Maths thread (2 Viewers)

drsabz101 · Jan 29, 2016

thankyou ^^

axwe7 · Jan 29, 2016

Drsoccerball said:
$u- \sqrt{4} +3 = 0 ?$

$u= \sqrt{4} -3$

lol ?

lolled

Green Yoda · Jan 29, 2016

Drsoccerball said:
pls learn Latex...

Here's one I made (No year 11 + help pls)
$Solve for x$

$4x^4+6x^3+3x^2+6x+4=0$

2x^3(2x+3)3x(x+2)
x= 0,0,-3/2,-2

InteGrand · Jan 29, 2016

leehuan said:
He let u=sqrt(x+1) ---> u^2=x+1

Reduced the equation to something more obviously a quadratic.

$\noindent Oh yeah, missed his square root part (probably wouldn't have missed it if it were in LaTeX). $\Big{(}$And it was $u=\sqrt{x-1}$, not +1.$\Big{)}$

Drsoccerball · Jan 29, 2016

Rathin said:
2x^3(2x+3)3x(x+2)
x= 0,0,-3/2,-2

if you sub in 0 you get 4=0

... try again.

axwe7 · Jan 29, 2016

Drsoccerball said:
pls learn Latex...

Here's one I made (No year 11 + help pls)
$Solve for x$

$4x^4+6x^3+3x^2+6x+4=0$

How would you do that?

axwe7 · Jan 29, 2016

Rathin said:
lel we finished basic arithmetic and algebra in 2 days (1.5 actually cuz we spent half a lesson getting textbooks)

we havent even started wth?

leehuan · Jan 29, 2016

Rathin said:
2x^3(2x+3)3x(x+2)
x= 0,0,-3/2,-2

What? By inspection x=0 is definitely not an answer so your factorisation is confusing

InteGrand said:
$\noindent Oh yeah, missed his square root part (probably wouldn't have missed it if it were in LaTeX). $\Big{(}$And it was $u=\sqrt{x-1}$, not +1.$\Big{)}$

True, got signs muddled again

InteGrand · Jan 29, 2016

axwe7 said:
How would you do that?

$$\noindent Observe that we have a symmetric polynomial, so hint: divide through by $x^2$ (which we can do as $x=0$ is not a solution) and form a quadratic in $x+\frac{1}{x}$.$

Green Yoda · Jan 29, 2016

Drsoccerball said:
if you sub in 0 you get 4=0 ... try again.

I cant find a factor of this...i would normally solve vis long division

Drsoccerball · Jan 29, 2016

InteGrand said:
$$\noindent Observe that we have a symmetric polynomial, so hint: divide through by $x^2$ (which we can do as $x=0$ is not a solution) and form a quadratic in $x+\frac{1}{x}$.$

Drsoccerball said:
pls learn Latex...

Here's one I made (No year 11 + help pls)
$Solve for x$

$4x^4+6x^3+3x^2+6x+4=0$

...

leehuan · Jan 29, 2016

Drsoccerball said:
...

Well to be fair, THIS early into the year a Yr 11 kid would experience difficulty with that question

Drsoccerball · Jan 29, 2016

leehuan said:
Well to be fair, THIS early into the year a Yr 11 kid would experience difficulty with that question

might as well ask them 10+9

leehuan · Jan 29, 2016

Drsoccerball said:
might as well ask them 10+9

Honestly the 2k16ers might want to give it a go but these 2k17ers I would even limit the hardest questions at slightly above elementary quadratics..

Green Yoda · Jan 29, 2016

lol, i give up..ive tried my best

leehuan · Jan 29, 2016

This is absolutely ridiculous

$\\ 4{ x }^{ 2 }+6x+3+\frac { 6 }{ x } +\frac { 4 }{ { x }^{ 2 } } =0\\ Let\quad u=x+\frac { 1 }{ x } \Rightarrow { u }^{ 2 }={ x }^{ 2 }+2+\frac { 1 }{ { x }^{ 2 } } \\ 4{ x }^{ 2 }+6x+8+\frac { 6 }{ x } +\frac { 4 }{ { x }^{ 2 } } =5\\ 4\left( { x }^{ 2 }+2+\frac { 1 }{ { x }^{ 2 } } \right) +6\left( x+\frac { 1 }{ x } \right) -5=0\\ 4{ u }^{ 2 }+6u-5=0\\ u=\frac { -6\pm \sqrt { 116 } }{ 8 } \\ x+\frac { 1 }{ x } =\frac { -3\pm \sqrt { 29 } }{ 2 } \\ { x }^{ 2 }+\left( \frac { 3\pm \sqrt { 29 } }{ 2 } \right) x+1=0\\ x=\frac { -\frac { 3+\sqrt { 29 } }{ 2 } \pm \sqrt { \frac { 11+3\sqrt { 29 } }{ 2 } } }{ 2 } ,\frac { -\frac { 3-\sqrt { 29 } }{ 2 } \pm \sqrt { \frac { 11-3\sqrt { 29 } }{ 2 } } }{ 2 }$

I put it on wolfram as well and some solutions were complex. Mate, don't scare off the new generation like that.

Edit #4: Fixed that mistake as well. Lmao

si2136 · Jan 29, 2016

How is that question Year 11? It contains complex numbers.

InteGrand · Jan 29, 2016

si2136 said:
How is that question Year 11? It contains complex numbers.

It's probably Year 12 level haha.

InteGrand · Jan 29, 2016

Drsoccerball said:
...

What did you mean by "No year 11"? You mean no Year 11's are allowed to do that Q.? (And I tried to put the hint in spoiler tags, but it appears that the spoilers are dodgy on this site.)

sakana · Jan 29, 2016

Rathin said:
I stumbled upon this question and have no idea how to do it.
Express in standard scientific notation form:
The number of cm^2 in 5 hectares, given that 1 hectare = 10^4m^2

Ayyyy cambridge bud

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