Isolating a radical and raising both sides to the index, a hurricane wind-speed model, solving equations with rational exponents using reciprocal exponents, extraneous solutions introduced by squaring, and equations with two radicals that require squaring twice.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 6 — Rational Exponents and Radical Functions
Solve Radical Equations
Objectives
Five outcomes. The fourth is the reason every answer in this lesson has to be checked.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 452-459 — the lesson these objectives are drawn from
Warm-up
Lesson 6.5 graphed radical functions. This lesson solves equations built from them.
Discussion prompt
Solve the cube root of the quantity 2x plus 7 equals 3. Then try squaring both sides of x equals negative 3 and see what solutions the result has.
Hint: Undo a cube root by cubing.
Answer:
\[ \sqrt[3]{2x+7} = 3 \;\Longrightarrow\; 2x+7 = 27 \;\Longrightarrow\; x = 10 \]
\[ x = -3 \;\Longrightarrow\; x^2 = 9 \;\Longrightarrow\; x = 3 \text{ or } x = -3 \]
The first is straightforward. The second is the warning: squaring turned an equation with one solution into one with two, because squaring loses the sign. Every equation solved by squaring in this lesson needs its answers checked.
Concept
A radical equation is solved by isolating the radical and raising both sides to the index. Raising to an odd power is safe, but raising to an even power can produce apparent solutions that satisfy the new equation and not the original, so every answer must be checked.
radical equation — An equation containing a radical whose radicand includes a variable. Solving one means isolating the radical and raising both sides to the index.
\[ \sqrt[n]{\,\cdot\,} = k \;\Longrightarrow\; (\,\cdot\,) = k^n \]
The checking step is not caution for its own sake. In Example 5 exactly half the apparent solutions are wrong, and in Example 6 the same is true, so skipping the check is not a risk but a mistake.
Figure (svg): Two columns separating operations that preserve solutions from those that can add them
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 452-454
Section
Section 1
Concept
Isolate the radical on one side, raise both sides to the power matching the index, and solve the polynomial equation that results. Then check every answer in the original equation.
\[ \sqrt[3]{2x+7} = 3 \;\Longrightarrow\; 2x+7 = 27 \;\Longrightarrow\; x = 10 \]
The index decides the power: a cube root needs cubing and a fourth root needs raising to the fourth. Squaring a cube root equation leaves a radical behind and makes matters worse.
Figure (svg): The three steps for solving a radical equation, applied to a cube root equation
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 452-452 — Solving Radical Equations
Picture it
Example 1: a cube root equation.
Figure (svg): The three steps for solving a radical equation, applied to a cube root equation
Cubing removed the radical in one move because the index was 3. The check confirmed the answer, and with an odd index it was guaranteed to.
Worked example
Example 1, with the book's check.
\[ \text{Solve } \sqrt[3]{2x+7} = 3. \]
Check whether the radical is isolated
Why: It already stands alone on the left.
Cube both sides
Why: The index is 3, so cubing removes the radical and 3 cubed is 27.
\[ 2 x + 7 = 27 \]
Solve the linear equation
Why: Subtracting 7 gives 2x equal to 20.
\[ x = 10 \]
Check in the original
Why: Two times 10 plus 7 is 27, whose cube root is 3.
Figure (svg): The solution to Worked example solve a cube root equation shown as a ladder of expressions, one row per algebraic move
\[ x = 10 \]
Verify: notice why no extraneous solution appeared
Why: Cubing is reversible: every real number has exactly one cube root, so the cubed equation has exactly the same solutions as the original. Raising to an odd power never adds solutions, which is why this check confirmed rather than eliminated.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 452-452
Matching
The exponents must multiply to 1.
Match the pairs
Why: In every case the exponent one over n is undone by raising to the nth power, since the two multiply to 1. The odd-index rows are safe; the even-index rows can introduce extraneous solutions, which is the subject of the fourth idea.
