Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root. Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.
Hypothesis: $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
Conclusion: $ax^2 + bx + c$ has no rational root

To form a proof by contradiction, we take the negation of the conclusion:
$\neg B$: $ax^2 + bx + c$ has a rational root.
We now have a suitable hypothesis and conclusion for proof by contradiction:
A (Hypothesis): $cx^2 + bx + a$ has no rational root where $a$, $b$, and $c$ are real numbers with $c \not = 0$.
A1: $ax^2 + bx + c$ has a rational root.
Given that this is a proof by contradiction, we can work forward from both the hypothesis and conclusion, as shown above.

My Workings
A2: Let $x = \dfrac{p}{q}$ where $p$ and $q \not = 0$ are integers. This is the definition of a rational number (in this case, $x$): A rational number is any number that can be expressed as the quotient/fraction of two integers.
A3: $a\left(\dfrac{p}{q}\right)^2 + b\left(\dfrac{p}{q}\right) + c = 0$
$\implies \dfrac{ap^2}{q^2} + \dfrac{bp}{q} + c = 0$ where $q \not = 0$.
$\implies ap^2 + bpq + cq^2 = 0$
A4: $ap^2 + bpq + cq^2 = 0$ where $c \not = 0$
$\implies ap^2 + bpq = -cq^2$ where $-cq \not = 0$ since $c \not = 0$ and $q \not = 0$.
A5: $ap^2 + bpq + cq^2 = 0$ where $ap^2 + bpq \not = 0$ and $cq^2 \not = 0$.
But $ap^2 + bpq + cq^2 = 0$? Contradiction. $Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof and provide feedback on its correctness.
 A: The roots of
$ax^2+bx+c = 0
$
are
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$
and
the roots of
$cx^2+bx+a = 0
$
are
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2c}
$.
If the first are rational
and the second are not,
then their sum
and ratio
are irrational.
Since the roots of the first equation are rational,
their sum and product are rational.
These are
$\dfrac{b}{a}$
and
$\dfrac{c}{a}$.
Since the ratio
of the two equations' roots is irrational,
$\dfrac{a}{c}$
is irrational.
Since their sum is irrational,
$\dfrac{b}{c}$
is irrational.
But these both contradict
the previously proven
rationality
of these two ratios.
Therefore the roots
of the second equation
are also rational.
A: To continue what you started 
$ap^2 + bpq + cq^2 = 0$ divide both sides by $p^2$.
$a + b\frac qp + c(\frac {q^2}{p^2}) = 0$
So $\frac pc$ is a rational solution to $cx^2 + bx + a = 0$.  A contradiction.

Worth noting: if $w$ is a solution to $ax^2 + bx + c = 0$ ($a \ne 0; c\ne 0$) then $\frac 1w$ is a solution to $cx^2 +bx +a=0$.  And if $w \ne 0$ then $w$ is rational if and only $\frac 1w $ is rational.
A: Since you have proved that $ap^2+bpq \neq 0$ and $cq^2 \neq 0$, you are very close to the answer.
Notice that $p(ap+bq)=-cq^2$. But $p(ap+bq) \neq 0 \Rightarrow p \neq 0$.
Divide both sides by $p^2$. Then you get $a+b\frac qp=-c\frac {q^2}{p^2}.$
This shows that $cx^2+bx+a=0$ has a rational root $\frac qp$. WHICH is contradiction.
