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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...
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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 1.5Then 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
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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 + 0.9916iI think both of the above methods would be easier than using and understanding theory of equations
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