Worked example
Guided Practice 1 and 2.
\[ \text{Solve } \sqrt[3]{x-9} = -1 \text{ and } \sqrt{x+25} = 4. \]
First: cube both sides
Why: Negative 1 cubed is negative 1.
\[ x - 9 = -1 \]
First: solve
Why: Adding 9 gives 8.
\[ x = 8 \]
Second: square both sides
Why: The index is 2, and 4 squared is 16.
\[ x + 25 = 16 \]
Second: solve and check
Why: Subtracting 25 gives negative 9; substituting back, the radicand is 16 and its square root is 4.
\[ x = -9 \]
Figure (svg): The solution to Worked example two more, one needing rearrangement shown as a ladder of expressions, one row per algebraic move
\[ x = 8; \qquad x = -9 \]
Verify: notice that a negative solution can be perfectly valid
Why: The second answer is negative 9, and there is nothing wrong with that: the radicand x plus 25 is positive there. What a square root forbids is a negative radicand, not a negative solution. Confusing the two is a common reason for rejecting a correct answer.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 452-452
Trap
\[ \sqrt[3]{2x+7} = 3 \]
Square both sides to remove the radical
Why: Squaring is the familiar move from Lesson 4.5, so it is applied here too.
\[ (\sqrt[3]{2x+7})^2 = 9 \;\Longrightarrow\; (2x+7)^{2/3} = 9 \quad \text{(radical still there)} \]
Squaring a cube root gives a two-thirds power, which is still a radical. Nothing has been gained.
\[ (\sqrt[3]{2x+7})^3 = 3^3 \;\Longrightarrow\; 2x+7 = 27 \]
Raise both sides to the power matching the index
Why: A cube root is undone by a cube, because the exponents one third and 3 multiply to 1.
\[ \tfrac{1}{3}\cdot 3 = 1 \]
The right power is always the index. Checking that the two exponents multiply to 1 is the quickest way to confirm you have chosen it correctly.
Fill the middle
Example 1.
Fill in the blanks
\sqrt[3]3 = 3 \;\Longrightarrow\; 2x+7 = 3^___} = 27
Why: The index is 3, so both sides are cubed and 3 cubed is 27. Raising the right side to the same power as the left is what keeps the equation balanced, and forgetting to do it to both sides is the commonest slip in the first step.
Ranking
Solving a radical equation.
Put in order
Why: Step one before step two matters: raising a sum containing a radical to a power leaves cross terms with radicals in them, as the fifth idea shows. Step five is compulsory rather than advisory whenever the power used was even.
Prediction
Commit before reasoning.
Predict first
What happens if you square both sides of the square root of x, plus 1, equals 4 without isolating first?
Correct: A radical survives in the cross term.
\[ (\sqrt{x}+1)^2 = x + 2\sqrt{x} + 1 \]
Why: Squaring a sum produces a middle term of twice the product, and that product still contains the radical. Isolating first means squaring a lone radical, which removes it completely. When an equation has two radicals this cannot be avoided entirely, and the fifth idea deals with that case by squaring twice.
Section
Section 2
Concept
A model with a radical answers one question directly and the reverse question by solving. Substitute the given output, isolate the radical, square both sides, and solve for the input.
\[ v(p) = 6.3\sqrt{1013-p} \]
The check is still worth doing, but a physical model rarely produces an extraneous solution: the quantities are usually positive and the equation usually linear once the radical is gone.
Figure (svg): Wind velocity plotted against central air pressure, with the pressure at fifty-four point five metres per second marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 453-453 — Solve a radical equation given a function
Picture it
Example 2: a hurricane's wind speed from its central pressure.
Figure (svg): Wind velocity plotted against central air pressure, with the pressure at fifty-four point five metres per second marked
A wind speed of 54.5 metres per second corresponds to a central pressure of about 938 millibars. The curve falls to the right because a higher pressure means a smaller radicand.
