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Polynomial Equations
A polynomial equation in two variables is an equation of the form p(x, y) = q(x, y)
where both p(x, y) and q(x, y) are polynomials in two variables.
Examples.
- xy + 2 = y 2 3 x 4 (xy + 2 is a quadratic polynomial. So is y 2 3 x 4.)
- y x = 2 (y x is a linear polynomial. 2 is a constant polynomial.)
- x 2 5 x + y 2 = 7 xy y + 2
- 3 x 2 xy + 4y 2 5 x + 6 7 = 0
Domains of polynomial equations
Because every polynomial in two variables has a domain of R 2 , the implied domain of any polynomial equation in two variables is R 2 , the entire plane.
Solutions of equations
If p(x, y) = q(x, y) is a polynomial equation in two variables, then a point in the plane (↵, ) 2 R 2 is a solution of the equation if the number p(↵, ) equals the number q(↵, ). That is, if p(↵, ) = q(↵, ).
Examples.
- The point in the plane (1, 2) is a solution of the equation x 2 + y 2 = x + y + 2 because 1^2 + 2 2 = 5 = 1 + 2 + 2.
- The point (0, 1) is a solution of the same equation x 2 +y 2 = x+y + because 0 2 + ( 1) 2 = 1 = 0 + ( 1) + 2.
- The point (3, 1) is not a solution of the same equation x 2 + y 2 = x + y + 2 because 3^2 + 1 2 = 10 6 = 6 = 3 + 1 + 2.
The set of every solution of an equation p(x, y) = q(x, y) is the set S = { (↵, ) 2 R 2 | p(↵, ) = q(↵, ) } Thus, if S is the set of solutions of the equation from the previous examples, x 2 + y 2 = x + y + 2, then (1, 2) 2 S and (0, 1) 2 S, while (3, 1) 2 / S.
Solutions as geometric objects
The set of solutions of a polynomial equation in one variable is always finite, so we can just write out a list of the solutions. In contrast, a polynomial equation in two variables can have infinitely many solutions. It would be impossible to write a list of infinitely many points in the plane, but it often is possible to make a drawing of every point in the plane that is a solution of a particular equation.
Examples.
- Let p(x, y) be the linear polynomial x and let q(x, y) be the constant polynomial 3. The solutions of the equation p(x, y) = q(x, y), the equation x = 3, are the points (↵, ) in the plane with the property that
↵ = p(↵, ) = q(↵, ) = 3
Thus, (3, 7), and (3, 2), and (3, 100) are solutions of the equation x = 3. The set of all solutions, of all pairs of numbers whose x-coordinates equal 3, forms a vertical line in the plane.
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- The point (^) (3, 1) is not a solution of the same equation
x 2 +y 2 =x+y+2 because 32+12^ = 106=3+i+2.
The set of solutions (^) of an equation p(x, (^) y) = q(x, (^) y) is the set S={() (^) ER 2 p(3)=q(3)} Thus, if S is the set of solutions of the (^) equation from the previous examples, = x+y+2, then (1,2) (^) E Sand (0,—i) (^) eS, while (3,1)ØS.
Solutions as geometric objects
of solutions of a polynomial (^) eauatirrtiiio variables is a subset of the plane. :aw the set of solutions (^) of a particular
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- Let (^) p(x, (^) y) be the linear polynomial x — 3 and let q(x, y)e constant polynomial (^) 0. The solutions of the equation p(r, (^) y) = q(x, (^) y), the+Zc cç equation x (^) — 3 = 0, are the points (^) (, 3) in the plane with the property thatC^4
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Lt wSS or more simply, (^) points (a, (^) /) in the plane with o = 3. Thus, (^) (3, 7), and (3, —2), and (^) (3, 100) (^) are solutions of the equation x — 0. The set of all solutions, of all pairs (^) of numbers whose x-coordina’qual 3, forms a vertical line in the plane.
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Examples.
- The set of solutions of the equation y 4 y 2 = x 4 2 x 2 is called the “devil’s curve”.
- The three previous examples were of equations that had infinitely many solutions. Sometimes a polynomial equation will have no solutions. If you square a number, the result cannot be negative. If you add two nonneg- ative numbers together, the result is still nonnegative. This is to say that the equation x 2 + y 2 = 1 has no solution. There are no pairs of numbers that you can square and then add together to get the negative number 1.
- Some equations in two variables have a finite number of solutions. The claim below displays one such equation.
Claim: The only solution of the equation x 2 + y 2 = 0 is the point (0, 0).
