Lesson 2 · from §107 of the 1910 text
Special Devices: Squares of Sums and Differences
Special devices
Systems of equations are often met which can be solved by substitution, but which are more conveniently solved as in the following examples. In every case the aim of the device is to replace the given system by one or more equivalent systems of linear equations.
Example 1. Solve the system \begin{cases} x + y = 5 & (1) \\ xy = 6 & (2) \end{cases}
| Step | Result | |
|---|---|---|
| Squaring (1), | x^2 + 2xy + y^2 = 25 | (3) |
| (2) times 4, | 4xy = 24 | (4) |
| (3) − (4), | x^2 - 2xy + y^2 = 1 | (5) |
| From (5), | x - y = \pm 1 | (6) |
From (6) and (1) we have two linear systems:
A \begin{cases} x + y = 5 \\ x - y = 1 \end{cases} \qquad B \begin{cases} x + y = 5 \\ x - y = -1 \end{cases}
Solving A and B, \begin{cases} x = 3 \\ y = 2 \end{cases} and \begin{cases} x = 2 \\ y = 3 \end{cases}
Example 2. Solve \begin{cases} x^2 + y^2 = 13 & (1) \\ xy = 6 & (2) \end{cases}
| Step | Result | |
|---|---|---|
| (2) times 2, | 2xy = 12 | (3) |
| (1) + (3), | x^2 + 2xy + y^2 = 25 | (4) |
| From (4), | x + y = \pm 5 | (5) |
| (1) − (3), | x^2 - 2xy + y^2 = 1 | (6) |
| From (6), | x - y = \pm 1 | (7) |
(5) and (7) combined give four pairs of linear equations:
A \begin{cases} x + y = 5 \\ x - y = 1 \end{cases} \quad B \begin{cases} x + y = 5 \\ x - y = -1 \end{cases} \quad C \begin{cases} x + y = -5 \\ x - y = 1 \end{cases} \quad D \begin{cases} x + y = -5 \\ x - y = -1 \end{cases}
The solution of A, B, C, and D is left to you: it is the check question below.
Quick check
Solve the four linear systems A, B, C, D of Example 2 (the system x^2 + y^2 = 13, xy = 6). Give all four sets of roots.
The key identities are
(x + y)^2 = x^2 + 2xy + y^2, \qquad (x - y)^2 = x^2 - 2xy + y^2,
so whenever the given equations let you build x^2 \pm 2xy + y^2 (or x^2 \pm 4xy + 4y^2, and so on) you can take a square root and drop to first-degree equations. Each choice of sign gives a separate linear system, and each linear system gives one pair of roots.
Typing your answers. Give every set of roots as a pair in parentheses, in the order the exercise names, for example (3, 2), (-2, -3). The pairs may come in any order. You may also write x = 3, y = 2; x = -2, y = -3. Radicals are typed sqrt(3), so 2\sqrt3 is 2sqrt(3).
From the history of algebra
Euclid, the Arabs, and the Hindus. Euclid (300 B.C.) solved systems of quadratic equations by means of geometry. But any solution that we would recognize as algebraic did not exist until the Arabs and the Hindus studied the subject. Although Diophantos solved many problems that lead to quadratic equations, he always expressed all of his unknowns in terms of one symbol, and so avoided systems of quadratics, much as we did with problems in one unknown in the earlier chapters.
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Solve in a manner similar to that of the two preceding examples, pair results, and check each rational set of roots.
\begin{cases} x - y = 1 \\ x y = 12 \end{cases}
\begin{cases} x + 2 y = 4 \\ x y + 6 = 0 \end{cases}
\begin{cases} x^{2} + 4 y^{2} = 41 \\ x y + 10 = 0 \end{cases}
\begin{cases} 3 x - y = 12 \\ 9 x^{2} + y^{2} = 72 \end{cases}
\begin{cases} x^{2} - 3 x y + 4 y^{2} = 11 \\ x y = 10 \end{cases}
\begin{cases} x^{2} + y^{2} = 25 \\ x^{2} + 2 x y + y^{2} = 49 \end{cases}
\begin{cases} x^{2} + y^{2} = 25 \\ x^{2} - 2 x y + y^{2} = 49 \end{cases}
\begin{cases} x^2 + y^2 = 15 \\ x + y = 3\sqrt3 \end{cases}
\begin{cases} x^{2} - x y = 4 \\ y^{2} - x y = -3 \end{cases}
\begin{cases} x^{2} = y^{2} + 5 \\ y^{2} + x y = 10 \end{cases}