Worked example
Example 2. Divide, square, then solve.
\[ \text{With } v(p) = 6.3\sqrt{1013-p}, \text{ find } p \text{ when } v = 54.5. \]
Substitute the wind speed
Why: The equation becomes 54.5 equals 6.3 times the square root.
\[ 54.5 = 6.3 \sqrt{1013 - p} \]
Isolate the radical
Why: Dividing both sides by 6.3 gives about 8.65.
\[ 8.65 = \sqrt{1013 - p} \]
Square both sides
Why: Eight point six five squared is about 74.8.
\[ 74.8 = 1013 - p \]
Solve for the pressure
Why: Subtracting 1013 and dividing by negative 1.
\[ p =\text{ about } 938 \]
Figure (svg): The solution to Worked example find the central pressure shown as a ladder of expressions, one row per algebraic move
\[ p \approx 938 \text{ millibars} \]
Verify: substitute the pressure back
Why: At p equal to 938, the radicand is 75, whose square root is about 8.66, and 6.3 times that is about 54.6 metres per second — matching the given speed to the accuracy of the rounding. The answer is also physically sensible: a central pressure below 940 millibars is typical of a major hurricane.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 453-453
Fill the middle
Example 2, at the first step.
Fill in the blanks
54.5 = 6.3\sqrt8.65 \;\Longrightarrow\; \sqrt___ = ___
Why: Fifty-four point five divided by 6.3 is about 8.65. Dividing first means the next step squares a lone radical rather than a product, which is both simpler and safer.
Worked example
Guided Practice 4.
\[ \text{Find } p \text{ when } v = 48.3 \text{ metres per second.} \]
Substitute and isolate
Why: Forty-eight point three over 6.3 is about 7.667.
\[ 7.667 = \sqrt{1013 - p} \]
Square both sides
Why: Seven point six six seven squared is about 58.8.
\[ 58.8 = 1013 - p \]
Solve
Why: Subtracting and negating gives about 954.
\[ p =\text{ about } 954 \]
Compare with the first
Why: A slower wind corresponds to a higher central pressure.
\[ 954\text{ against } 938 \]
Figure (svg): The solution to Worked example a slower hurricane shown as a ladder of expressions, one row per algebraic move
\[ p \approx 954 \text{ millibars} \]
Verify: check the direction of the relationship
Why: The wind speed fell from 54.5 to 48.3, about 11 percent, and the pressure rose from 938 to 954. That is the right direction: deeper hurricanes have lower central pressure and faster winds. Note also that the small percentage change in wind speed corresponds to a very small change in pressure, because the pressure enters under a square root.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 453-453
Error analysis
A student solves the hurricane model.
Annotate
On: \( 54.5 = 6.3\sqrt{1013-p} \;\Longrightarrow\; 54.5^2 = 6.3(1013-p) \)
Isolate the radical before raising to a power. Coefficients left attached get raised too, and forgetting that is easier than remembering it.
Ranking
Solving a radical model for its input.
Put in order
Why: Step three is the one that distinguishes a careful solution from a lucky one. Step five's units matter especially here: the answer is a pressure in millibars, and a number without units would not identify which quantity had been found.
Prediction
Commit before reasoning.
Predict first
For what pressures does the model give a real wind speed?
Correct: Pressures at most 1013 millibars.
\[ 1013 - p \ge 0 \;\Longrightarrow\; p \le 1013 \]
Why: The radicand 1013 minus p must be non-negative, so p is at most 1013. That is physically sensible: 1013 millibars is roughly ordinary sea-level pressure, and a hurricane's centre always has lower pressure than its surroundings. At exactly 1013 the model gives a wind speed of zero, which is the calm case.
Comparison
Fill the blanks. The model runs both ways.
Comparison matrix
| Question | Pressure to wind speed | Wind speed to pressure |
|---|---|---|
| What you substitute | p | v |
| The operation | subtract, root, multiply | divide, square, subtract |
| Is a check needed? | no, it is an evaluation | yes, squaring was used |
| The answer's units | metres per second | millibars |
Going backwards undoes each operation in reverse order, which is the inverse function of Lesson 6.4 written out one step at a time.
