Quadratic functions - using substitution

Question

If $\alpha$ and $\beta$ are roots of the quadratic equation $ax^2+2bx+c=0$, find the quadratic equation with the roots $\alpha+\frac{1}{\alpha}$ and $\beta+\frac{1}{\beta}$ using transformation method or by substitution.

Answer 1

Hint. Note that $\alpha+\beta=-\frac{2b}{a}$ and $\alpha\beta=-\frac{c}{a}$. Hence $$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{-2b/a}{c/a}=-\frac{2b}{c}$$ which implies that $$S:=\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}=-\frac{2b}{a}-\frac{2b}{c}.$$ Are you able to find the product $$P:=\left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right)=\alpha\beta +\frac{(\alpha+\beta)^2}{\alpha\beta}-2+\frac{1}{\alpha\beta}?$$ Then the quadratic equation that you are looking for is $x^2-Sx+P=0$.

Answer 2

Let $y=x+\dfrac1x$ $$\implies x^2+(-y)x+1=0\ \ \ \ (1)$$

We already have $$ax^2+bx+c=0\ \ \ \ (2)$$

Solve the simultaneous equations $(1),(2)$ for $x^2,x$

and use $$x^2=(x)^2$$ to eliminate $x$

Answer 3

If we denote $S=p+q$ and $P=pq$ then $p,q$ are the roots of the quadratic $X^2-Sx +P=0$. We have $\alpha+\beta=-2b/a$ and $\alpha\beta=c/a$. Let's compute

$$\alpha+{1\over\alpha}+\beta+{1\over \beta}=\alpha+\beta+{\alpha+\beta\over \alpha\beta}=-2\left({b\over a}+{b\over c}\right)$$

$$\left(\alpha+{1\over\alpha}\right)\left(\beta+{1\over \beta}\right)=\alpha\beta+{\alpha^2+\beta^2\over \alpha\beta}+{1\over\alpha\beta}$$

On top we can check that $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$. We finish by substituting