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How do you solve the cubic equations like 8x^3+4x^2-4x-1=0 and y^3+4y^2+3y-1=0 withour a calculator? 

$$\\8x^3+4x^2-4x-1=0\\$$

this one is not jumping out at me.    

Here is the graph showing the 3 real solutions

ROF LOL 

i think i will tell about it in a post!overall.....it wasnt a disasterjust something "embarrasing " happened!

its good zegroes is here......but how can he be back...he never left did he.....i mean he didnt tell us and go!haha

I'm laughing my head off BJ  :))

I like your answer too Badinage, I am sure many husbands would agree with you :))

Well now you can ask her yourself CPhill :))))

f(t)=10+76e^-.693(1/6)

maybe

10+76 x [shift] [ln] -0.693^(1/6)=

Yes, maybe you came back at a good time Rosala.

Although yesterday moring (my time) was a lot of fun and you were not here :/

Hi Rosala,

I trust there were no disasters at the wedding :))

Zegroes is back, how about that, you will have to be around more too now :))))

Where is the original thread?         

I already did Zac :)

Yes this conversation is entertaining :)

I would teach permutations when i teach probability.  

I would put it with probability

Yes and 3 goes into 3 once

Thanks MG,

I didn't know that there were dummy books for penguins to learn how to fly.

This is a very useful thing for me to know  

Hi Zac,

Thanks for answering these questions :)

There is a couple of things here.

1) I think it should be       $$10t^2+5t-2=0\qquad (NOT -5t)$$

2) I am not sure what the asker INTENDED the question to be BUT this was the question.

1/2+2/5t-1=1/5t+t

Technically this should be interpreted as

$$\\\frac{1}{2}+\frac{2}{5}t-1=\frac{1}{5}t+t\\\\
\frac{1}{2}+\frac{2t}{5}-1=\frac{1t}{5}+t\qquad $The t is on the top$\\\\
\frac{-1}{2}+\frac{2t}{5}=\frac{6t}{5}\\\\
\frac{-1}{2}=\frac{4t}{5}\\\\
\frac{-1*5}{2*4}=t\\\\
t=\frac{-5}{8}\\\\$$

Also, I would like you to consider putting in some blank lines.  

Sometimes mine is spread out too much, I realize that, but I personally find it a bit hard to read when there are no spaces.  This may just be a peronal preference.  I am not sure :/

I'm not sure if it's right, but I ended up with a quadratic by the end. Here goes:

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,-\,}}{\mathtt{1}} = {\frac{{\mathtt{1}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,\small\textbf+\,}}{\mathtt{t}}$$ Take the 1/5t off both sides.

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{t}}$$ Multiply everything by 10t.

$${\frac{\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}\right)}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}} = {\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}$$ Simplify.

$${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{10}}{\mathtt{\,\times\,}}{\mathtt{t}} = {\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}$$

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{0}}$$ Now using the quadratic formula,

$$t=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ or $$t=\frac{-b-\sqrt{b^2-4ac}}{2a}$$

a=10, b=-5, c=-2

$$t=\frac{5+\sqrt{5^2-4\times{10}\times{-2}}}{2\times{10}}$$ or $$t=\frac{5+\sqrt{5^2-4\times{10}\times{-2}}}{2\times{10}}$$

$$t=\frac{5+\sqrt{25+80}}{20}$$ or $$t=\frac{5-\sqrt{25+80}}{20}$$

$$t=\frac{5+\sqrt{105}}{20}$$ or $$t=\frac{5-\sqrt{105}}{20}$$

which works out to $$\approx$$ 0.76235 or -0.26235.

P.S. I don't know how to say "plus minus" something.

$$(x+12)\times{4}=x+60$$

$$4x+48=x+60$$

$$3x=12$$

$$x=4$$

If it's an exam question, often it has been designed so you'll be able to see one solution by inspection, e.g., try various easy possibilities such as ½, ¼, 1, 2, 3, 5, 10, and their negatives. Once you know one solution, call it k, then use division by (x-k) to end up with a quadratic and solve that using the memorized formula.

If the problem hasn't been designed for easy solution, then I'm afraid you're SOOL.

☹   ☹ ☹

You should have added this question to the original thread.

A plate with +++ charges has the higher potential than one with - - -  charges; "higher" means more positive potential.

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I can only make some sense of your question. I'm assuming he can paint a room with 3 fiths of a can of paint.

P=paint cans, r=rooms

$$\frac{3}{5}p:1r$$  Now we need to make p=1. We can do this by dividing both sides by 3/5.

(Note that $$\frac{3}{5}p:1r$$ can also be written as $$3p:5r$$)

$$1p:1\frac{2}{3}r$$ So if we have 33 paint cans (multiply both sides by 33), he can paint. . .

$$33p:55r$$

55 rooms.

I'm assuming this is what you mean . . .

$$19x=2+55000$$

$$19x=55002$$ 

Now divide each side by 19 to make x the subject.

$$x=2894.842$$ (3 decimal places) or $$2894\frac{16}{19}$$

Wow everyone's busy I just returned how is this forum going! Crazy becoz a godzilla just arrived!

hi, zegroes!