how to form a quadratic equation If $\alpha$ and  $\beta$ are the root of the equation  $x^2 - 4x +6 =0$ , find the equation whose roots are $\alpha + 1/\beta$ and $\beta + 1/\alpha$.
 A: Hint: Use Viete's Formulas
$$\alpha+\frac1\beta=\frac{\alpha\beta+1}\beta=\frac7\beta$$
$$\beta+\frac1\alpha=\frac{\alpha\beta+1}\alpha=\frac7\alpha$$
An equation's thus
$$0=\left(x-\frac7\alpha\right)\left(x-\frac7\beta\right)=x^2-7\left(\frac{\alpha+\beta}{\alpha\beta}\right)x+\frac{49}{\alpha\beta}$$
and, again, Viete's handy here.
A: let us solve following equation
$x^2-4*x+6=0$
roots of this equation is  
$x_1=(4+\sqrt{8}*i)/2$
and  
$x_2=(4-\sqrt{8}*i)/2$
where  $i=\sqrt{-1}$
this may help you
http://answers.yahoo.com/question/index?qid=20081211060741AAUAcYJ
A: $x^2 - \frac{28}{6}x + \frac{49}{6}$.
$\alpha + \beta = 4$ and $\alpha\beta = 6$. Expand $(x - (\alpha + \frac{1}{\beta}))(x-(\beta + \frac{1}{\alpha}))$. This gives us $x^2 - (\alpha + \beta)(1 + \frac{1}{\alpha\beta})x + (\alpha\beta + 2 + \frac{1}{\alpha\beta})$ So, doing the necessary substitutions gives us: $x^2 - (4\times \frac{7}{6}) x + (6 + 2 + \frac{1}{6})$ which is equal to $x^2 - \frac{28}{6}x + \frac{49}{6}$  
A: Let $\displaystyle y=\alpha+\frac1{\beta}$ which is $\displaystyle\frac{\alpha\beta+1}{\beta}=\frac{6+1}{\beta}$ as $\displaystyle\alpha\beta=\frac61$
$\displaystyle\implies\beta=\frac7y$ which is  a root of $\displaystyle x^2-4x+6=0$
$\displaystyle\implies\left(\frac7y\right)^2-4\left(\frac7y\right)+6=0$
Multiply either sides by $y^2$
We shall get the same equation if we start with $\displaystyle \beta+\frac1{\alpha}$ which is $\displaystyle =\frac7{\alpha}$
