Recent content by Darken1

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    MHB Simplification of surds and powers

    Can't see the image clearly but I think this is what I would do. For the multiplication portion [5^(6/24)][(5^(24/5))/(5^5)] = [5^(6/24)][(5^(24/5))(5^-5)] Hint: (x^2)*(x^3) =? vs (x^2)^3?
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    MHB Complex number as a root and inequality question

    I can't see the url outside of me replying. Anyways what about dividing the original equation by (x^2 + 1) and then attempting to factor whatever it is?
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    MHB How Do You Solve These Complex Logarithmic and Polynomial Equations?

    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...
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    MHB Complex number as a root and inequality question

    For x^4+... I just try to convert it to (x^2+...)(x^2+...) format by guessing tbh. I know how to factor cubes, the quadratic formula, and how to divide one polynomial by another.
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    MHB Complex number as a root and inequality question

    Question 1: (a) Show that the complex number i is a root of the equation x^4 - 5x^3 + 7x^2 - 5x + 6 = 0 (b) Find the other roots of this equation Work: Well, I thought about factoring the equation into (x^2 + ...) (x^2+...) but I couldn't do it. Is there a method for that? Anyways the reason I...
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    MHB How Do You Solve These Complex Logarithmic and Polynomial Equations?

    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...
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