Cubic equation

[TheLivingMartyr]

I have worked for pages and pages, but I just cannot find a way to work out this question. it goes:

 

Determine all pairs of values of the real numbers p and q, where 1 + i is one root of the equation:

 

z³ + pz² + qz - pq = 0

 

I know things such as one of the other two roots is 1 - i, and that the 3rd root is real, and also that an equation with roots a and b is (x - a)(x - b) = 0, but please believe me, I have tried using all my knowledge of this to work out the answers, and all my efforts have come to nought.

 

Please help me, I would like to know how one goes about tackling a problem like this.

[imatfaal]

Hello Martyr - my maths is way below most of the guys on the forum so I look forward to an answer from them -cos mine is screwed.

 

Whenever I try this I get an non-real value for p.

 

My method which is clearly wrong (cos I dont get a real value for p) is to set up an function as follows

 

[math] (x-(1-i))(x-(1+i))(x+a)=0 [/math]

ie roots are

[math](1-i) (1+i) (-a) [/math]

multiply this out

[math] (x^2 -x -ix -x +ix +2)(x+a)=0[/math]

[math] (x^2 -2x +2)(x+a)=0[/math]

[math]x^3 -2x^2+2x+ax^2-2ax+2a=0[/math]

put like terms together

[math]x^3 +x^2(a-2)+x(2-2a)+2a =0[/math]

 

you now have two cubics - the one in the question, and the one you have constructed. the coefficients must be the same for the question to work (i hope) so

 

coeffs x^3 term 1=1 ok

coeffs x^2 term p=a-2

coeffs x^1 term q = 2-2a

coeffs x^0 term pq = 2a

 

that gives you three equations with three unknowns - which should be solvable, if a real solution exists. But pretty soon - after rearranging you end up with

[math] 2p^2 +4p+4=0 [/math]

but the problem is that per the quadratic solution equation

[math] b^2-4ac = 16-4.2.4=-16[/math]

ie unreal solutions

 

I really hope someone comes in an give a decent answer - cos I cannot see it . sorry to be so useless

[Daedalus]

Perhaps the information provided by Wolfram will help you: Cubic Formula .

[TheLivingMartyr]

Anyone? there have been 36 views but no help i know this question isn't impossible, so there must be someone who knows what to do. I'll show you my working so far if it helps

 

Our task is to find all possible values of p and q in the equation below, where p and q are real numbers.

 

z³ + pz² + qz - pq = 0

 

let z³ + pz² + qz - pq = T

 

this equation, with roots a, b and c, can be written as:

 

(z - a)(z - b)(z - c) = T

 

z(z - a)(z - b) - c(z - a)(z - b) = T

 

z(z² - az - bz +ab) - c(z² - az - bz + ab) = T

 

z³ - az² - bz² + abz - cz² + acz + bcz - abc = T

 

z³ + z²(-a - b - c) + z(ab + ac + bc) - abc = z³ + pz² + qz - pq

 

Now I shall compare and equate the coefficients

so,

 

p = -a - b - c

 

q = ab + ac + bc

 

pq = abc

 

Now I can form a quadratic equation out of these terms

 

(-a - b - c)(ab + ac + bc) = abc

 

-a²b - a²c - ab² - b²c - ac² - bc² - 4abc = 0

 

We have already been given two roots to our original equation, which are:

 

a = (1 + i)

b = (1 - i)

c = ?

 

So we substitute in values for a and b:

 

-[(1 + i)²(1 - i)] - c(1 + i)² - [(1 + i)(1 - i)²] - c(1 - i)² - c²(1 + i) - c²(1 - i) - 4c(1 + i)(1 - i) = 0

 

-2 - 2i - 2ci - 2 + 2i + 2ci - c² - c²i - c² + c²i - 8c = 0

 

-2c² - 8c - 4 = 0

 

Now I solve the quadratic. I know that the value MUST be real, so if the discriminant is negative, I know that I have made a mistake.

 

Discriminant = (-8)² - (4 * -2 * -4)

 

= 64 - 32

 

= 32 ==> The solutions will be real

 

now input into quad formula

 

c = (8 (+-)(32)^1/2)/-4

 

c = (8 (+-)4(2)^1/2)/-4

 

c = -(2 (+-) (2)^1/2)

 

so c = {-2 -(2)^1/2 or -2 + (2)^1/2}

 

Now we know all the roots of original equation T: a = 1 + i, b = 1 - i, c = -2 + (2)^1/2 or -2 - (2)^1/2

 

so we can refer back to our definitions of p and q in terms of a, b and c, which are:

 

p = -a - b - c

 

q = ab + ac + bc

 

now we substitute in values for a, b and c

 

first, take c = -2 + (2)^1/2

 

p = -(1 + i) - (1 - i) - (-2 + (2)^1/2)

 

p = -1 - i - 1 + i + 2 - (2)^1/2

 

first value of p = -(2)^1/2

 

now for q

 

q = (1 + i)(1 - i) + (1 + i)(-2 + (2)^1/2) + (1 - i)(-2 + (2)^1/2)

 

q = 2 - 2 + (2)^1/2 - 2i + 2i(2)^1/2 - 2 + (2)^1/2 + 2i - 2i(2)^1/2

 

first value of q = -2 + 2(2)^1/2

 

Now, take c = -2 - (2)^1/2

 

p = -(1 + i) - (1 - i) - (-2 - (2)^1/2)

 

p = -1 - i - 1 + i + 2 + (2)^1/2

 

second value of p = (2)^1/2

 

now for q

 

q = (1 + i)(1 - i) + (1 + i)(-2 - (2)^1/2) + (1 - i)(-2 - (2)^1/2)

 

q = 2 - 2 - (2)^1/2 - 2i - 2i(2)^1/2 - 2 - (2)^1/2 + 2i + 2i(2)^1/2

 

second value of q = -2 - 2(2)^1/2

 

......ohh. I seem to have worked it out for myself... ahh..oh well!

 

oh damn, I was so busy posting my own reply i didn't notice the other ones, I kinda worked it out myself whilst typing out my working. well, thanks anyway `

 

sorry about that!

[Schrödinger's hat]

......ohh. I seem to have worked it out for myself... ahh..oh well!

 

oh damn, I was so busy posting my own reply i didn't notice the other ones, I kinda worked it out myself whilst typing out my working. well, thanks anyway `

 

sorry about that!

 

No worries. This way we can take credit for helping you out without actually having to do anything

[TheLivingMartyr]

ahhhaah, nice one