M muttiah Banned Joined Aug 18, 2006 Messages 138 Gender Male HSC 2007 Dec 5, 2006 #1 5x^2 - 12x + 17=0 x^2 -2xcostheta + 1 =0 x^2 + 2ix + 1=0
P pLuvia Guest Dec 5, 2006 #2 1. Use the quadratic formula, then change the sqrt{-1} into i then you will get complex roots 2. Use quadratic formula and notice that 4cos2x-4=4sin2x, and since there is a square root then the sin2x becomes sinx 3. Use quadratic formula like any other normal equation you are going to solve
1. Use the quadratic formula, then change the sqrt{-1} into i then you will get complex roots 2. Use quadratic formula and notice that 4cos2x-4=4sin2x, and since there is a square root then the sin2x becomes sinx 3. Use quadratic formula like any other normal equation you are going to solve
B bos1234 Member Joined Oct 9, 2006 Messages 491 Gender Male HSC 2007 Dec 6, 2006 #3 how do i do this? ix^2 - x + 4i = 0 i got the answer but i got it as positive but the book has it as negative.. i took -b as 1.. i think the book took -b as -1.. any1 know y?
how do i do this? ix^2 - x + 4i = 0 i got the answer but i got it as positive but the book has it as negative.. i took -b as 1.. i think the book took -b as -1.. any1 know y?
A acmilan I'll stab ya Joined May 24, 2004 Messages 3,989 Location Jumanji Gender Male HSC N/A Dec 6, 2006 #4 easiest way to know which is right is to sub it back in, whichever gives 0 is right
P pLuvia Guest Dec 6, 2006 #5 Did you rationalise the denominator, as in not leaving it in a complex form? You should have got something like this x=[1+sqrt{17}]/2i =[-i-+isqrt{17}]/2
Did you rationalise the denominator, as in not leaving it in a complex form? You should have got something like this x=[1+sqrt{17}]/2i =[-i-+isqrt{17}]/2
B bos1234 Member Joined Oct 9, 2006 Messages 491 Gender Male HSC 2007 Dec 6, 2006 #6 ahh kk thanks.. yeah when i rationalised it i got denom as negative.,, k thanks!
