basic algebra and arithmetic "marathon". (9 Viewers)

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". (9 Viewers)

hey

Retiring

Premium Member

hey

hey

Member

Administrator

hey

Member

Active Member

Active Member

hey

Active Member

hey

Active Member

Active Member

hey

Active Member

Active Member

hey

Users Who Are Viewing This Thread (Users: 0, Guests: 9)