Unit 2, lesson 21

Solving Rational and Radical Equations

How to clear fractions and remove square roots without losing track of the domain, why these steps can create false solutions, and how to test every candidate in the original equation.

  • Theory part 41
  • Practice part 42
  • Quadratics, Circles, and Roots

The graphs of y=x+2y = \sqrt{x + 2} and y=xy = x cross at exactly one point. Solving the equation x+2=x\sqrt{x + 2} = x by squaring, though, produces two numbers, 22 and −1-1. Only one of them is a crossing point. The other was made by the algebra, and the steps that make such false answers, clearing fractions and squaring, are exactly the steps these equations need.

Key terms

Original domain

Definition The original domain of an equation in a real variable xx is the set of real numbers at which every expression in the equation, on both sides, is defined as a real number. For rational and radical equations this means that no denominator is 00 and that the radicand of every even root (square root, fourth root and so on) is nonnegative. Odd roots such as x3\sqrt[3]{x} add no restriction. A solution of the equation is a number in the original domain that makes the two sides equal.

In plain words Before any algebra, write down which numbers are allowed, the same way you find a natural domain. A number outside that set can never be a solution, whatever the later steps say.

Example Consider the equation

x+3=4x−1.\sqrt{x + 3} = \frac{4}{x - 1}.

The square root needs x+3≥0x + 3 \ge 0, so x≥−3x \ge -3, and the fraction needs x≠1x \ne 1. The original domain is [−3,1)∪(1,∞)[-3, 1) \cup (1, \infty). If the algebra later produces x=1x = 1 or x=−5x = -5, that number is rejected at once.

Equivalent transformations

Definition Two equations are equivalent on a set DD if they have exactly the same solutions in DD. A step that turns an equation into one that is equivalent to it on the original domain DD is an equivalent transformation. The following steps are equivalent transformations.

  1. Adding the same expression to both sides, or subtracting it, when the expression is defined at every point of DD.
  2. Multiplying or dividing both sides by the same nonzero number.
  3. Multiplying both sides by an expression g(x)g(x) that is defined and nonzero at every point of DD.

Equivalence only speaks about numbers in DD. The new equation may also be defined and true at numbers outside DD, and those numbers are never solutions. Other common steps are not always equivalent. Multiplying both sides by an expression that is 00 at some point of DD, or squaring both sides, can add solutions. Dividing both sides by an expression that is 00 at some point of DD can lose solutions.

In plain words Safe steps keep the solution set exactly as it was. After a risky step you hold candidates, not answers, and you have to sort them.

Example The equation x2=3xx^2 = 3x has two solutions. Subtracting 3x3x is safe and gives x(x−3)=0x(x - 3) = 0, so the solutions are 00 and 33. Dividing both sides by xx instead gives x=3x = 3, and the solution 00 disappears, because dividing by xx is not allowed when x=0x = 0.

Least common denominator

Definition Factor every denominator completely. The least common denominator (LCD) of several rational expressions is the product of the least common multiple of the numerical factors and every distinct polynomial factor that appears, each raised to the highest power with which it occurs in any one denominator. Multiplying both sides of a rational equation by the LCD clears every fraction. The LCD is nonzero at every point of the original domain, so on that domain this is an equivalent transformation. The cleared equation, however, is defined for all real xx, and it may have solutions where the LCD is 00, which lie outside the original domain.

In plain words Use the smallest expression that every denominator divides into. A larger common denominator also clears the fractions, but it raises the degree and brings in extra candidates to reject.

Example Solve the equation

3x+2−1x−2=4x2−4.\frac{3}{x + 2} - \frac{1}{x - 2} = \frac{4}{x^2 - 4}.

Since x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2), the original domain is x≠2x \ne 2 and x≠−2x \ne -2, and the LCD is (x−2)(x+2)(x - 2)(x + 2). Multiplying both sides by it gives

3(x−2)−(x+2)=4,2x−8=4,x=6.\begin{gathered} 3(x - 2) - (x + 2) = 4, \\[0.4em] 2x - 8 = 4, \\[0.4em] x = 6. \end{gathered}

The candidate 66 is in the domain and checks, since

38−14=18=432.\frac{3}{8} - \frac{1}{4} = \frac{1}{8} = \frac{4}{32}.