Section
Section 3
Concept
When the variable carries a rational exponent, isolate the power and then raise both sides to the reciprocal of that exponent. The two exponents multiply to 1, leaving the variable alone.
\[ x^{2/3} = 16 \;\Longrightarrow\; (x^{2/3})^{3/2} = 16^{3/2} \;\Longrightarrow\; x = 64 \]
This is the same idea as raising to the index, since a radical is a rational exponent. Isolating the power first is as necessary here as isolating the radical was.
Figure (svg): An equation with a rational exponent solved by raising both sides to the reciprocal exponent
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 453-454 — Examples 3 and 4
Picture it
Example 3, one line at a time.
Figure (svg): An equation with a rational exponent solved by raising both sides to the reciprocal exponent
Sixteen to the three halves is 4 cubed, or 64, computed by taking the root before the power as Lesson 6.1 recommends.
Worked example
Examples 3 and 4.
\[ \text{Solve } 3x^{2/3} = 48 \text{ and } (x+2)^{3/4} - 1 = 7. \]
First: isolate the power
Why: Dividing by 3 gives x to the two thirds equal to 16.
\[ x ^{\frac{2}{3}} = 16 \]
First: raise to the reciprocal
Why: The reciprocal of two thirds is three halves, and 16 to the three halves is 4 cubed.
\[ x = 64 \]
Second: isolate the power
Why: Adding 1 gives the bracket to the three quarters equal to 8.
\[ (x + 2) ^{\frac{3}{4}} = 8 \]
Second: raise to the reciprocal
Why: The reciprocal of three quarters is four thirds, and 8 to the four thirds is 2 to the fourth.
\[ x + 2 = 16 \]
Second: solve for x
Why: Subtracting 2 gives 14.
\[ x = 14 \]
Figure (svg): The solution to Worked example two rational-exponent equations shown as a ladder of expressions, one row per algebraic move
\[ x = 64; \qquad x = 14 \]
Verify: substitute both back
Why: For the first, 64 to the two thirds is 4 squared, or 16, and 3 times 16 is 48. For the second, 16 to the three quarters is 2 cubed, or 8, minus 1 is 7. Both check. Note that the intermediate powers were evaluated by taking the root first, which kept every number small.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 453-454
Matching
The two must multiply to 1.
Match the pairs
Why: The reciprocal simply turns the fraction upside down, and a whole-number exponent is the special case where the reciprocal is a unit fraction. Evaluating the right side is then easiest by taking the root before the power, as Lesson 6.1 established.
Worked example
Guided Practice 5 to 10. The last has two solutions.
\[ \text{Solve } 3x^{3/2}=375, \; -2x^{3/4}=-16, \; -\tfrac{2}{3}x^{1/5}=-2, \; (x+3)^{5/2}=32, \; (x-5)^{4/3}=81, \; (x+2)^{2/3}+3=7. \]
The first three: isolate, then use the reciprocal
Why: The powers become 125, 8 and 3, and the reciprocals are two thirds, four thirds and 5.
\[ 25, 16, 243 \]
Fourth: reciprocal of five halves
Why: Thirty-two to the two fifths is 2 squared, so the bracket is 4.
\[ x = 1 \]
Fifth: reciprocal of four thirds
Why: Eighty-one to the three quarters is 3 cubed, so the bracket is 27.
\[ x = 32 \]
Sixth: isolate first
Why: Subtracting 3 gives the bracket to the two thirds equal to 4, and the reciprocal is three halves.
\[ x + 2 = +- 8 \]
Sixth: note both signs
Why: An even numerator in the exponent means both signs work.
\[ x = 6\text{ or } x = -10 \]
Figure (svg): The solution to Worked example six more shown as a ladder of expressions, one row per algebraic move
\[ 25, \; 16, \; 243, \; 1, \; 32, \; \{6, -10\} \]
Verify: check the two solutions of the last
Why: At x equal to 6 the bracket is 8, and 8 to the two thirds is 4. At x equal to negative 10 the bracket is negative 8, and negative 8 to the two thirds is the square of the cube root of negative 8, which is negative 2 squared, or 4. Both give 4, so both are genuine — the even numerator is what allows the negative case.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 454-454
Trap
\[ x^{2/3} = 16 \]
Divide both sides by two thirds
Why: The exponent is treated as a coefficient.
