help with this cubic equation

brokenluv

29 Jun 08, 23:34

1.5 = ( 1 + k ) ^3
find k.

i got to this step : k^3 + 2k^2 + 3k - 0.5 = 0
and then i got stuck. please help thx!

^tamago^

29 Jun 08, 23:45

1.5 = (1+k)Â³
3 = 2(kÂ³+3kÂ²+3k+1)
2kÂ³+6kÂ²+6k-1=0
k = 0.1447 â‰ˆ 0.145 (3.s.f)

^tamago^

29 Jun 08, 23:49

brokenluv

29 Jun 08, 23:52

Originally posted by ^tamago^:

1.5 = (1+k)³
3 = 2(k³+3k²+3k+1)
2k³+6k²+6k-1=0
k = 0.1447 ≈ 0.145 (3.s.f)

i dont understand this step

^tamago^

29 Jun 08, 23:55

2kÂ³+6kÂ²+6k-1=0

is no different from

kÂ³+3kÂ²+3k-Â½=0

which you have gotten. now, this equation cannot be solved directly.

brokenluv

29 Jun 08, 23:58

Originally posted by ^tamago^:

2k³+6k²+6k-1=0

is no different from

k³+3k²+3k-½=0

which you have gotten. now, this equation cannot be solved directly.

erm its 2k^2, not 3k^2

eagle

29 Jun 08, 23:59

Originally posted by ^tamago^:

1.5 = (1+k)³
3 = 2(k³+3k²+3k+1)
2k³+6k²+6k-1=0
k = 0.1447 ≈ 0.145 (3.s.f)

got simpler method

1.5 = (1+k)³
cube root(1.5) = (k+1)

k = cube root(1.5) -1 = 0.145

brokenluv

30 Jun 08, 00:03

Originally posted by eagle:

got simpler method

1.5 = (1+k)³
cube root(1.5) = (k+1)

k = cube root(1.5) -1 = 0.145

wow thx! i totally forgot my amath after i left sec school

^tamago^

30 Jun 08, 00:16

Originally posted by eagle:

got simpler method

1.5 = (1+k)Â³
cube root(1.5) = (k+1)

k = cube root(1.5) -1 = 0.145

yeah, but he want solve cubic. :(

jay_rocks

30 Jun 08, 00:22

This one not function meh?

brokenluv

30 Jun 08, 00:30

Originally posted by ^tamago^:

yeah, but he want solve cubic. :(

haha doesnt matter as long as i get the answer right

jiaxing2

2 Jul 08, 21:10

there should be 2 more roots which are complex numbers...

can be solved by theory of eqn

sum of the 3 roots = -3

product of 3 roots = 0.5

since there are no complex coefficients to the cubic eqn...then if complex roots exist, there exist in conjugate pairs...ie in the form of a + ib and a - ib where a and b are real numbers and i = root of ( -1 )

let k be the real root. 2a + k = -3

solving a = - [ 1 + 0.5(1.5)^1/3 ]

similarly ( a^2 + b^2 ) * k =0.5 (product of 3 roots)

go solve for b...and u get ur 2 complex roots

for ur info only...unless this qn carries alot of marks...i dun think u are really required to find out what the complex roots are...

eagle

3 Jul 08, 15:20

Originally posted by jiaxing2:

there should be 2 more roots which are complex numbers...

can be solved by theory of eqn

sum of the 3 roots = -3

product of 3 roots = 0.5

since there are no complex coefficients to the cubic eqn...then if complex roots exist, there exist in conjugate pairs...ie in the form of a + ib and a - ib where a and b are real numbers and i = root of ( -1 )

let k be the real root. 2a + k = -3

solving a = - [ 1 + 0.5(1.5)^1/3 ]

similarly ( a^2 + b^2 ) * k =0.5 (product of 3 roots)

go solve for b...and u get ur 2 complex roots

for ur info only...unless this qn carries alot of marks...i dun think u are really required to find out what the complex roots are...

No need do so hard to solve the complex roots also

Simplified version

(k+1)^3 - (cube root[1.5])^3 = 0

you can use the formula x^3 - y^3 = (x^2 + xy + y^2)(x-y)
where x = k+1 and y = cube root of 1.5

Then you can do the normal use equation to solve for quadratic equations method to determine the final complex roots, in which your discriminant should be -ve

eagle

3 Jul 08, 15:25

Another easy method to solve for k+1 for the complex roots:

You draw the Im-Real axis. You can also see that one of the roots is cube root of 1.5, which is a real number.

Then you rotate the point around the origin by 120 degrees (360/3)

one of the complex roots of k+1 would be:
- 1.145 sin 30 + 1.145 cos 30 = -0.5725 + 0.9916i

I think both of the above methods would be easier than using and understanding theory of equations