Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $\alpha$ is a root of the equation $4x^2+2x-1=0$, then prove that $4\alpha^3-3\alpha$ is the other root. How do I proceed? The sum of the roots, the product of the roots lead me nowhere. Should I find the roots of the equation and substitute in the given two expressions of $\alpha$ and check whether manipulating the first root gives me the second root (which seems much complicated), or is there any other way?

share|improve this question
    
Check $4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1 = 0$ and $4\alpha^3-3\alpha \ne \alpha$. –  njguliyev Nov 3 '13 at 12:03

2 Answers 2

up vote 2 down vote accepted

Set $x=4\alpha^2-3\alpha$ in the original polynomial equation, and check to see that the left side reduces to zero under the assumption that $4\alpha^2+2\alpha-1=0$, or equivalently that $\alpha^2 = \frac 14(-2\alpha+1)$.

share|improve this answer

The other root $\beta$ is determined by the properties $\alpha+\beta=-\frac 12$ and $\alpha\beta=-\frac14$. Use polynomial division to show that $(4\alpha^3-3\alpha)+\alpha = [???]\cdot(4\alpha^2+2\alpha-1)-\frac12$ and $(4\alpha^3-3\alpha)\cdot\alpha = [???]\cdot(4\alpha^2+2\alpha-1)-\frac14$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.