If roots of equation $ax^2+bx+c=0$ is $\alpha,\beta$ find the roots of equation $bcx^2+a(bc+a^2)x+ac^2=0$ in terms of $\alpha,\beta$
My Try:
We know that
$$\alpha+\beta=-\frac ba \quad \text{ and } \quad \alpha \beta=\frac ca$$
I have tried to substitute $\alpha,\beta$ into the new equation, but I have faced some trouble how to separate roots,
So far I got;
$$x^2+\bigg(\frac {1}{\alpha\beta}-\frac {1}{(\alpha+\beta)(\alpha\beta)^2}\bigg)x+\frac{1}{\alpha+\beta}=0$$
How to proceed? If you suggest a completely different method that is fine. Thanks in advance!
The given quadratic equation has the roots $\alpha$ and $\beta$. Then we have that $$ \alpha + \beta = -\frac{b}{a} \quad \text{ and } \quad \alpha \cdot \beta = \frac{c}{a} $$ The new quadratic $$ bc \cdot x^2 + a(bc + a^2)x + ac^2 = 0 $$ $\blacksquare~$What I have so far:
Let the roots of the new quadratic be $m, n$. Then we have that $$ m + n = -a \cdot \frac{bc + a^2}{bc} = - a\bigg(1 + \frac{a^2}{bc}\bigg) = - a \cdot {\bigg(1 - \color{red}{ \frac{1}{\alpha \beta (\alpha + \beta)}}\bigg)} $$ and $$ mn = \frac{ac^2}{bc} = \frac{a}{b} \cdot c = - a \cdot \bigg( \frac{\alpha \beta}{ \alpha + \beta} \bigg) $$ Hence the quadratic has become $$ x^2 + a \cdot \bigg(1 - \color{red}{\frac{1}{\alpha \beta (\alpha + \beta)}}\bigg) x - a \cdot \bigg( \frac{\alpha \beta}{ \alpha + \beta} \bigg) = 0 $$ I haven't found out any way to calculate and substitute $a$ yet (it seems too that we can't determine the roots just by $\alpha$ and $\beta$).