Symmetric systems: solving with sum and product
Learning Objectives
- Recognize a symmetric system: one whose equations are unchanged when \(x\) and \(y\) trade places.
- Rewrite symmetric expressions in terms of \(s = x + y\) and \(p = xy\), using \(x^2 + y^2 = s^2 - 2p\) and its companions.
- Solve the system \(x + y = s,\; xy = p\) by constructing the quadratic \(t^2 - st + p = 0\).
- Count the solutions from the sign of \(s^2 - 4p\), and discard \((s, p)\) pairs that fail \(s^2 \geq 4p\).
- Explain why solutions of symmetric systems come in swapped pairs \((u, v)\), \((v, u)\).
The garden, revisited
Lesson 1 opened with a rectangular garden of perimeter \(20\) m and area \(21\) m². Back then we eliminated one unknown by hand and reached the equation \(x^2 - 10x + 21 = 0\). But the honest, unprocessed form of that problem has two unknowns — the two sides — and two equations:
\[x + y = 10, \qquad xy = 21.\]
Notice something about this system: if \(x\) and \(y\) trade places, both equations stay exactly the same. The problem cannot tell the two unknowns apart — and that indifference is not a nuisance but a powerful piece of structure.
Lesson 2 ended with precisely the tool such a system calls for: two numbers are completely determined by their sum and their product, as the roots of \(t^2 - St + P = 0\). This lesson turns that theorem into a general method, and with it Unit 1 comes full circle.
Symmetric systems
A system of equations in the unknowns \(x\) and \(y\) is symmetric if swapping \(x\) and \(y\) leaves every equation unchanged.
The two simplest symmetric expressions,
\[s = x + y \qquad\text{and}\qquad p = xy,\]
are called the elementary symmetric expressions in \(x\) and \(y\).
For instance, \(x + y = 10,\; xy = 21\) is symmetric, and so is \(x^2 + y^2 = 13,\; x + y = 5\). By contrast, \(x - y = 1,\; xy = 6\) is not: swapping turns the first equation into \(y - x = 1\), a different condition.
Everything symmetric speaks the language of \(s\) and \(p\)
The strategy will be to trade the unknowns \(x, y\) for the new unknowns \(s, p\). For that we must rewrite symmetric expressions in terms of \(s\) and \(p\), and a few identities do most of the work — each verified by simply expanding the right-hand side:
\[x^2 + y^2 = (x+y)^2 - 2xy = s^2 - 2p,\]
\[(x - y)^2 = (x+y)^2 - 4xy = s^2 - 4p,\]
\[x^3 + y^3 = (x+y)^3 - 3xy(x+y) = s^3 - 3sp,\]
\[\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = \frac{s}{p} \qquad (x, y \neq 0).\]
These four are instances of a general fact, proved in higher algebra: every symmetric polynomial expression in \(x\) and \(y\) can be written in terms of \(s\) and \(p\) alone. For this course, the identities above are all we need.
Keep an eye on the second identity: \((x-y)^2 = s^2 - 4p\) is exactly the discriminant of the quadratic about to appear. It measures how far apart the two unknowns are — and it will decide how many solutions the system has.
Sum–product systems reduce to a quadratic
Let \(s\) and \(p\) be given real numbers. A pair \((u, v)\) is a solution of the system
\[x + y = s, \qquad xy = p\]
if and only if \(u\) and \(v\) are the roots of the quadratic equation
\[t^2 - st + p = 0\]
in the unknown \(t\). Consequently, with \(\Delta = s^2 - 4p\):
if \(s^2 - 4p > 0\), the system has exactly two solutions, \((t_1, t_2)\) and \((t_2, t_1)\), where \(t_1 \neq t_2\) are the two roots;
if \(s^2 - 4p = 0\), it has exactly one solution, \(x = y = \tfrac{s}{2}\);
if \(s^2 - 4p < 0\), it has no real solutions.
If \(u + v = s\) and \(uv = p\), then by the converse theorem of Lesson 2 (forming an equation from its roots), \(u\) and \(v\) are exactly the roots of \(t^2 - st + p = 0\). Conversely, if \(u\) and \(v\) are the roots of that equation, Vieta's relations give \(u + v = s\) and \(uv = p\), so \((u, v)\) solves the system.
The count of solutions now follows from Lesson 1 applied to \(t^2 - st + p = 0\), whose discriminant is \(\Delta = s^2 - 4p\). If \(\Delta > 0\) there are two distinct roots \(t_1, t_2\), and the solutions of the system are the two ordered pairs \((t_1, t_2)\) and \((t_2, t_1)\). If \(\Delta = 0\) the double root \(\tfrac{s}{2}\) forces \(u = v = \tfrac{s}{2}\), a single pair. If \(\Delta < 0\) the quadratic has no real roots, so no real pair \((u, v)\) exists.
The garden, solved as a system
For \(x + y = 10,\; xy = 21\): here \(s = 10\), \(p = 21\), and \(s^2 - 4p = 100 - 84 = 16 > 0\). The unknowns are the roots of
\[t^2 - 10t + 21 = 0,\]
which is precisely the equation from Lesson 1 — with roots \(3\) and \(7\). The system therefore has two solutions, \((3, 7)\) and \((7, 3)\), describing the same garden with its sides listed in the two possible orders.
