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

Drongoski · Mar 1, 2010

My Question:

$(x+1)(x-2)(3+x)\leq0$

Very easy: a quick sketch of the cubic polynomial shows:

x <= -3 or -1 =< x =< 2

ohexploitable · Mar 2, 2010

Drongoski said:
My Question:

$(x+1)(x-2)(3+x)\leq0$

Very easy: a quick sketch of the cubic polynomial shows:

x <= -3 or -1 =< x =< 2

The idea of the 'marathon' is to answer a question and post another.

ohexploitable · Mar 2, 2010

Drongoski said:
Maths Tutoring(Yr Place) - Epping/Beecroft/Eastwood/Carlingford area. PM if interested.

lol, this thread is for Year 11 students...

ohexploitable · Mar 2, 2010

New Question:

$4\leq x^{2}<9$

Drongoski · Mar 2, 2010

ohexploitable said:
lol, this thread is for Year 11 students...

I'm sorry. I forgot it's a marathon.

OK!

Factorise: m⁴ + m² + 25

ohexploitable · Mar 5, 2010

stevey6404 said:
My question
Prove that x^2 + xy + y^2 > 0 for any non-zero values of x and y

$\\a=1 \\b=y \\c=y^{2} \\ \\$Using the quadratic formula:$ \\\begin{align*}b^2-4ac&=y^{2}-4(1)(y^{2}) \\&=y^{2}-4y^{2} \\&=-ve\end{align*} \\ \\$So the parabola is concave up and doesn't cut the x-axis.$ \\\therefore x^{2}+xy+y^{2}>0 \\ \\$amirite?$$

Drongoski · Mar 5, 2010

You are right. An equivalent solution:

$x^{2}\ +\ xy\ +\ y^{2}\ \ =\ \ \left ( x+\frac{y}{2} \right )^{2}\ +\ \frac{3}{4}y^{2}\ > \ 0\ \ \ (since\ x,\ y\ \neq \ 0)$

Now factorise: m⁴ + m² + 25

ohexploitable · Mar 7, 2010

Drongoski said:
Now factorise: m⁴ + m² + 25

$\begin{align*}m^{4}+m^{2}+25&=(m^{2}+5)^{2}-9m^{2}\\&=(m^{2}+5-3m)(m^{2}+5+3m)\\&=(m^{2}-3m+5)(m^{2}+3m+5)\end{align*}$

ohexploitable · Mar 7, 2010

New Question

$x^{3}\geq5x^{2}-4x$

thongetsu · Mar 9, 2010

x^3-5x^2+4x >(+ equal to) 0
x(x^2-5x+4) >(+ equal to) 0
x(x-4)(x-1) >(+ equal to) 0
x= 0,1,4

i think.

thongetsu · Mar 9, 2010

xD :

x^8 = 1

ohexploitable · Mar 10, 2010

thongetsu said:
xD :

x^8 = 1

x=1 or x=-1

ohexploitable · Mar 10, 2010

thongetsu said:
x^3-5x^2+4x >(+ equal to) 0
x(x^2-5x+4) >(+ equal to) 0
x(x-4)(x-1) >(+ equal to) 0
x= 0,1,4

i think.

Nah,

$0\leq x\leq \frac{4}{5},x\geq4$

mirakon · Mar 10, 2010

ohexploitable said:
nah,

$0\leq x\leq \frac{4}{5},x\geq4$

4/5?

ohexploitable · Mar 10, 2010

mirakon said:
4/5?

Oh wait lol...

$0\leq x\leq 1,x\geq4$

How did I get 4/5???

thongetsu · Mar 10, 2010

lol i didnt wanna do further working

ohexploitable · Mar 10, 2010

New Q:

Rationalise:

$\frac{1}{\sqrt[3]{2}-1}$

mirakon · Mar 10, 2010

ohexploitable said:
new q:

rationalise:

$\frac{1}{\sqrt[3]{2}-1}$

(2^(2/3) - 2^(1/3) +1)

ohexploitable · Mar 10, 2010

You're meant to post a new question.

New Q.

Simplify:

(x – a)(x – b)(x – c) . . . (x – z)

mirakon · Mar 10, 2010

Ans: 0

Also can't be bothered to post my own question, someone pls do it for me Thanks!

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

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