Prove that, If two quadratic equations have same roots then ratio of their coefficients are proportional

Question

Let $ax^2+bx+c=0$ and $px^2+qx+r=0$ be two quadratic equations having same roots (where $a,p≠0$) then prove that,

$\frac{a}{p}=\frac{b}{q}=\frac{c}{r}$

Proof: for the first equation we have,

sum of roots $=-\frac{b}{a}$

Product of roots $= c/a$

and for the second equation we have,

Sum of roots $=-\frac{q}{p}$

Product of roots $=\frac{r}{p}$

Hence, as both equations have same roots we have, sum of roots of first equation is same as sum of the roots of second equation! and similarly for the product!

Hence, $-\frac{b}{a}=-\frac{q}{p}$

and $\frac{c}{a}=\frac{r}{p}$

From which we have, $\frac{a}{p}=\frac{b}{q}$ and $\frac{c}{r}=\frac{a}{p}$

From this we have, $\frac{a}{p}=\frac{b}{q}=\frac{c}{r}$

Is the proof correct?

What if we have $q=0$

Please help

Answer 1

You are correct till "Hence, $...-\frac{b}{a}=-\frac{q}{p}$ and $\frac{c}{a}=\frac{r}{p}$ "

Then, You have written "$\frac{a}{p}=\frac{b}{q}$ and $\frac{c}{r}=\frac{a}{p}$" So, you are asuming that $q\not= 0$ if $q=0$ you cannot write "$\frac{b}{q}$". Hence, for $q= 0$ you have to use $\frac{b}{a}=\frac{q}{p}$ and $\frac{c}{a}=\frac{r}{p}$

Answer 2

Let's work it out again.

$$ax^2 + bx + c = 0$$ $$px^2+qx+r = 0$$ having same roots implies that:

$$\frac{-b}{a} = \frac{-q}{p}$$

$$\frac{c}{a} = \frac{r}{p}$$

So it would be more appropriate to say that:

$$bp = qa$$ $$cp = ar$$

The equation you claim only holds if $q, r\neq 0$. Similarly, you can interpret the past two equations to form the inverse of your result, and then you would have to ensure $b,c \neq 0$.

So more generally it is not a good idea to use fractions and ratios in something like this where you aren't restricting what values the denominators can take. It is much better to say that two quadratics always have two roots common when the coefficient of $x^k$ of one equation times the leading coefficient of the other, is equal to the coefficient of $x^k$ of the second equation times the leading coefficient of the first. $(k \in {0,1,2})$

You can play around with these two relations and work out what you might find convenient. When $q,r\neq 0$ you get the more useful relation that you also worked out. But when either of them is zero, the equation doesn't hold. When $b, c \neq 0$ you get $\frac{p}{a} = \frac{q}{b} = \frac{r}{c}$.