Proof: Suppose that (↵, ) is a solution of the equation x 2 + y 2 = 0. We’ll show that (↵, ) = (0, 0). That means that (0, 0) is the only solution. If (↵, ) is a solution of the equation x 2 + y 2 = 0, then ↵ 2 + 2 = 0. The square of a number can’t be negative. Thus ↵ 2 0 and 2 0. Since neither ↵ 2 nor 2 are negative, the only way they could sum to 0 is if they both equal zero. That is, ↵ 2 + 2 = 0 implies that ↵ 2 = 0 and 2 = 0. The only number 122
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- (^) The three previous examples were of (^) equations that had infinitely many solutions. Sometimes a polynomial equation will (^) have no solutions. If you square a number, the result cannot be negative. (^) If you add two nonneg ative numbers together, the result is still nonnegative. (^) This is to say that the equation x (^2) + y 2 —1 has no solution. There are no pairs of (^) numbers that you can square and then add together to get the negative number (^) —1.
- Some equations in two variables have a finite number of (^) solutions. The claim (^) below displays one such equation.
Claim: The only solution (^) of the equation x (^2) + y 2 = 0 is the point (^) (0,0).
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Proof: Suppose (^) that (^) (a’, ,B) is a solution of the equation x (^2) + y 2 = 0. We’ll show that (^) (a’, 3) = (0,0). That means that (0,0) is the only solution. i (^) ta’, i’ is a solution (^) of the equation (^2) + y 2 = 0, then (^2) + = 0. The square of a number can’t he (^) negative. Thus a’ 2 0 and /32^ 0. Since neither a’ 2 nor ,2^ are negative, the (^) only way they (^) could sum to 0 is if they both equal zero. That is, a’ (^2) + /32^ = 0 implies that a’ (^2) = 0 and 32 = 0. The only number
i. • The set of solutions of the equation (^) y (^4) —^ =^ —^ 2x2^ is callee 4 devil’s^ curve’
that you can square to get 0, is 0 itself. In other words, since ↵ 2 = 0, we must have that ↵ = 0. Since 2 = 0, we must have that = 0. We’ve now shown what we wanted to. If (↵, ) is a solution of the equation x 2 + y 2 = 0, then ↵ 2 + 2 = 0, which implies that (↵, ) = (0, 0). Thus, (0, 0) is the only solution of the equation x 2 + y 2 = 0. ⌅
Equivalent equations
Similar to equations in one variable, there are rules for when equations in two variables are equivalent, and equivalent equations have the same solu- tions. When it comes to polynomial equations in two variables, there are only two rules for equivalent equations that we’ll need.
Equivalent by addition:
The equation p(x, y) + h(x, y) = q(x, y) is equivalent to p(x, y) = q(x, y) h(x, y).
Equivalent by multiplication:
If c 6 = 0 is a number, then the equation cp(x, y) = q(x, y) is equivalent to p(x, y) = q(x,y c ).
Consequence of equivalent equations:
Equivalent equations have the same set of solutions.
Example.
- The equation 8 x 2 + 2y 2 + 6x = 4xy + 14y + 10
Claim: Suppose T : R 2! R 2 is a planar transformation — either an addition function or an invertible matrix — and that S ✓ R 2 is the set of solutions of the polynomial equation p(x, y) = q(x, y). Then T (S) is the set of solutions of the equation p T ^1 (x, y) = q T ^1 (x, y).
Proof: Let’s take a point in the set T (S). That is, a point of the form T (↵, ) where (↵, ) 2 S, which means that (↵, ) is a solution of the equation p(x, y) = q(x, y), or in other words, that
p(↵, ) = q(↵, )
We want to show that this point, T (↵, ), in the set T (S) is a solution of the equation p T ^1 (x, y) = q T ^1 (x, y)
To see this, we need to check that p T ^1 (T (↵, )) equals q T ^1 (T (↵, )). We’ll do this in a moment, but before we do, remember two things: that the definition of inverse functions is that T ^1 (T (x, y)) = (x, y), and that p(↵, ) = q(↵, ). Now we check that p T ^1 (T (↵, )) equals q T ^1 (T (↵, )):
p T ^1 (T (↵, )) = p(T ^1 (T (↵, ))) = p(↵, ) = q(↵, ) = q(T ^1 (T (↵, ))) = q T ^1 (T (↵, )) Thus, points in the set T (S), such as T (↵, ), are solutions of the equation p T ^1 (x, y) = q T ^1 (x, y). ⌅
The previous claim is important to stress. We’ll call it the Principle of Transforming Solutions (POTS). If S is the set of solutions of p(x, y) = q(x, y), then T (S) is the set of solutions of p T ^1 (x, y) = q T ^1 (x, y).
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Example.
- Let p(x, y) = x and q(x, y) = 3. If S is the set of solutions of the equation p(x, y) = q(x, y), which is the equation x = 3, then we’ve seen that S is a vertical line. The addition function A(2,0) : R 2! R 2 moves points in the plane right by 2 units, so the set A(2,0) (S) is the line of solutions of x = 3 shifted right by 2 units. It’s the vertical line of points whose x-coordinates equal 5.
