HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)
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someth1ng
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Re: 2012 HSC MX1 Marathon
Errrr...well, I did 2U and 3U - the 2U was fun. The 3U got a bit less fun but 4U...definitely not fun.
bleakarcher
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Re: 2012 HSC MX1 Marathon
Why coz its harder? ^
Why coz its harder? ^
seanieg89
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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.
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.
GingerKid1
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Re: 2012 HSC MX1 Marathon
fukin dumb cunts
fukin dumb cunts
zeebobDD
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Carrotsticks
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Re: 2012 HSC MX1 Marathon
Then solve for m.
Find the discriminant of it, and make it > 0.
Then solve for m.
deswa1
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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
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
Carrotsticks
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Carrotsticks
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Re: 2012 HSC MX1 Marathon
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.
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.
nightweaver066
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Re: 2012 HSC MX1 Marathon
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
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