\[ x = 16 \div \tfrac{2}{3} = 24 \quad \text{(wrong)} \]
Twenty-four to the two thirds is about 8.3, not 16. The exponent is not multiplying anything, so dividing does not remove it.
\[ (x^{2/3})^{3/2} = 16^{3/2} \;\Longrightarrow\; x = 64 \]
Raise both sides to the reciprocal exponent
Why: The two exponents multiply to 1, which is what leaves x alone.
\[ \tfrac{2}{3}\cdot\tfrac{3}{2} = 1 \]
A power is undone by another power, never by division. Checking that the exponents multiply to 1 confirms the right choice every time.
Fill the middle
Example 4.
Fill in the blanks
(x+2)^4/3 = 8 \;\Longrightarrow\; x+2 = 8^___} = 16
Why: The reciprocal of three quarters is four thirds, and 8 to the four thirds is the cube root of 8, which is 2, raised to the fourth, giving 16. Taking the root first turned a potentially awkward calculation into 2 to the fourth.
Sorting
Look at the numerator of the exponent.
Sort into buckets
Sort each equation by its number of real solutions.
It is the numerator that decides, because the numerator is the power applied after the root. The textbook's Example 3 has an even numerator and the negative solution is worth looking for.
Prediction
Commit before reasoning.
Predict first
Why does raising x to the two thirds, all to the three halves, leave x?
Correct: Because the exponents multiply to 1.
\[ (x^{2/3})^{3/2} = x^{(2/3)(3/2)} = x^1 = x \]
Why: The power of a power rule from Lesson 6.2 says the exponents multiply, and two thirds times three halves is 1, leaving x to the first. This is exactly the same mechanism as undoing a power with a root, since the nth root is the one-over-n power — one rule, two notations.
Section
Section 4
Concept
Raising both sides to an even power destroys sign information, so the new equation can have solutions the original does not. Every apparent solution must be substituted into the original equation and rejected if it fails.
extraneous solution — An apparent solution that satisfies an equation obtained during the solving process but not the original equation. Raising both sides to an even power is the usual source.
\[ x+1 = \sqrt{7x+15} \;\Longrightarrow\; x = 7 \text{ only} \]
Squaring solves the original equation and its sign-flipped twin at the same time. The check is what separates the two sets of answers again.
Figure (svg): A line and a square root curve meeting once, with the reflected line showing where the extraneous solution came from
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 454-454 — Solve an equation with an extraneous solution
Picture it
Example 5, drawn: the line, the curve and the reflected line.
Figure (svg): A line and a square root curve meeting once, with the reflected line showing where the extraneous solution came from
The solid line meets the curve only at x equal to 7. The dashed reflected line meets it at x equal to negative 2, and squaring cannot tell the two lines apart.
Worked example
Example 5, with both checks shown.
\[ \text{Solve } x+1 = \sqrt{7x+15}. \]
Square both sides
Why: The left expands to x squared plus 2x plus 1.
\[ x ^{2} + 2 x + 1 = 7 x + 15 \]
Write in standard form
Why: Collecting terms gives x squared minus 5x minus 14 equals zero.
\[ x ^{2} - 5 x - 14 = 0 \]
Factor and solve
Why: The trinomial factors as x minus 7 times x plus 2.
\[ x = 7\text{ or } x = -2 \]
Check x equal to 7
Why: The left is 8 and the radicand is 64, whose root is 8.
Check x equal to negative 2
Why: The left is negative 1 and the radicand is 1, whose root is positive 1.
Figure (svg): The solution to Worked example one solution of two survives shown as a ladder of expressions, one row per algebraic move
\[ x = 7 \]
Verify: see what the failed check actually says
Why: At x equal to negative 2 the two sides are negative 1 and positive 1 — equal in size and opposite in sign. Squaring made them agree, which is exactly how the false solution entered. Every extraneous solution from squaring has this shape.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 454-454
Sorting
Substitute into the original equation.
