Problem:
If the ratio of the two roots of the equation $a_1x^2+b_1x+c_1=0$ is equal to the ratio of the two roots of $a_2x^2+b_2x+c_2=0$, show that $\dfrac{(b_1)^2}{(b_2)^2}=\dfrac{a_1c_1}{a_2c_2}$.
My book's solution:
Let the roots of the equations $a_1x^2+b_1x+c_1=0$ and $a_2x^2+b_2x+c_2=0$ are $\alpha$ & $\beta$ and $\gamma$ & $\delta$ respectively. Now, according to the question,
$$\frac{\alpha}{\beta}=\frac{\gamma}{\delta}$$
$$\frac{\alpha+\beta}{\alpha-\beta}=\frac{\gamma+\delta}{\gamma-\delta}$$
$$\left(\frac{\alpha+\beta}{\alpha-\beta}\right)^2=\left(\frac{\gamma+\delta}{\gamma-\delta}\right)^2$$
$$\frac{(\alpha+\beta)^2}{(\gamma+\delta)^2}=\frac{(\alpha-\beta)^2}{(\gamma-\delta)^2}$$
$$\frac{(\alpha+\beta)^2}{(\gamma+\delta)^2}=\frac{(\alpha+\beta)^2-4\alpha\beta}{(\gamma+\delta)^2-4\gamma\delta}$$
$$\frac{(\alpha+\beta)^2}{(\gamma+\delta)^2}=\frac{(\alpha+\beta)^2-((\alpha+\beta)^2-4\alpha\beta)}{(\gamma+\delta)^2-((\gamma+\delta)^2-4\gamma\delta)}\tag{1}$$
$$\frac{(\alpha+\beta)^2}{(\gamma+\delta)^2}=\frac{\alpha\beta}{\gamma\delta}$$
$$\frac{\frac{b_1^2}{a_1^2}}{\frac{b_2^2}{a_2^2}}=\frac{\frac{c_1}{a_1}}{\frac{c_2}{a_2}}$$
$$\frac{{b_1}^2}{{b_2}^2}=\frac{a_1c_1}{a_2c_2}(\text{showed})$$
Questions:
- Isn't $(1)$ wrong? $4\alpha\beta\neq((\alpha+\beta)^2-4\alpha\beta)$ and $4\gamma\delta\neq((\gamma+\delta)^2-4\gamma\delta)$. So, isn't $(1)$ wrong?
- If $(1)$ is wrong, could you please provide me with an alternative correct solution to the problem?
Best Answer
Statement $(1)$ is correct. Your statement, $4\alpha\beta\ne(\alpha+\beta)^2-4\alpha\beta\;$ and $\;4γδ≠(γ+δ)^2−4γδ$ is also correct.
Hint: $$\frac{p}{q}=\frac{r}{s}\implies \frac{p}{q}=\frac{p-r}{q-s}$$