Simple Math Practice Problems

[MarkT]

It is soluble; there are two solutions.

Go for it, Tim

[Mystic-Al-Bob]

good luck at all!!

complex number: a + b*i; i^2 = -1; a and b are real numbers

[Argalius]

I hate working with I

[Klaas]

a^2 + b^2 = -10

3a - b = 0

(a^2 + b^2)/10 = -1

=> (a^2 + b^2)/10 = i^2

3a - b = 0

=> b = 3a; a = b/3

=> (a^2 + 9a^2)/10 = i^2

=> a^2 = i^2

=> a = i

=> b = 3i

or

=> a = -i

=> b = -3i

Not sure if that's right, the only thing I remembered is that i^2 = -1, so I went further on that.

[Mystic-Al-Bob]

that's right. good work.

[Klaas]

Cool

Well I'm not good at asking questions, so anyone else who likes to ask one, go ahead

[DarkAngelBGE]

Nice one. Too bad I read the solution before I got my hand at finding the solution. Maybe you ask a similar question, Klaas ?

[Mystic-Al-Bob]

I have another one:

a^2 + b^2 = 0

c^2 - a = 3

4*(c^2) - 3b = 16i

[DarkAngelBGE]

Gah, I won't do this till the end, but here is the stuff that I have now:

a^2 + b^2 = 0 (1)

c^2 - a = 3 <=> a = c^2 / 3 (2)

4*(c^2) - 3b = 16i <=> b = ( 4c^2 - 16i ) / 3 (3)

(1) ^ (2) ^ (3) =>

( c^2 / 3 )^2 + ( (4c^2 - 16i ) / 3 ) ^2 = 0

<=> c^4 / 9 + ( 16c^4 - 128c^2 i^2 + 256i^2 ) / 9 = 0

<=> 17c^4 + 128c^2 - 256 = 0

<=> c^4 + 128/17 c^2 - 256/17 = 0

<=> (c^2 + 64/17)^2 - 8448/17^2 = 0

<=> (c^2 + 64/17)^2 = 8448/17^2

Now do sqrt(), then find out what c is. Put the value that you get for c into (2) and (3) and you get a and b.

Not enough time to complete, sorry.

[Mystic-Al-Bob]

looks very difficult

[DarkAngelBGE]

Actually it is not very difficult. Well okay, let me complete the thing then.

....

<=> (c^2 + 64/17)^2 = 8448/17^2

<=> c^2 + 64/17 = sqrt( 8448/17^2 ) V c^2 + 64/17 = -sqrt( 8448/17^2 )

<=> c = sqrt( sqrt( 8448/17^2 ) -64/17 ) V c^2 = sqrt( -sqrt ( 8448/17^2 ) -64/17 )

=> c = (rounded) 1.28 V c = sqrt( -9.17 )

Well, next time I find some time I will post all the squareroots of -9.17.

[Mystic-Al-Bob]

that looks wrong

c=2+i

maybe you get the other one (a and

[Klaas]

I don't really find it either, I just end up with the same long thing as Tim

Btw you made mistake in the beginning, so if you fix that your sollution could be right?

c^2 - a = 3 <=> a = c^2 / 3 (2)

it's:

c^2 - a = 3 <=> a = c^2 - 3

[Mystic-Al-Bob]

so tim should try again. maybe he will do it

[DarkAngelBGE]

Stupid mistake Yeah I will prolly try again.

[Mystic-Al-Bob]

should I tell you the right solution?

[chichigrande]

I guess you shall. It would be a good idea.

[MarkT]

And yet:

a^2 + b^2 = 0

=> a^2 = -b^2; b = +/- i.a

c^2 - a = 3

=> c^2 = a + 3

4c^2 - 3b = 16i

=> 4(a+3) - 3b = 16i

=> 4a - 3b = 16i - 12

=> 4a +/- 3i.a = 16i - 12

=> a = ( 16i - 12 ) / ( 4 +/- 3i )

Either: a = -3.84 + 1.12i, b = +/- (3.84.i + 1.12), c = +/- ( 0.529 + 1.058.i ) (approx).

Or: a = 4i, b = +/- 4, c = +/- ( 2 + i )

Eight solutions total.

*chants* TeX, TeX, TeX, TeX, ....

Next question, anyone?

[chichigrande]

Sure, go ahead and ask a new question.

[Mystic-Al-Bob]

very nice, mark

[Mystic-Al-Bob]

NEW QUESTION:

How many possibilities are there to write the number 2004 as a difference of two square numbers of natural numbers??

QUESTION 2:

0 = 4*x^3 + 9*x^2 - 1

x=?