Sort into buckets
For x + 1 = sqrt(7x + 15) and sqrt(10x + 9) = x + 3, sort each apparent solution.
Both extraneous cases share the same signature: a negative left side against a non-negative square root. Spotting that pattern is faster than substituting.
Worked example
Guided Practice 11 and 12. The check does different work in each.
\[ \text{Solve } x - \tfrac{1}{2} = \sqrt{\tfrac{1}{4}x} \text{ and } \sqrt{10x+9} = x+3. \]
First: square both sides
Why: The left gives x squared minus x plus one quarter; the right gives x over 4.
\[ x ^{2} - x + \frac{1}{4} = \frac{x}{4} \]
First: solve
Why: Clearing fractions gives 4x squared minus 5x plus 1 equals zero, which factors.
\[ x = \frac{1}{4}\text{ or } x = 1 \]
First: check both
Why: At x equal to 1 both sides are one half; at one quarter the left is negative one quarter and the right is positive one quarter.
\[ x = 1\text{ only} \]
Second: square and solve
Why: Ten x plus 9 equals x squared plus 6x plus 9, so x squared minus 4x equals zero.
\[ x = 0\text{ or } x = 4 \]
Second: check both
Why: At 0 both sides are 3; at 4 both sides are 7.
Figure (svg): The solution to Worked example two more, one with both solutions valid shown as a ladder of expressions, one row per algebraic move
\[ x = 1; \qquad x = 0 \text{ or } 4 \]
Verify: notice that the check is not always destructive
Why: The second equation's two solutions both survive, because the right side x plus 3 is positive at both of them. An extraneous solution appears only when the two sides have opposite signs, so checking is not a formality that always removes something — it is a genuine test.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 455-455
Trap
\[ x+1 = \sqrt{7x+15} \;\Longrightarrow\; x = 7 \text{ or } x = -2 \]
Report both roots of the quadratic
Why: The quadratic was solved correctly, so both its roots are taken as answers.
\[ \text{two solutions} \quad \text{(one is false)} \]
Substituting negative 2 gives negative 1 on the left and positive 1 on the right. The two are not equal, so negative 2 solves the squared equation only.
\[ \text{check } x=7: \; 8 = \sqrt{64} = 8 \quad \checkmark \]
Substitute every apparent solution into the ORIGINAL equation
Why: The squared equation has more solutions than the original, so only the original can decide.
\[ \text{check } x=-2: \; -1 \neq \sqrt{1} = 1 \quad \times \]
A square root is never negative, so any apparent solution making the other side negative is automatically extraneous. Noticing that can save the substitution.
Fill the middle
Example 5.
Fill in the blanks
x=-2: \; -2+1 = -1, \quad \sqrt1 = \sqrt___ = ___
Why: The square root of 1 is positive 1, and the left side is negative 1. The two are not equal, so negative 2 is extraneous. The principal square root is never negative, which is exactly why the left side being negative doomed this candidate.
Prediction
Commit before reasoning.
Predict first
Before substituting, what quick test spots an extraneous solution of an equation with an isolated square root?
Correct: Check whether the non-radical side is negative there.
\[ \sqrt{\,\cdot\,} \ge 0 \;\Longrightarrow\; \text{the other side must be } \ge 0 \]
Why: A principal square root is never negative, so if the other side comes out negative the equation cannot hold. In Example 5, x equal to negative 2 makes the left side negative 1, which settles it without evaluating the radical at all. A negative solution is not itself a problem — x equal to negative 9 was perfectly valid in the first idea — but a negative side opposite a square root always is.
Comparison
Fill the blanks. Only one of them is risky.
Comparison matrix
| Question | Cubing both sides | Squaring both sides |
|---|---|---|
| Reversible? | yes | no: the sign is lost |
| Can add solutions? | no | yes |
| Is a check required? | no, though it is good practice | yes, always |
| Why | every real has one cube root | a positive number has two square roots |
The same parity distinction as everywhere else in this chapter, appearing here as the difference between a safe step and a step that must be audited.
