How do I transform the equation based on the condition? If $q$ and $w$ are the roots of the equation $$2x^2-px+7=0$$ 
Then $q/w$ is a root of ? 
P.s:- It is an another question of How do I transform the equation based on this condition?
 A: Let $y=\frac{q}{w}$.
Clearly, q,w are $\frac{p ±\sqrt{p^2-56}}{4}$
If we take q to be $\frac{p +\sqrt{p^2-56}}{4}$, w is $\frac{p -\sqrt{p^2-56}}{4}$
So, $y=\frac{q}{w}=\frac{p +\sqrt{p^2-56}}{p -\sqrt{p^2-56}}$
Applying componendo dividendo, $\frac{y+1}{y-1}=\frac{p}{\sqrt{p^2-56}}$
On squaring, 
 $\frac{y^2+1+2y}{y^2+1-2y}=\frac{p^2}{p^2-56}$
Applying componendo dividendo again, $\frac{y^2+1}{2y}=\frac{p^2-28}{28}$
or, $28y^2-2(p^2-28)y+28=0$ 
$=>14y^2-(p^2-28)y+14=0$ is the desired equation  whose one root is $\frac{q}{w}$.
If we take q to be $\frac{p -\sqrt{p^2-56}}{4}$, w is $\frac{p +\sqrt{p^2-56}}{4}$
So, $y=\frac{q}{w}=\frac{p -\sqrt{p^2-56}}{p +\sqrt{p^2-56}}$
Applying componendo dividendo, $\frac{y+1}{y-1}=-\frac{p}{\sqrt{p^2-56}}$
On squaring we shall get the desired quadratic equation whose one root is $\frac{q}{w}$.
Just observe that both the derived equations are same.
This is the minimal equation with rational coefficients for any integer p.
If we  allow irrational coefficients, $\frac{q}{w}$ will be a root y=$\frac{p -\sqrt{p^2-56}}{p +\sqrt{p^2-56}}$  or y=$\frac{p + \sqrt{p^2-56}}{p - \sqrt{p^2-56}}$.
A: The following result (and its generalizations to higher degree polynomials) is very useful. Consider the equation $ax^2+bx+c=0$, where $a\ne 0$.  Then the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$. In our case, we have 
$$q+w=\frac{p}{2}\quad\text{and}\quad qw=\frac{7}{2}.$$ 
We will find a quadratic equation whose roots are $\frac{q}{w}$ and $\frac{w}{q}$ without calculating $\frac{q}{w}$ or $\frac{w}{q}$. 
Note that 
$$\frac{q}{w}+\frac{w}{q}=\frac{q^2+w^2}{qw}=\frac{(q+w)^2-2qw}{qw}=\frac{(q+w)^2}{qw}-2.$$
Substitute our known values of $q+w$ and $qw$. We get
$$ \frac{q}{w}+\frac{w}{q}=\frac{p^2}{14}-2=\frac{p^2-28}{14}.$$
Note that trivially the product of $\frac{q}{w}$ and $\frac{w}{q}$ is $1$.
So the numbers $\frac{q}{w}$ and $\frac{w}{q}$ are roots of the quadratic
$$x^2 -\left(\frac{p^2-28}{14}\right)x +1=0.$$
If we wish, we can change to the equivalent 
$14x^2+(28-p^2)x+14=0$.
