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

ohexploitable · Apr 6, 2010

New Question:

Factorise:

$a^{4}+4b^{4}$

Aquawhite · Apr 6, 2010

Answer to above:

$(a^2-2 a b+2 b^2) (a^2+2 a b+2 b^2)$

Sorry, I havn't got a question to continue on

AAEldar · Apr 6, 2010

Aquawhite said:
Answer to above:

$(a^2-2 a b+2 b^2) (a^2+2 a b+2 b^2)$

Sorry, I havn't got a question to continue on

Mind if I post one then? It's not overly difficult, but it can be rather confusing:

Simplify:

$\frac{2x+y}{x-3}+\frac{x-y}{x+3}-\frac{3x+2y}{x^2+x-6}$

ohexploitable · Apr 6, 2010

AAEldar said:
Simplify:

$\frac{2x+y}{x-3}+\frac{x-y}{x+3}-\frac{3x+2y}{x^2+x-6}$

cbf...

I'd do the first 2 bits first and make the denominator (x-3)(x+3)(x-2) etc.

ohexploitable · Apr 6, 2010

New Question:

For what values of x is it true that:

$x\cdot x\cdot x\leq x+x+x$

nonsenseTM · Apr 6, 2010

Aquawhite said:
Answer to above:

$(a^2-2 a b+2 b^2) (a^2+2 a b+2 b^2)$

Sorry, I havn't got a question to continue on

would any one tell me HOW are you supposed to get that?

Trebla · Apr 6, 2010

nonsenseTM said:
would any one tell me HOW are you supposed to get that?

ohexploitable said:
New Question:

Factorise:

$a^{4}+4b^{4}$

Aquawhite said:
Answer to above:

$(a^2-2 a b+2 b^2) (a^2+2 a b+2 b^2)$

Sorry, I havn't got a question to continue on

a⁴+4b⁴ = a⁴+4a²b²+4b⁴ - 4a²b²
= (a² + 2b²)² - 4a²b²
= (a² + 2b² - 2ab)(a² + 2b² + 2ab)

ohexploitable · Apr 6, 2010

nonsenseTM said:
would any one tell me HOW are you supposed to get that?

Complete the square.

nonsenseTM · Apr 6, 2010

thank you so much trebla

hscishard · Apr 7, 2010

ohexploitable said:
New Question:

For what values of x is it true that:

$x\cdot x\cdot x\leq x+x+x$

That questions confusing.

$x^{3}\leq 3x$
$x(x^{2}-3)= 0$
$0, -\sqrt{3} \sqrt{3}$
Numberline testing thing.
$x\leq -\sqrt{3},0\leq x\leq \sqrt{3}$

hscishard · Apr 7, 2010

Simplify

(3x+3)(x+2) - 3(x+2)(x+1)

Lol.

ohexploitable · Apr 7, 2010

hscishard said:
simplify

(3x+3)(x+2) - 3(x+2)(x+1)

lol.

0

hscishard · Apr 7, 2010

ohexploitable said:
0

How long did it take you to do that? Lol

ohexploitable · Apr 7, 2010

New Question:

$\alpha\text{ and }\beta\text{ are the roots of the equation }3x^{2}+3x+(2-p)=0\text{.}\\\text{ Find the value of }p\text{ when }\alpha^{2}+\beta^{2}=\frac{4}{9}\text{.}$

hscishard · Apr 7, 2010

AAEldar said:
Mind if I post one then? It's not overly difficult, but it can be rather confusing:

Simplify:

$\frac{2x+y}{x-3}+\frac{x-y}{x+3}-\frac{3x+2y}{x^2+x-6}$

That better be a 4 mark question. That would take 1 full minute

.

hscishard · Apr 7, 2010

ohexploitable said:
New Question:

$\alpha\text{ and }\beta\text{ are the roots of the equation }3x^{2}+3x+(2-p)=0\text{.}\\\text{ Find the value of }p\text{ when }\alpha^{2}+\beta^{2}=\frac{4}{9}\text{.}$

I did it quickly but is it -1/3?

Edit Ok is it 7/6. Lol didnt see the 2 bit.-.-

Lol final answer. 7/3. Shame. 2 edits.

I give up. Next question please.

ohexploitable · Apr 7, 2010

I got p=2

hscishard · Apr 7, 2010

Second "goes" are always correct. Lol. Following that rule from now on.

hscishard · Apr 7, 2010

ohexploitable said:
I got p=2

It can't be. Lol Then the producy of roots is 0.

I'm always better at maths at night. Weird huh?

ohexploitable · Apr 7, 2010

I wish I paid attention in class

Did you get:

$\begin{align*}\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta&=\frac{4}{9}\\\left(-\frac{b}{a}\right)^{2}-2\left(\frac{c}{a}\right)&=\frac{4}{9}\end{align*}$

???

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

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