Preliminary mathematics marathon (1 Viewer)

[DNETTZ]

Man how'd you guys get the x in the denom.

Wolfram told me it was so and so it was.

 

[hscishard]

OMG. Lol

Entrant how did you differentiate y=|x|. SHOWING STEPS.

 

[hscishard]

Fucking made me google signum.
So what. You differentiate y=x and then y=-x.

Differentiate y=|x^3+3x+2|
Good luck finding the signum.

 

[DNETTZ]

y=|x|
Piecemeals a good idea perhaps?

x>1
y=x
y'=1

x<1
y=-x
y'=-1

x=0
y'=?

so
y=(x^2)^1/2 thats abolsute value y=|x|

y'=((2x)*1/2)(y)
y'= y/x
y'=((x^2)^1/2)/x
Thats my proof for wolfram's answer
I just d/dx 'd wrong.

I mean y' = 2x*1/2*(x^2)^(-1/2)
y'=x*1/x
note numerator is special somehow, because i expanded something and it all died. Above the 2x and 1/2 should both have the abby vallies attached to them, so that explains it's special.

so y'=|x|/x?

Perhaps the professionals should take over now before i commit the mathematical holocaust of dividng by 0.

 

[DNETTZ]

Fucking made me google signum.
So what. You differentiate y=x and then y=-x.

Differentiate y=|x^3+3x+2|
Good luck finding the signum.

d/dx (sqrt(x^3+3x+2)^2)

thats the differiencial by above working, should be right i think.
Denominator gives crit value i think (speculating, correct me if I'm wrong?)

Never mind, I AM wrong.

 

[edmundsung]

a simple question which i cannot answer

Show that x^2 + kx + k + 2 =0 cannot have real roots if k lies between 2-2[FONT=新細明體]√3 and 2+2[FONT=新細明體]√3.[/FONT][/FONT]
[FONT=新細明體][FONT=新細明體][/FONT][/FONT]
[FONT=新細明體][FONT=新細明體]thxx[/FONT]
[/FONT]

 

[hscishard]

Re: a simple question which i cannot answer

Show that x^2 + kx + k + 2 =0 cannot have real roots if k lies between 2-2[FONT=新細明體]√3 and 2+2[FONT=新細明體]√3.[/FONT][/FONT]

[FONT=新細明體][FONT=新細明體]thxx[/FONT]
[/FONT]

Unreal roots if discriminant is <0

b^2-4ac < 0
=k^2 -4(k+2) < 0
=k^2-4k -8 < 0

I don't know how to do it the original way but my way..
Completing the sqaure:
(k-2)^2 -8 -4 =0
(k-2)^2 = 12


Since it was a < sign,

^ Values of k where it won't have real roots.

 

[hscishard]

y=|x|
Piecemeals a good idea perhaps?

x>1
y=x
y'=1

x<1
y=-x
y'=-1

x=0
y'=?

so
y=(x^2)^1/2 thats abolsute value y=|x|

y'=((2x)*1/2)(y)
y'= y/x
y'=((x^2)^1/2)/x
Thats my proof for wolfram's answer
I just d/dx 'd wrong.

I mean y' = 2x*1/2*(x^2)^(-1/2)
y'=x*1/x
note numerator is special somehow, because i expanded something and it all died. Above the 2x and 1/2 should both have the abby vallies attached to them, so that explains it's special.

so y'=|x|/x?

Perhaps the professionals should take over now before i commit the mathematical holocaust of dividng by 0.

Isn't signum piecey thing? Lol is it on the syllabus?

Confusing man

 

[edmundsung]

Re: a simple question which i cannot answer

Unreal roots if discriminant is <0

b^2-4ac < 0
=k^2 -4(k+2) < 0
=k^2-4k -8 < 0

I don't know how to do it the original way but my way..
Completing the sqaure:
(k-2)^2 -8 -4 =0
(k-2)^2 = 12


Since it was a < sign,

^ Values of k where it won't have real roots.

yeah i still want a more correct proof
but anyway thank you so much

 

[hscishard]

I wonder if teachers will accept that last few lines. That's how I solve inequations.

I think the usual way goes like this:
Find roots of x axis(Quad forumla, you should get the same 2+/- 2root3)
Test points. Lol very gay method IMO.

Or

Sketch k^2-4k-8<0 (remember dotted lines) Then shade the area where it's below the y axis

Or
Add to my last lines "solving inequation"

Or
Shit. How do you do it the original way

 

[ohexploitable]

Graphing method is best, after you've done lots of them you can visualise the graph and its x intercepts in your head and get the solution immediately.

 

[edmundsung]

another question

If a, b and c are consecutive positive integers show that
ax^2 + bx + c = 0 cannnot have real roots. (hint: write b and c in terms of 'a' and consider the sign of the discriminant as the sign of a quadratic.

my solution so far:

Let b=a+1
c=a+2
therefore delta=b^2 - 4ac
=-3a^2 + 2a - 7

but i don't know what to write after that ><

 

[roadrage75]

Re: another question

If a, b and c are consecutive positive integers show that
ax^2 + bx + c = 0 cannnot have real roots. (hint: write b and c in terms of 'a' and consider the sign of the discriminant as the sign of a quadratic.

my solution so far:

Let b=a+1
c=a+2
therefore delta=b^2 - 4ac
=-3a^2 + 2a - 7

but i don't know what to write after that ><

i don't think you calculated delta correctly....

i think delta = -3a^2 -6a + 1.....

and now all you have to prove is that delta <0 for all positive integers 'a', and ther're many ways to do that!

 

[ohexploitable]

Re: another question

i think delta = -3a^2 -6a + 1.....

Try again.

 

[roadrage75]

Re: another question

Try again.

why don't you?

 

[roadrage75]

Re: another question

Δ=-(3a²-2a+7)

consider, 3a²-2a=a(3a-2)

if a is a positive integer then 3a-2>0

and since a is +ve then a(3a-2)=(+ve)(+ve)=(+ve)

so,

Δ=-(+ve+7)

Δ=-(+ve)

Δ=-ve

.: it doesn't have real roots.

it's b^2 -4ac, not b^2 - 4bc

 

[hscishard]

Re: another question

why don't you?

Ya I got what you got.

 

[roadrage75]

Re: another question

Ya I got what you got.

cool

 

[Entrant]

Re: another question

My Question.

 

[shaon0]

Re: another question

My Question.

Duble integrals aren't in HSC.
If I = S ln(x)/x dx
I=1/2 (ln(x))^2 +C
Use u=ln(x)