Vieta twice over
Solve the system
\[x + y + xy = 11, \qquad (x + y)\,xy = 30.\]
Both equations are symmetric, so pass to \(s = x + y\) and \(p = xy\):
\[s + p = 11, \qquad s\,p = 30.\]
Look closely: this is itself a sum–product system — in the unknowns \(s\) and \(p\)! We know their sum (\(11\)) and their product (\(30\)), so \(s\) and \(p\) are the roots of
\[w^2 - 11w + 30 = 0,\]
whose roots are \(5\) and \(6\). Two cases, each a sum–product system for \(x\) and \(y\):
Case \(s = 5,\; p = 6\): check \(s^2 - 4p = 25 - 24 = 1 \geq 0\); solve \(t^2 - 5t + 6 = 0\), roots \(2\) and \(3\), giving \((2, 3)\) and \((3, 2)\).
Case \(s = 6,\; p = 5\): check \(s^2 - 4p = 36 - 20 = 16 \geq 0\); solve \(t^2 - 6t + 5 = 0\), roots \(1\) and \(5\), giving \((1, 5)\) and \((5, 1)\).
Four solutions in all: \((2,3),\; (3,2),\; (1,5),\; (5,1)\) — verify one: \(1 + 5 + 5 = 11\) and \(6 \cdot 5 = 30\). One theorem, applied at two levels of the same problem.
A pair that promises more than it delivers
Solve \(x + y + xy = 5,\; (x + y)\,xy = 6\). As before, \(s + p = 5\) and \(sp = 6\), so \(s\) and \(p\) are the roots of \(w^2 - 5w + 6 = 0\): the pairs \((s, p) = (2, 3)\) or \((s, p) = (3, 2)\).
Now the existence check earns its keep:
\((s, p) = (2, 3)\): \(s^2 - 4p = 4 - 12 = -8 < 0\). No real numbers have sum \(2\) and product \(3\) — this case yields nothing, and we discard it.
\((s, p) = (3, 2)\): \(s^2 - 4p = 9 - 8 = 1 \geq 0\); solve \(t^2 - 3t + 2 = 0\), roots \(1\) and \(2\).
The system has exactly two solutions, \((1, 2)\) and \((2, 1)\). A valid pair \((s, p)\) at the upper level does not guarantee real \(x, y\) at the lower level — existence must be checked at every step, just as the critical examination of Lesson 2 warned.
Connection to Computer Science
The encryption that protects online banking and shopping (RSA) hides two enormous prime numbers, publishing only their product \(n = pq\); its security rests entirely on how hard it is to split \(n\) back into \(p\) and \(q\). But if a careless implementation also leaks a second symmetric quantity — in effect the sum \(p + q\) — the secret is gone at once: attack programs recover the primes as the two roots of \(t^2 - st + n = 0\), exactly this lesson's construction. Knowing the product is safe; knowing sum and product is knowing everything.
Symmetric problem, symmetric answer?
The system cannot tell \(x\) from \(y\) — so it is tempting to conclude that the answer cannot either, and write "by symmetry, \(x = y\)". The garden refutes this immediately: its solutions are \((3, 7)\) and \((7, 3)\), and in neither of them is \(x\) equal to \(y\).
What symmetry actually guarantees is subtler and worth stating precisely: if \((u, v)\) solves a symmetric system, then so does \((v, u)\) — swapping the unknowns changes no equation, so it cannot change what solves them. The symmetry lives in the set of solutions, which must be closed under swapping; it does not live inside each individual solution. A symmetric question can have asymmetric answers, provided they come in mirrored pairs.
When is there a solution with \(x = y\)? Our identity \((x - y)^2 = s^2 - 4p\) answers exactly: \(x = y\) happens precisely when \(s^2 - 4p = 0\) — when the discriminant of \(t^2 - st + p\) vanishes and the two roots collapse into one. The discriminant, once again, is the whole story: positive means a genuinely split pair, zero means the perfectly balanced one, negative means no pair at all.
Exercises
Exercise 1
Two numbers satisfy \(x + y = 6\) and \(xy = 3\). Without finding \(x\) and \(y\), compute:
Exercise 2
Solve each system by constructing the quadratic \(t^2 - st + p = 0\).
Exercise 3
For each system, choose the complete set of real solutions \((x, y)\).
Exercise 4
Decide whether each statement about symmetric systems is true or false.
Exercise 5
Swap \(x\) and \(y\) in every equation. Does each equation stay the same?
Summary
A system is symmetric when swapping \(x\) and \(y\) leaves every equation unchanged; its solution set is then closed under swapping — solutions come in pairs \((u, v)\), \((v, u)\).
Symmetric expressions rewrite in terms of \(s = x + y\) and \(p = xy\): in particular \(x^2 + y^2 = s^2 - 2p\), \(x^3 + y^3 = s^3 - 3sp\), and \((x - y)^2 = s^2 - 4p\).
The method: reduce the system to values of \(s\) and \(p\), then recover the unknowns as the roots of \(t^2 - st + p = 0\) — Lesson 2's converse theorem at work.
Existence check: a candidate pair \((s, p)\) produces real solutions only when \(s^2 \geq 4p\); pairs failing the check are discarded. Equality \(s^2 = 4p\) is exactly the case \(x = y = \tfrac{s}{2}\).
With the equation (Lesson 1), Vieta's relations (Lesson 2), and symmetric systems (this lesson), Unit 1 is complete: next, the quadratic function and its graph.