basic algebra and arithmetic "marathon". (1 Viewer)

ohexploitable · Apr 7, 2010

hscishard said:
Should I buy the Year 11 ext book? Or keep hogging it by borrowing from the local library?

I kept renewing the book for 3 months. They didn't catch me until last week

Here you go:

3unitYR11.pdf

hscishard · Apr 7, 2010

ohexploitable said:
Here you go:

3unitYR11.pdf

OMFG. TY,Thankyou,Merci,Xie Xie,Grazie,Arigato. Thats all I know, but you get the point

.

. This is not piracy

ohexploitable · Apr 7, 2010

New Question:

$\text{Find }x\text{ in terms of }a\text{ and }b\text{ if }\frac{a^{2}b}{x^{2}}+\left(1+\frac{b}{x}\right)a=2b+\frac{a^{2}}{x}\text{.}$

Also has anyone noticed that the Cambridge textbooks use the same font as LaTeX?

Amogh · Apr 8, 2010

ohexploitable said:
New Question:

$\text{Find }x\text{ in terms of }a\text{ and }b\text{ if }\frac{a^{2}b}{x^{2}}+\left(1+\frac{b}{x}\right)a=2b+\frac{a^{2}}{x}\text{.}$

Also has anyone noticed that the Cambridge textbooks use the same font as LaTeX?

LOL they do too
Btw, I reckon you should take the book off before you get into some copyright bs

ohexploitable · Apr 8, 2010

Amogh said:
LOL they do too
Btw, I reckon you should take the book off before you get into some copyright bs

Yar, it's set to private.

NewiJapper · Apr 9, 2010

Any chance of 2 Unit questions in here?

ohexploitable · Apr 9, 2010

NewiJapper said:
Any chance of 2 Unit questions in here?

This is 2 unit.

NewiJapper · Apr 9, 2010

hscishard · Apr 9, 2010

He gets the hardest questions in two unit.
It's been two days, noone has still answered that question.

I'm a senior now. Damn, I sound old.

ohexploitable · Apr 9, 2010

New Question:

$\text{Find }x\text{ in terms of }a\text{ and }b\text{ if }\frac{a^{2}b}{x^{2}}+\left(1+\frac{b}{x}\right)a=2b+\frac{a^{2}}{x}\text{.}$

Also has anyone noticed that the Cambridge textbooks use the same font as LaTeX

Since no one has answered it...

$\begin{align*} \frac{a^{2}b}{x^{2}}+\left(1+\frac{b}{x}\right)a&=2b+\frac{a^{2}}{x} \\\frac{a^{2}b}{x^{2}}+a+\frac{ab}{x}&=2b+\frac{a^{2}}{x} \\a^{2}b+ax^{2}+abx&=2bx^{2}+a^{2}x \\a^{2}b+ax^{2}+abx-2bx^{2}-a^{2}x&=0 \\ax^{2}-2bx^{2}+abx-a^{2}x+a^{2}b&=0 \\(a-2b)x^{2}+(ab-a^{2})x+a^{2}b&=0\text{ (in the form } ax^{2}+bx+c) \end{align*}$

$\begin{align*} \\x&=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \\x&=\frac{-(ab-a^{2})\pm\sqrt{(ab-a^{2})^{2}-4(a-2b)(a^{2}b)}}{2(a-2b)} \\x&=\frac{a^{2}-ab\pm\sqrt{a^{2}(b-a)^{2}-4a^{2}b(a-2b)}}{2(a-2b)} \\x&=\frac{a(a-b)\pm\sqrt{a^{2}\left[(b-a)^{2}-4b(a-2b)\right ]}}{2(a-2b)} \\x&=\frac{a(a-b)\pm a\sqrt{(b-a)^{2}-4b(a-2b)}}{2(a-2b)} \\x&=\frac{a(a-b)\pm a\sqrt{b^{2}-2ab+a^{2}-4ab+8b^{2}}}{2(a-2b)} \\x&=\frac{a(a-b)\pm a\sqrt{a^{2}-6ab+9b^{2}}}{2(a-2b)} \\x&=\frac{a(a-b)\pm a\sqrt{(a-3b)^{2}}}{2(a-2b)} \\x&=\frac{a(a-b)\pm a(a-3b)}{2(a-2b)} \end{align*}$

$\begin{align*} \\\text{so} \\x&=\frac{a(a-b)+a(a-3b)}{2(a-2b)} \\x&=\frac{a(a-b+a-3b)}{2(a-2b)} \\x&=\frac{a(2a-4b)}{2(a-2b)} \\x&=\frac{a(2a-4b)}{2a-4b} \\x&=a \end{align*}$

$\begin{align*} \\\text{or} \\x&=\frac{a(a-b)-a(a-3b)}{2(a-2b)} \\x&=\frac{a(a-b-a+3b)}{2(a-2b)} \\x&=\frac{a(2b)}{2(a-2b)} \\x&=\frac{2ab}{2(a-2b)} \\x&=\frac{ab}{a-2b} \end{align*}$

$\boxed{\therefore x=a\text{ or }\frac{ab}{a-2b}}$

AAEldar · Apr 9, 2010

Wow, that's why I couldn't do it, I didn't even think of the quadratic formula. A lot of working though! Haha.

ohexploitable · Apr 9, 2010

AAEldar said:
Wow, that's why I couldn't do it, I didn't even think of the quadratic formula. A lot of working though! Haha.

I did like full working, it could've been condensed into half that.

ohexploitable · Apr 9, 2010

New Question:

Not nearly as hard as before lol

$\frac{3t}{t^{2}-6}=\sqrt{3}$

hscishard · Apr 9, 2010

Exploit, how bothered are you? Lol

ohexploitable · Apr 9, 2010

hscishard · Apr 9, 2010

ohexploitable said:
New Question:

Not nearly as hard as before lol

$\frac{3t}{t^{2}-6}=\sqrt{3}$

Gay, why couldn't it be +6.

Wait it's not an inequality. Lol

hscishard · Apr 9, 2010

It's 2 root 3 and -root 3.

Why doesn't Latex work

Bring everything to one side.
Make it into one single fraction.
Expand brackets
Factorise with negative root 3
Divide negative root 3 both sides
Use quadratic forumla to solve
Simplify both answers.

Stuff you latex

bouncing · Apr 9, 2010

hscishard said:
It's 2 root 3 and -root 3.

Why doesn't Latex work
Bring everything to one side.
Make it into one single fraction.
Expand brackets
Factorise with negative root 3
Divide negative root 3 both sides
Use quadratic forumla to solve
Simplify both answers.

Stuff you latex

are you sure.. i didnt get that

exploit wahts the answer

i got 11/2root3 and -6/2root3
using quadratic formula....

mioumiou · Apr 9, 2010

hscishard said:
It's 2 root 3 and -root 3.

Why doesn't Latex work
Bring everything to one side.
Make it into one single fraction.
Expand brackets
Factorise with negative root 3
Divide negative root 3 both sides
Use quadratic forumla to solve
Simplify both answers.

Stuff you latex

Well, I got the same answer as you.....

hscishard · Apr 9, 2010

bouncing said:
are you sure.. i didnt get that

exploit wahts the answer

i got 11/2root3 and -6/2root3
using quadratic formula....

Your second one is -root 3 btw. Lol Just rationalise and simplified.
When subbing 11/2root3.. did you try that?

basic algebra and arithmetic "marathon". (1 Viewer)

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