Showing that the roots of an equation are always real and evaluating positive indices

Question

I am stuck on finding the roots, I cannot seem to work around this with $\lambda$ can someone explain the derivation for this?

If the roots of the equation $x^2 + bx + c = 0$ are $\alpha$, $\beta$ and the roots of the equation $x^2 + \lambda bx + \lambda^2 c = 0$ are $\gamma, \delta$ show that the equation whose roots are $\alpha \gamma + \beta \delta$ and $\alpha \delta + \beta \gamma$ is:

$$x^2 - \lambda b^2x+ 2\lambda^2 c(b^2-2c)=0$$

Show that the roots of this equation are always real.

How is the following equation derived?

Express with positive indices $$\frac{2b^-3x^2}{7c^-4y^2} = \frac{2x^2c^4}{7b^3y^2}$$

Answer 1

The roots of $x^2 + \lambda b x + \lambda^2 c$ are $\lambda \alpha$ and $\lambda \beta$. I take it you mean $\alpha \gamma + \beta \delta$, not $\alpha \lambda + \beta \delta$. Making the substitution for $\gamma$ and $\delta$, we are looking to show that the roots of the equation are $\lambda(\alpha^2 + \beta^2)$ and $2\lambda\alpha\beta$.

It suffices to show that $\alpha^2 + \beta^2$ and $2\alpha \beta$ are roots of $x^2 - b^2x + 2c (b^2 - 2c) = (x - 2c)(x - (b^2 - 2c))$

We know that $c = \alpha \beta$ and $b = -\alpha - \beta$, since $(x - \alpha) (x - \beta) = x^2 + b x + c$. So $2c = 2 \alpha \beta$ and $b^2 - 2c = \alpha^2 + \beta^2$.

Answer 2

Hint: $x^2+bx+c=(x-\alpha)(x-\beta)$ and $x^2 + \lambda bx + \lambda^2 c = (x-\delta)(x-\gamma)$. So, we have the following: (1) $\alpha+\beta=-b$, (2) $\alpha\beta=c$, (3) $\delta+\gamma=-\lambda b$ and (4) $\delta\gamma=\lambda^2c$.

Then consider the sum $(\alpha \lambda + \beta \delta)+(\alpha \delta + \beta \gamma)$ and the product $(\alpha \lambda + \beta \delta)(\alpha \delta + \beta \gamma)$, and try to express these in terms of the four equations I mentioned above.