep just got A, interesting question (Did you want solutions?). I'm suprised that is even year 9 level. Drew the diagram wrong which means I wasted like 10mins lol.. (forgot the part between petal and corner of box). Moving onto two!

Edit: And q2 wasn't hard to get after you did q1.

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