Roots of polynomial equations — practice questions

First, identify the coefficients of the cubic $ax^3+bx^2+cx+d=0$. Here, $a=2, b=-7, c=4, d=1$.

(i) The sum of the roots is given by $\Sigma \alpha = -\frac{b}{a}$. $\alpha + \beta + \gamma = -\frac{-7}{2} = \frac{7}{2}$.

(ii) The sum of the product of roots in pairs is given by $\Sigma \alpha\beta = \frac{c}{a}$. $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{4}{2} = 2$.

(iii) The product of the roots is given by $\alpha\beta\gamma = -\frac{d}{a}$. $\alpha\beta\gamma = -\frac{1}{2}$.

The equation is a quartic with $a=1, b=2, c=-5, d=1, e=-3$.

First, we need the values of $\Sigma \alpha$ and $\Sigma \alpha\beta$. Using Vieta's formulas: $\Sigma \alpha = -\frac{b}{a} = -\frac{2}{1} = -2$. $\Sigma \alpha\beta = \frac{c}{a} = \frac{-5}{1} = -5$.

Now, we use the identity $\Sigma \alpha^2 = (\Sigma \alpha)^2 - 2(\Sigma \alpha\beta)$. This identity holds for polynomials of any degree.

Substitute the values we found: $\Sigma \alpha^2 = (-2)^2 - 2(-5)$ $= 4 - (-10)$ $= 4 + 10 = 14$.

So, the sum of the squares of the roots is 14.

Let the original equation be $P(x) = x^3 - 2x^2 + 5x - 4 = 0$, where $x$ can be $\alpha, \beta,$ or $\gamma$.

We want a new equation in a variable, let's say $y$, where the roots are $y = \alpha+1, y = \beta+1, y = \gamma+1$. This gives us a general relationship between the new roots ($y$) and the old roots ($x$): $y = x+1$.

To find the new equation, we rearrange this relationship to make $x$ the subject: $x = y-1$.

Now, we substitute this expression for $x$ back into the original polynomial equation: $(y-1)^3 - 2(y-1)^2 + 5(y-1) - 4 = 0$

Next, we expand the brackets: $(y^3 - 3y^2 + 3y - 1) - 2(y^2 - 2y + 1) + 5(y-1) - 4 = 0$ $y^3 - 3y^2 + 3y - 1 - 2y^2 + 4y - 2 + 5y - 5 - 4 = 0$

Finally, collect like terms to get the new equation in $y$: $y^3 + (-3-2)y^2 + (3+4+5)y + (-1-2-5-4) = 0$ $y^3 - 5y^2 + 12y - 12 = 0$

This is the required equation. It's conventional to write the final answer using the variable $x$, so the equation is $x^3 - 5x^2 + 12x - 12 = 0$.