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Simple Math Practice Problems


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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.

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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. :P

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

....

<=> (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. :P

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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?

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