Evaluating expressions involving roots of cubic equations

Question

I'm having trouble understanding the concept of "roots"; could someone please solve this problem and explain the logic behind their method?

If $\alpha$, $\beta$ and $\gamma$ are the roots of $x^3+2x^2-3x+4$, find:

1. $\alpha\beta\gamma,$
2. $\alpha+\beta+\gamma,$
3. $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma},$
4. $\alpha^2+\beta^2+\gamma^2.$

Answer 1

Given $x=\alpha,\beta,\gamma$ are the roots of the equation $x^3+2x^2-3x+4=0$

So using factor theorem, $x-\alpha,x-\beta,x-\gamma$ are the factors of given equation.

So $$x^3+2x^2-3x+4 = (x-\alpha)\cdot (x-\beta)\cdot (x-\gamma)$$

$$\displaystyle x^3+2x^2-3x+4=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot \alpha)x-\alpha\cdot \beta\cdot \gamma$$

Now Camparing Coefficients, We get

$$\alpha+\beta+\gamma = -2$$ and $$\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot \alpha=-3$$ and $$\alpha\beta\gamma = -4$$

So $$\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot \alpha) =10$$

So $$\displaystyle \frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma} = \frac{\alpha\cdot \beta+\beta\cdot \gamma+\gamma\cdot \alpha}{\alpha\beta\gamma}=\frac{3}{4}$$