Exercise 1.1.5

Proof.

(a)

The substitution $x=y-b∕2$ in the equation $x^2+\mathit{bx}+c=0$ gives $$\begin{array}{llll}\hfill 0& =x^2+\mathit{bx}+c& \hfill & \\ \hfill & ={\left(y-\frac{b}{2}\right)}^2+b\left(y-\frac{b}{2}\right)+c& \hfill & \\ \hfill & =y^2-\frac{b^2}{4}+c.& \hfill & \end{array}$$

This last equation has two solutions

$$y=\pm \frac{1}{2}\sqrt{b^2-4c}.$$

The two solutions of the equation $x^2+\mathit{bx}+c=0$ are so

$$x=\frac{-b\pm \sqrt{b^2-4c}}{2}.$$

(b)

The substitution $x=y-b∕4$ in $x^4+b{x}^3+c{x}^2+\mathit{dx}+e=0$ gets rid of the coefficient of $y^3$.

(c)

More generally, the substitution $x=y-a_{n-1}∕n$ in the equation $$x^n+a_{n-1}{x}^{n-1}+\cdots +a_0.$$

gets rid of the coefficient of $y^{n-1}$ in the transformed equation.