Polynomial + Compx question (1 Viewer)

[deepulse]

Hey everyone, i just got my trial results from 4u back today..not so good 59/120

Oh well, i have this question which i think i did rite, but i got 0/3

a) If x is real and (x + i)^4 is imaginary, find the possible values of x in surd form. - 3

What i did was expanded the (x + i)^4 to x^4 + 4x^3i - 6x^2 - 4xi + 1

and since its imaginary, Re[(x+i)^4] = 0

so x^4 - 6x^2 + 1 = 0

solving ladida

i get x = +- sqrt[3 +- 2sqrt(2)]

am i doing it right? is my answer wrong? i showed all working and i got a big fat 0

 

[turtle_2468]

Yeah... you solved the quadratic wrong. Because if you solve the above you should get x^2= (3 +- sqrt(35)) but you divided the 6 by 2 then squared and subtracted one so instead of 6^2-1=35 you got 3^2-1=8 which simplifies to 2sqrt(2). Which means you should get at least 1/3, because that was the only wrong step...
hang on, it's not the only wrong step. If you have 3-sqrt(35) then you'll see that that's a negative number so you can't sqrt it, so the correct answer would be x= +- sqrt[3+sqrt(35)]. But then if you did have sqrt 8 you could have both + and - in the brackets... so 1/3 I'd say..

 

[deepulse]

i got, using quad formula

x^4 - 6x^2 + 1


x^2 = [6 +- sqrt ( 32) ]/ 2

= [6 +- 4sqrt(2) ] / 2

= 3 +- 2sqrt(2)

whic is +ve for +- since 2sqrt2 is like 2.8 ish

then i took the sqrt of that, so

x = +- sqrt [ 3 +- 2sqrt(2)]

i just graphed x^4 - 6x^2 + 1 using graphmatica, and the roots are that above (x = +- sqrt [ 3 +- 2sqrt(2)]) so im confused now

 

[wogboy]

Your solution looks quite acceptable to me. Ask your teacher to find out why you lost marks for that, you might be able to regain them. Good luck

 

[deepulse]

Hehe cool thankx, that means my 2 other friends also get the marks (they did the same as me but also got 0 ) bad day for teacher? hehe thankx heaps guys!!

 

[deepulse]

UPDATE: stupid dam teacher wont give me (or any of the other 4u guys..which ALL did it the same way) marks... He's just like... oh..ur suppose to do it demoivre's way...then he says he'll think about it..4 days later: oh pete, did u look at the solution i gave you... yea i did...ok..~!?

Then he said something about coming back in march next year asking for more marks aswell...i was like... ...ok...*walks out of class*

lol...oh wells..thanks everyone anyways~! hope u all went well in ur trials!!

 

[ND]

Well i haven't actually gone through any of the calculations or anything, but if you got the correct solution, and if the question didn't specify a certain method, i don't see how he can not give you full makrs.

 

[Richard Lee]

(x+i)^4=x^4+4ix^3-6x^2-4ix+1
ie: x^4-6x^2+1=0 and x^3-x><0
Solve x^4-6x^2+1=0
x^2=3+/-2*sqrt(2)=[sqrt(2)+/-1]^2
Therefore:
x1=sqrt(2)+1
x2=sqrt(2)-1
x3=-sqrt(2)-1
x4=-sqrt(2)+1

If something wrong. Inform me!
Thanx!