MHB How Do You Solve These Complex Logarithmic and Polynomial Equations?

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The discussion revolves around solving a system of logarithmic equations and a polynomial equation. For the logarithmic equations, participants attempted to manipulate the equations but faced challenges in factoring and deriving solutions, ultimately noting that the system lacked clear equals signs. In the polynomial section, it was established that complex roots imply the presence of their conjugates, leading to confusion about the correct factors and roots of the polynomial. The conversation highlighted the importance of proper factoring techniques and the application of the distributive rule. Overall, the participants grappled with the complexities of both mathematical problems, seeking clarity and correct solutions.
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First Question:
Solve the following system of equations
log{x+1}y=2
log{y+1}x=1/4


Work:
Turned them into equations
(x+1)^2=y (y+1)^(1/4)=x

Substituted second equation into the first equation
((y+1)^(1/4)+1)^2=y

factored out and eventually got
((y+1)^1/4)^2+2((y+1)^1/4)+1=y
Tried to use quadratic formula (a=1,b=2,c=1) and got y=-1. -1=-1 Didn't seem relevant.

Tried to factor it out by substituting x = (y+1)^1/4
(x+1)(x+1)=x^4-1 x^4-1=(x^2+1)(x^2-1)=(x^2+1)(x+1)(x-1)
(x+1)(x+1)=(x^2+1)(x+1)(x-1)
(x+1)=(x^2+1)(x-1)
(x+1)=(x^2+1)(x-1)
-(x+1) = -(x+1)
0=(x^2+1)(x-1)-(x+1)
x^3-x^2-2=0
Couldn't factor it out.

Also noticed
(x+1)(x+1)=y
x=-1,y=0
Second Question:
Consider the polynomial p(x)=x^4+ax^3+bx^2+cx+d, where a,b,c,d are real numbers.

Given that 1+i and 1-2i are zeroes of p(x), find the values of a,b,c,d.

Work:
x=1+i and x=1-2i are zeroes
(x-1-i)(x-1+2i)=0
Factored it out
x^2-2x+ix+3
Seems to be missing another factor to get to the original equation.
Perhaps I could divide the original equation by the equation I factored out to get the last factor?
No idea at all

Edit: I need sleep geez...
 
Last edited:
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Darken said:
First Question:
Solve the following system of equations
log{x+1}_{2} log{y+1}_{¼}
I don't understand the problem here, because the "system of equations" has no equals signs and therefore doesn't consist of equations.

Darken said:
Eventually
0=x^3-x^2+x-1
Couldn't factor it out.
However, if you are correct in getting to the equation $0 = x^3 - x^2 + x - 1$ then you can solve it by writing it as $0 = x^2(x-1) + x-1 = (x^2+1)(x-1).$ That has just one real solution $x=1$.

Darken said:
Second Question:
Consider the polynomial p(x)=x^4+ax^3+bx^2+cx+d, where a,b,c,d are real numbers.

Given that 1+i and 1-2i are zeroes of p(x), find the values of a,b,c,d.

Work:
x=1+i and x=1-2i are zeroes
(x-1-i)(x-1+2i)=0
Factored it out
x^2-2x+ix+3
Seems to be missing another factor to get to the original equation.
Perhaps I could divide the original equation by the equation I factored out to get the last factor?
No idea at all
Hint: If a polynomial with real coefficients has a complex zero $u+iv$, then the complex conjugate $u-iv$ is also a zero of the polynomial.
 
Opalg said:
I don't understand the problem here, because the "system of equations" has no equals signs and therefore doesn't consist of equations.

However, if you are correct in getting to the equation $0 = x^3 - x^2 + x - 1$ then you can solve it by writing it as $0 = x^2(x-1) + x-1 = (x^2+1)(x-1).$ That has just one real solution $x=1$.Hint: If a polynomial with real coefficients has a complex zero $u+iv$, then the complex conjugate $u-iv$ is also a zero of the polynomial.

My bad, fixed the problem.My equation was wrong.
So from the previous equation
(x+1)=(x^2+1)(x-1)
I just factored out right hand side and minused (x+1) from both sides to get x^3 - x^2 + x - 1-(x+1)=0.
Supposed to have been
x^3-x^2-2=0
Don't know how to factor this one out.
Side Note: How did you convert x^2(x−1)+x−1 into (x^2+1)(x−1)? Is there an equation for it?

Hm so I guess that means
x=1+i and x=1-2i
x=1-i and x=1+2i
are all zeroes? Think I can go from there.
 
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Darken said:
My bad, fixed the problem.My equation was wrong.
So from the previous equation
(x+1)=(x^2+1)(x-1)
I just factored out right hand side and minused (x+1) from both sides to get x^3 - x^2 + x - 1-(x+1)=0.
Apparently you multiplied rather than "factored". There was no point in that - the standard method for solving such equations is to factor so if the expression is already factored, leave it that way!

Supposed to have been
x^3-x^2-2=0
Don't know how to factor this one out.
It had already been factored for you!

Side Note: How did you convert x^2(x−1)+x−1 into (x^2+1)(x−1)? Is there an equation for it?
The "distributive rule": ab+ ac= a(b+ c). Here, the "a" is x-1: x^2(x- 1)+ 1(x- 1)= (x^2+ 1)(x- 1).
Hm so I guess that means
x=1+i and x=1-2i
x=1-i and x=1+2i
are all zeroes? Think I can go from there.
No, if (x^2+ 1)(x- 1)= 0 then x^2+ 1= 0 or x- 1= 0. The first of those is the same as x^2= -1 which has roots i and -i. x- 1= 0 only for x= 1. The three roots of the cubic equation are i, -i, and 1. I have no idea where you got "1+i", "1- 2i", "1- i" and "1+ 2i".
 
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