Section
Section 5
Concept
When an equation contains two radicals, one squaring is not enough: the cross term of the expansion still contains a radical. Isolate that remaining radical and square a second time.
\[ \sqrt{x+2}+1 = \sqrt{3-x} \;\Longrightarrow\; x = -1 \]
Every squaring is a chance to introduce an extraneous solution, so an equation squared twice needs its answers checked with particular care.
Figure (svg): Two radical graphs meeting at a single point, the only solution of an equation needing two squarings
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 455-455 — Solve an equation with two radicals
Picture it
Example 6: the two sides graphed.
Figure (svg): Two radical graphs meeting at a single point, the only solution of an equation needing two squarings
The graphs meet exactly once, at the point where x is negative 1 and y is 2. The algebra produced two candidates and the graph confirms which one is real.
Worked example
Example 6, method 1.
\[ \text{Solve } \sqrt{x+2}+1 = \sqrt{3-x}. \]
Square both sides
Why: The left expands to x plus 2, plus twice the root, plus 1.
\[ x + 3 + 2 \sqrt{x + 2} = 3 - x \]
Isolate the remaining radical
Why: Collecting the non-radical terms leaves twice the root equal to negative 2x.
\[ 2 \sqrt{x + 2} = -2 x \]
Divide, then square again
Why: Dividing by 2 and squaring gives x plus 2 equal to x squared.
\[ x + 2 = x ^{2} \]
Solve the quadratic
Why: Standard form gives x squared minus x minus 2 equals zero, which factors.
\[ x = 2\text{ or } x = -1 \]
Check both
Why: At 2 the sides are 3 and 1; at negative 1 they are 2 and 2.
\[ x = -1\text{ only} \]
Figure (svg): The solution to Worked example square twice shown as a ladder of expressions, one row per algebraic move
\[ x = -1 \]
Verify: confirm with the graph
Why: Graphing both sides shows a single intersection, at the point where x is negative 1 and y is 2 — the book's second method. The graph agrees with the algebra and would have flagged the extraneous solution immediately, which is why sketching both sides is a useful habit for these equations.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 455-455
Fill the middle
Example 6, at the first squaring.
Fill in the blanks
(\sqrt2+1)^2 = (x+2) + ___\sqrt___ + 1
Why: The middle term is twice the product of the square root and 1, which is twice the square root. That cross term is the reason a single squaring does not finish the job, and it is exactly the term the perfect-square pattern of Lesson 5.3 predicts.
Worked example
Example 6, examined at the second squaring.
\[ \text{Explain why } x=2 \text{ appeared as a solution.} \]
Look at the equation before the second squaring
Why: It reads twice the square root of x plus 2 equals negative 2x.
\[ 2 \sqrt{x + 2} = -2 x \]
Note what the left side can be
Why: A square root is non-negative, so the left is at least zero.
\[ \text{left } \ge 0 \]
Note what the right side requires
Why: For the equation to hold, negative 2x must be at least zero, so x must be at most zero.
\[ x \le 0 \]
Test the two candidates
Why: Negative 1 satisfies that condition; 2 does not.
\[ 2\text{ is ruled out} \]
Figure (svg): The solution to Worked example where the extraneous solution came from shown as a ladder of expressions, one row per algebraic move
\[ x \le 0 \;\Longrightarrow\; x = -1 \]
Verify: compare with the substitution check
Why: Substituting 2 into the original gives 3 on the left and 1 on the right, so it fails — the same conclusion reached a different way. Noticing the sign condition before squaring is quicker, and it explains why the false solution appeared rather than merely detecting it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 455-455
Error analysis
A student squares an equation with two radicals.
Annotate
On: \( \sqrt{x+2}+1 = \sqrt{3-x} \;\Longrightarrow\; (x+2) + 1 = 3-x \)
Squaring a sum always produces a cross term, exactly as in Lesson 5.3. The cross term still contains a radical, which is why a second squaring is needed.