POTS tells us that the line A(2,0) (S) is the set of solutions of the equation p A (2^1 ,0) (x, y) = q A (2^1 ,0) (x, y)
Because A (2^1 ,0) (x, y) = (x 2 , y), you can check that the equation
p A (2^1 ,0) (x, y) = q A (2^1 ,0) (x, y) is the same as the equation x 2 = 3 In summary, POTS tells us that the line A (^) (2,0) (S) is the set of solutions of x 2 = 3, or equivalently, of x = 5.
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The addition function A( 2 ,o) : R 2 —+ R moves ponts in the plane right (^) by 2 units, so the set A( 2 .o)(S) is the line of so1utions — (^3) = 0 shifted right (^) by 2 units. It’s the vertical line of points whose x-coordinatqua1.’5.
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Namely, A( 2 ,o) (S), is the set of solutions of the (^) equation poA’O)(x,y) = In order to progress further, (^) we’d need to know what the inverse of the planar transformation (^) A( 2 ,o) is, and we do: A’ 0 ) = A_( 2 ,o). Thus, our new line, A( 2 ,o) (8), is^ the^ set of solutions of the equation p o^ A_( 2 ,o)(x, (^) y) = 0 We want to write the above (^) equation more simply. To this end, notice that p o^ A(_^2 ,o)(x,^ y) = p(x — 2, (^) y) = (x — 2) — 3 = x — 5 Thus, (^) theVA( 2 ,o)(S)is the set of solutions (^) of the equation x — 5 = 0. We had seen lier in this chapter that (^) this is indeed the case. The solutions of x — (^5) = 0 are a vertical line (^) of points whose x-coordinate equals 5.
The statement of the previous claim (^) is important -po4ii- to stress. We’ll use it to help us determine what sorts of solutions (^) there are to equations p(x, y) =^0 when^ p(x,^ y) is^ a^ quadratic^ polynomial.
If S is the set of solutions (^) of p(x, (^) y) = 0, then T(S) is the set of solutions (^) of (^) p o T 1 (x, (^) y) = 0.
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Exercises
Let’s look at the equation 5x + y 2 = 3x 2 xy. Determine whether the points given in #1-4 are solutions of this equation.
5.) Are the equations x 2 + 2 = x and x 2 x + 2 = 0 equivalent by addition?
6.) Are the equations xy = y 2 and xy x = y 2 y equivalent by addition?
7.) Are xy + 3 = y 2 x 2 and 3 = y 2 x 2 xy equivalent by addition?
8.) Are the equations 2x 3 = x 2 and 2x = x 2 equivalent by addition?
9.) Are the equations 2xy 3 x 2 = y 2 x y and xy 3 x 2 = y 2 x y equivalent by multiplication?
10.) Are 5x 2 5 x + 10 = 20y 2 and x 2 x + 2 = 4y 2 equivalent by multi- plication?
We saw in this chapter that x 2 +y 2 = 0 has a single solution, the point (0, 0). In #11-14, let p(x, y) = x 2 + y 2 and q(x, y) = 0, so that p(x, y) = q(x, y) is an equation whose only solution is (0, 0).
11.) What is A (^) (2,3) (0, 0)?
12.) What are p A (2^1 ,3) (x, y) and q A (2^1 ,3) (x, y)?
13.) What’s the only solution of the polynomial equation p A (2^1 ,3) (x, y) = q A (2^1 ,3) (x, y)? (You can use POTS and #11 to answer this.)
14.) Use POTS and the polynomials p(x, y) = x 2 + y 2 and q(x, y) = 0 (as in #13) to write a polynomial equation in two variables that has only one solution: the point (a, b) in the plane.
log (^) e (x) is the most common logarithm used in math. There are lots of benefits to using logarithms base e, and these benefits will be explained in calculus. Because of these benefits, some call logarithm base e the “natural logarithm”, and they write it as ln(x). (Scientists often prefer to write ln(x). Mathematicians often write log (^) e (x) as log(x). To make matters more con- fusing, if a calculator has a button for log(x), it probably means log 10 (x).) Because plenty of people write log (^) e (x) as ln(x), we should practice seeing and writing the logarithm base e in this way. Find the values asked for in #15-30.
15.) ln ^1 (2) 16.) ln(e) 17.) ln(e 2 ) 18.) ln(e 3 )
19.) ln(e 4 ) 20.) ln( (^1) e ) 21.) ln( (^) e^1 2 ) 22.) ln( (^) e^1 3 )
23.) ln(
p e) 24.) ln( 3
p e) 25.) ln( 4
p e) 26.) ln( p^1 e )
27.) ln( p (^31) e ) 28.) ln( p (^41) e ) 29.) ln(1) 30.) ln ^1 (e)
Find the solutions of the following equations in one variable.
31.) ln(2x 3) = 0 32.) ln(3x 5) = 2 33.) 3 ln(7 x) 4 = 5
