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

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

• I'm not sure but is there a typo here? Isn't $\alpha\lambda+\beta\delta$ be $\alpha\gamma+\beta\delta$? – Habagat Maliksi Dec 5 '19 at 3:46

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 $$b^2 - 2c = \alpha^2 + \beta^2$$.
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.