Worksheet03: Dickson 011
Current Work (editable)
CHAPTER II Elementary Theorems on the Roots of an Equation 12. Quadratic Equation. If a, b, c are given numbers, a $$ 0, (1) ax^{2} + bx + c = 0 (a $$ 0) is called a quadratic equation or equation of the second degree. The reader is familiar with the following method of solution by "completing the square." Multiply the terms of the equation by 4a, and transpose the constant term; then 4a^{2}x^{2} + 4abx = -4ac. Adding b^{2} to complete the square, we get (2ax + b)^{2} = \Delta, \Delta = b^{2} - 4ac, (2) x_{1} = -b + $${\Delta} / 2a x_{2} = -b - $${\Delta} / 2a By addition and multiplication, we find that (3) x_{1} + x_{2} = -b/a, x_{1} x_{2} = c/a. Hence for all values of the variable x, (4) a(x - x_{1})(x - x_{2}) $$ ax^{2} - a(x_{1} + x_{2})x + ax_{1} x_{2} $$ ax^{2} + bx + c, the sign $$ being used instead of = since these functions of x are identically equal, i.e., the coefficients of like powers of x are the same. We speak of a(x - x_{1})(x - x_{2}) as the factored form of the quadratic function ax^{2} + bx + c, and of x - x_{1} and x - x_{2} as its linear factors. In (4) we assign to x the values x_{1} and x_{2} in turn, and see that 0 = ax_{1}^{2} + bx_{1} + c, 0 = ax_{2}^{2} + bx_{2} + c. Hence the values (2) are actually the roots of equation (1). We call \Delta = b^{2} - 4ac the discriminant of the function ax^{2} + bx + c or of the corresponding equation (1). If \Delta = 0, the roots (2) are evidently equal, so that, by (4), ax^{2} + bx + c is the square of $${a}(x - x_{1}), and con-*
Answer Key (non-editable)
\Chapter{II}{Elementary Theorems on the Roots of an Equation} \Par{12. Quadratic Equation.} If $a$, $b$, $c$ are given numbers, $a \neq 0$, \[ \Tag{(1)} ax^{2} + bx + c = 0 \quad (a \neq 0) \] is called a \emph{quadratic equation} or equation of the second degree. The reader is familiar with the following method of solution by ``completing the square.'' Multiply the terms of the equation by $4a$, and transpose the constant term; then \[ 4a^{2}x^{2} + 4abx = -4ac. \] Adding $b^{2}$ to complete the square, we get \begin{gather*} (2ax + b)^{2} = \Delta,\qquad \Delta = b^{2} - 4ac, \\ \Tag{(2)} x_{1} = \frac{-b + \sqrt{\Delta}}{2a}\qquad x_{2} = \frac{-b - \sqrt{\Delta}}{2a} \end{gather*} By addition and multiplication, we find that \[ \Tag{(3)} x_{1} + x_{2} = \frac{-b}{a},\qquad x_{1} x_{2} = \frac{c}{a}. \] Hence for all values of the variable $x$, \[ \Tag{(4)} a(x - x_{1})(x - x_{2}) \equiv ax^{2} - a(x_{1} + x_{2})x + ax_{1} x_{2} \equiv ax^{2} + bx + c, \] the sign $\equiv$ being used instead of $=$ since these functions of $x$ are \emph{identically equal}, i.e., the coefficients of like powers of $x$ are the same. We speak of $a(x - x_{1})(x - x_{2})$ as the \emph{factored form} of the quadratic function $ax^{2} + bx + c$, and of $x - x_{1}$ and $x - x_{2}$ as its \emph{linear factors}. In \Eq{(4)} we assign to $x$ the values $x_{1}$ and $x_{2}$ in turn, and see that \[ 0 = ax_{1}^{2} + bx_{1} + c,\qquad 0 = ax_{2}^{2} + bx_{2} + c. \] Hence the values \Eq{(2)} are actually the roots of equation \Eq{(1)}. We call $\Delta = b^{2} - 4ac$ the \emph{discriminant} of the function $ax^{2} + bx + c$ or of the corresponding equation \Eq{(1)}. If $\Delta = 0$, the roots \Eq{(2)} are evidently equal, so that, by \Eq{(4)}, $ax^{2} + bx + c$ is the square of $\sqrt{a}(x - x_{1})$, and con-*
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