If the ratio of roots of $ax^2+bx+c=0$ be equal to that of $a'x^2+b'x+c'=0$ then

Question

If the ratio of roots of $ax^2+bx+c=0$ be equal to that of $a'x^2+b'x+c'=0$ then prove that: $$\dfrac {b^2}{b'^2} = \dfrac {ac}{a'c'}$$

My Attempt: Let $\alpha $ and $\beta $ be the roots of the first equation and $\alpha' $ and $\beta' $ be the roots of the second equation. Then, $$\alpha + \beta = \dfrac {-b}{a}$$ $$\alpha. \beta = \dfrac {c}{a}$$ $$\alpha' + \beta'=\dfrac {-b'}{a'}$$ $$\alpha'.\beta'=\dfrac {c'}{a'}$$ According to Question: $$\dfrac {\alpha }{\beta}=\dfrac {\alpha' }{\beta'}$$

Answer 1

I would say:

$\frac {\alpha'}{\alpha'} = \frac {\alpha}{\beta}$ implies that there exists a constant $k$ such that $\alpha' = k\alpha, \beta' = k\beta$

$\frac {b'}{a'} = -(\alpha' + \beta') = -k(\alpha + \beta) = k\frac {b}{a}$

$\frac {c'}{a'} = \alpha'\beta' = k^2\alpha\beta = k^2\frac {c}{a}$

$\frac {a'b}{ab'} = k$

$\frac {a'^2b^2}{a^2b'^2} = k^2 = \frac {a'c}{ac'}$

$\frac {b^2}{b'^2}= \frac {ac}{a'c'}$

Answer 2

Let $$\frac{\alpha}{\beta} = \frac{\alpha'}{\beta'} = k.$$ Now, $\alpha+\beta = -\frac{b}{a} \implies (k+1)a\beta=-b$. Similarly, $(k+1)a'\beta'=-b'$. Therefore, $$\frac{b^2}{b'^2}=\frac{a(a\beta^2)}{a'(a'\beta^2)}.$$ But $\alpha \beta = \frac{c}{a} \implies c = ka\beta^2$. Similarly, $c' = ka'\beta'^2$. Therefore, $$\frac{c}{c'}=\frac{a \beta^2}{a' \beta'^2} \implies \frac{b^2}{b'^2}=\frac{ac}{a'c'}.$$

Answer 3

Using your identities:

\begin{align} \frac{b^2}{b'^2}&=\frac{a^2(\alpha+\beta)^2}{a'^2(\alpha'+\beta')^2}\\ &=\frac{a^2\alpha\left(1+\frac{\beta}{\alpha}\right)\beta\left(\frac{\alpha}{\beta}+1\right)}{a'^2\alpha'\left(1+\frac{\beta'}{\alpha'}\right)\beta'\left(\frac{\alpha'}{\beta'}+1\right)}\\ &=\frac{a^2\alpha\beta}{a'^2\alpha'\beta'}\\ &=\frac{a^2\frac{c}{a}}{a'^2\frac{c'}{a'}}\\ &=\frac{ac}{a'c'} \end{align}