Multiplying by the larger common denominator (x2−4)2(x^2 - 4)^2 instead leads to (x2−4)(2x−12)=0(x^2 - 4)(2x - 12) = 0, with the extra candidates 22 and −2-2, both outside the domain.

Squaring

Definition For real numbers aa and bb, if a=ba = b then a2=b2a^2 = b^2, but a2=b2a^2 = b^2 only says that a=ba = b or a=−ba = -b. So squaring both sides of A(x)=B(x)A(x) = B(x) keeps every solution and may add the solutions of A(x)=−B(x)A(x) = -B(x). When A(x)≥0A(x) \ge 0 and B(x)≥0B(x) \ge 0 for every xx under consideration, squaring is an equivalent transformation. For an isolated square root this gives an exact rule,

A(x)=B(x)exactly whenA(x)=B(x)2 and B(x)≥0.\begin{gathered} \sqrt{A(x)} = B(x) \\[0.4em] \text{exactly when} \\[0.4em] A(x) = B(x)^2 \ \text{and} \ B(x) \ge 0. \end{gathered}

The condition A(x)≥0A(x) \ge 0 need not be added, because A(x)=B(x)2A(x) = B(x)^2 already forces it.

In plain words Isolate the root first, then square. Squaring cannot tell BB from −B-B, so it also solves the equation with the opposite sign, and those extra answers must be weeded out. If a second root is left, isolate it and square again.

Example To solve x+2=x\sqrt{x + 2} = x, square both sides.

x+2=x2,x2−x−2=0,(x−2)(x+1)=0.\begin{gathered} x + 2 = x^2, \\[0.4em] x^2 - x - 2 = 0, \\[0.4em] (x - 2)(x + 1) = 0. \end{gathered}

The candidates are 22 and −1-1. The condition x≥0x \ge 0 keeps 22 and rejects −1-1, and indeed −1+2=1≠−1\sqrt{-1 + 2} = 1 \ne -1. The number −1-1 solves −x+2=x-\sqrt{x + 2} = x instead. On a graph, the squared equation finds where the line y=xy = x meets the whole sideways parabola y2=x+2y^2 = x + 2. The upper half y=x+2y = \sqrt{x + 2} gives the point (2,2)(2, 2) and the lower half y=−x+2y = -\sqrt{x + 2} gives (−1,−1)(-1, -1).

Extraneous solutions

Definition An extraneous solution is a number that solves an equation reached later in the work but does not solve the original equation. Extraneous solutions have two sources. A step that is not an equivalent transformation, mainly squaring both sides, can add numbers inside the original domain. And a later equation may be defined at numbers outside the original domain and hold there. Clearing fractions does this, because the cleared equation is defined for every real xx and may hold where the LCD is 00, and those are exactly the numbers a denominator excludes. Extraneous solutions are not arithmetic mistakes, so redoing the algebra carefully will not make them go away.

In plain words An extraneous solution is a false answer that the method itself creates. It satisfies the equation you ended with but not the one you started with.

Example Solve the equation

xx−1+1x+1=2x2−1.\frac{x}{x - 1} + \frac{1}{x + 1} = \frac{2}{x^2 - 1}.

The original domain is x≠1x \ne 1 and x≠−1x \ne -1. Multiplying by the LCD (x−1)(x+1)(x - 1)(x + 1) gives

x(x+1)+(x−1)=2,x2+2x−3=0,(x+3)(x−1)=0.\begin{gathered} x(x + 1) + (x - 1) = 2, \\[0.4em] x^2 + 2x - 3 = 0, \\[0.4em] (x + 3)(x - 1) = 0. \end{gathered}

The candidate 11 is extraneous, since it lies outside the original domain. The candidate −3-3 checks, because

−3−4+1−2=14and29−1=14.\frac{-3}{-4} + \frac{1}{-2} = \frac{1}{4} \quad \text{and} \quad \frac{2}{9 - 1} = \frac{1}{4}.

The only solution is x=−3x = -3.

