HSC 2012 MX1 Marathon #1 (archive) (2 Viewers)

[IamBread]

Re: 2012 HSC MX1 Marathon

I would have to disagree with this bit - Extension 2 isn't fun. lol.

D: Ext 2 is very fun!

 

[someth1ng]

Re: 2012 HSC MX1 Marathon

D: Ext 2 is very fun!

Errrr...well, I did 2U and 3U - the 2U was fun. The 3U got a bit less fun but 4U...definitely not fun.

 

[bleakarcher]

Re: 2012 HSC MX1 Marathon

Why coz its harder? ^

 

[seanieg89]

Re: 2012 HSC MX1 Marathon

Wouldn't necessarily call any of the school courses "fun" but at least 4U had the potential for some of the questions to be interesting. Higher mathematics is a lot more "fun" because there is room for creativity and originality.

 

[Examine]

Re: 2012 HSC MX1 Marathon

Though what do you do with tan260?

 

[Examine]

Re: 2012 HSC MX1 Marathon

nvm I geddit now

 

[Examine]

Re: 2012 HSC MX1 Marathon

Hmm, I seem to be having a lot of trouble with these. How did you guys learn how to do addition, subtraction, multiplication and division of trigonometry?

 

[GingerKid1]

Re: 2012 HSC MX1 Marathon

fukin dumb cunts

 

[zeebobDD]

Re: 2012 HSC MX1 Marathon

fukin dumb cunts

ahaha your a hektik brah

 

[Examine]

Re: 2012 HSC MX1 Marathon

Find the values of m for which the equation 4x^2-mx+9=0 has real roots.

^Need help

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Find the values of m for which the equation 4x^2-mx+9=0 has real roots.

^Need help

Find the discriminant of it, and make it > 0.

Then solve for m.

 

[deswa1]

Re: 2012 HSC MX1 Marathon

You have real roots when the disriminant is greater than or equal to zero...

<a href="http://www.codecogs.com/eqnedit.php?latex=\Delta =m^2-4(4)(9) \\ \Delta \geq 0\\ \therefore m^2-144\geq 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\Delta =m^2-4(4)(9) \\ \Delta \geq 0\\ \therefore m^2-144\geq 0" title="\Delta =m^2-4(4)(9) \\ \Delta \geq 0\\ \therefore m^2-144\geq 0" /></a>

You can solve the rest

 

[Examine]

Re: 2012 HSC MX1 Marathon

x^2+(k+6)x-2k=0 Find the discriminant.

This is my working out

(k+6)^2-4(1)(-2k)

k^2+12k+36-4+8k

k^2+20k+32=0 <-Is this the discriminant?

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

x^2+(k+6)x-2k=0 Find the discriminant.

This is my working out

(k+6)^2-4(1)(-2k)

k^2+12k+36-4+8k

k^2+20k+32=0 <-Is this the discriminant?

Check your working out in this line.

 

[Examine]

Re: 2012 HSC MX1 Marathon

Oh I see its k^2+12k+36=0

 

[Examine]

Re: 2012 HSC MX1 Marathon

Progress making = win

 

[Examine]

Re: 2012 HSC MX1 Marathon

Find the values of k for which the quad eqn x^2-(k+3)x+(k+6)=0

My working
(K+3)^2-4(1)(K+6)
k^2+6k+9-4k-24
K^2+2k-15
(k+5)(k-3)

Answers are -5,3

Right/wrong?

 

[Carrotsticks]

Re: 2012 HSC MX1 Marathon

Find the values of k for which the quad eqn x^2-(k+3)x+(k+6)=0

My working
(K+3)^2-4(1)(K=6)
k^2+6k+9-4k-24
K^2+2k-15
(k+5)(k-3)

Answers are -5,3

Right/wrong?

Not quite, you have to use the quadratic equation to solve for x.

Then make the inside of the square root (from the quadratic equation) > or equal to 0.

You just found the values of K such that the quadratic eqn has 1 real root.

 

[Examine]

Re: 2012 HSC MX1 Marathon

Not quite, you have to use the quadratic equation to solve for x.

Then make the inside of the square root (from the quadratic equation) > or equal to 0.

You just found the values of K such that the quadratic eqn has 1 real root.

I'm not sure if I follow :/

 

[nightweaver066]

Re: 2012 HSC MX1 Marathon

I'm not sure if I follow :/

You had the discriminant, but you let it = 0 and solved for values of k.

When the discriminant = 0, you find a quadratic equation with 1 real root.

If discriminant is > 0, you look for values of k such that it has real roots.

So you are solving the equation:
(k + 5)(k - 3) > 0