Checking candidates in the original equation

Definition Once the work includes squaring, clearing fractions or any other step that can bring in extraneous solutions, the numbers you find at the end are only candidates. A candidate cc is a solution exactly when cc lies in the original domain and substituting cc into the original equation, not into a later one, makes both sides equal. Substituting into the equation where a candidate appeared cannot expose it, because every candidate satisfies that equation. When the equation is A(x)=B(x)\sqrt{A(x)} = B(x) and a candidate cc satisfies A(c)=B(c)2A(c) = B(c)^2, testing B(c)≥0B(c) \ge 0 is an exact check. This helps when cc is irrational, where a decimal substitution only gives approximate evidence.

In plain words Go back to the very first line and test each candidate there. Checking in the squared or cleared equation proves nothing, since the false answers pass that test too.

Example Solve x+x=6x + \sqrt{x} = 6. Isolating the root gives x=6−x\sqrt{x} = 6 - x, and squaring gives

x=(6−x)2,x2−13x+36=0,(x−4)(x−9)=0.\begin{gathered} x = (6 - x)^2, \\[0.4em] x^2 - 13x + 36 = 0, \\[0.4em] (x - 4)(x - 9) = 0. \end{gathered}
CandidateSquared equation x=(6−x)2x = (6 - x)^2Original equation x+x=6x + \sqrt{x} = 6
444=44 = 4, true4+2=64 + 2 = 6, true
999=99 = 9, true9+3=129 + 3 = 12, false

Both candidates pass the squared equation, and only 44 passes the original one, so the only solution is x=4x = 4. The sign test agrees, since 6−4=2≥06 - 4 = 2 \ge 0 while 6−9=−3<06 - 9 = -3 < 0.

The sign test is handy for irrational candidates. Squaring x+4=x\sqrt{x + 4} = x gives x2−x−4=0x^2 - x - 4 = 0, with candidates

x=1±172.x = \frac{1 \pm \sqrt{17}}{2}.

Only the positive one, 1+172≈2.56\frac{1 + \sqrt{17}}{2} \approx 2.56, satisfies x≥0x \ge 0, so it is the only solution. The other candidate, about −1.56-1.56, lies in the original domain x≥−4x \ge -4, so the domain alone does not reject it. The sign test does.

Check yourself

  1. Solve the equation
2x−3+1x=6x(x−3).\frac{2}{x - 3} + \frac{1}{x} = \frac{6}{x(x - 3)}.
Show the answer

The original domain is x≠0x \ne 0 and x≠3x \ne 3. Multiplying by the LCD x(x−3)x(x - 3) gives 2x+(x−3)=62x + (x - 3) = 6, so 3x=93x = 9 and x=3x = 3. That candidate is outside the original domain, so the equation has no solution.

  1. Solve 3x+1=x−3\sqrt{3x + 1} = x - 3.
Show the answer

Squaring both sides gives

3x+1=x2−6x+9,x2−9x+8=0,(x−1)(x−8)=0.\begin{gathered} 3x + 1 = x^2 - 6x + 9, \\[0.4em] x^2 - 9x + 8 = 0, \\[0.4em] (x - 1)(x - 8) = 0. \end{gathered}

At x=1x = 1 the left side is 22 and the right side is −2-2, so 11 is extraneous. At x=8x = 8 both sides equal 55. The only solution is x=8x = 8.

  1. Solve 2x+3−x+1=1\sqrt{2x + 3} - \sqrt{x + 1} = 1.
Show the answer

The original domain is x≥−1x \ge -1. Isolate one root as 2x+3=1+x+1\sqrt{2x + 3} = 1 + \sqrt{x + 1} and square.

2x+3=1+2x+1+(x+1),x+1=2x+1.\begin{gathered} 2x + 3 = 1 + 2\sqrt{x + 1} + (x + 1), \\[0.4em] x + 1 = 2\sqrt{x + 1}. \end{gathered}

Squaring again gives (x+1)2=4(x+1)(x + 1)^2 = 4(x + 1), so (x+1)(x−3)=0(x + 1)(x - 3) = 0 and the candidates are −1-1 and 33. Both check, since 1−0=1\sqrt{1} - \sqrt{0} = 1 and 9−4=1\sqrt{9} - \sqrt{4} = 1. On the domain x≥−1x \ge -1 both sides were nonnegative before each squaring, so each squaring was an equivalent transformation there and every candidate in the domain had to be a solution. Dividing x+1=2x+1x + 1 = 2\sqrt{x + 1} by x+1\sqrt{x + 1} would have lost the solution −1-1.

How this lesson connects

Where this leads in Calculus