Ranking
Solving an equation with two radicals.
Put in order
Why: Step one reduces how much survives the first squaring, and step three is the move that makes the second squaring effective. With two squarings the check in step five is doubly necessary, since each one could have introduced a false solution.
Prediction
Commit before reasoning.
Predict first
After squaring both sides of an equation with two radicals, why does a radical remain?
Correct: Because squaring a sum produces a cross term.
\[ (\sqrt{u}+1)^2 = u + 2\sqrt{u} + 1 \]
Why: The square of a plus b is a squared plus 2ab plus b squared, and when a is a radical the middle term 2ab still contains it. Only the squared terms lose their radicals. Arranging so that each side has a single radical before squaring avoids the problem entirely when it is possible, which is why that is step one.
Two truths and a lie
All three are about squaring twice.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Example 6 squared twice and produced exactly one extraneous solution out of two candidates. There is no rule about how many appear: it depends entirely on the signs the original equation requires, and the only way to find out is to check.
Comparison
Fill the blanks. The undoing operation depends on what is doing the covering.
Comparison matrix
| Equation contains | Undo with | Check needed? |
|---|---|---|
| A cube root | cubing both sides | no, cubing is reversible |
| A square root | squaring both sides | yes, always |
| A rational exponent | the reciprocal exponent | yes if the numerator is even |
| Two radicals | squaring twice | yes, with particular care |
The pattern is the parity of the power being undone, which has decided the behaviour of roots throughout Chapter 6.
Pattern
One routine for any equation with a radical or a rational exponent.
A quick pre-check: a principal root is never negative, so any candidate making the other side negative is extraneous before any substitution is done.
OpenStax Algebra and Trigonometry 2e, §2.6 Other Types of Equations §2.6
Check
The right power to use.
Check your understanding
Solve the cube root of (2x + 7) = 3.
Answer: A
Why: Cubing gives 2x + 7 = 27, so 2x = 20 and x = 10.
Check
A rational exponent.
Check your understanding
Solve 3x^(2/3) = 48.
Answer: A
Why: Dividing by 3 gives x^(2/3) = 16, and raising to the 3/2 power gives 4^3 = 64.
Check
Extraneous solutions.
Check your understanding
Solve x + 1 = sqrt(7x + 15).
Answer: A
Why: Squaring gives x^2 - 5x - 14 = 0 with roots 7 and -2, but -2 makes the left side negative while a square root cannot be.
Real world
An athlete's hang time t, in seconds, relates to the jump height h, in feet, by t equals 0.5 times the square root of h.
Discussion prompt
Find the jump height for a hang time of 0.9 seconds, then find the hang time that would need a 4-foot jump, and say why doubling the hang time is so much harder than it sounds.
Hint: Isolate the radical, then square.
Answer:
\[ 0.9 = 0.5\sqrt{h} \;\Longrightarrow\; \sqrt{h} = 1.8 \;\Longrightarrow\; h = 3.24 \text{ feet} \]
\[ t = 0.5\sqrt{4} = 1.0 \text{ second} \]
A 0.9 second hang time needs a jump of about 3.24 feet, and a 4-foot jump gives 1.0 second.
Doubling the hang time from 0.9 to 1.8 seconds would need a jump of 12.96 feet — four times the height for twice the time, because the time depends on the square root of the height. That is the same square-root scaling as the pendulum in Lesson 6.5 and the coral cod in Lesson 6.1, and it explains why elite hang times cluster so tightly: the height required grows with the square of the time.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does cubing both sides of an equation ever introduce an extraneous solution?
Correct: No — cubing is reversible.
\[ a = b \;\Longleftrightarrow\; a^3 = b^3, \quad \text{but} \quad a^2 = b^2 \text{ allows } a = -b \]
Why: Every real number has exactly one real cube root, so cubing and taking a cube root undo each other completely and the cubed equation has exactly the solutions the original had. Squaring is different because a positive number has two square roots, and squaring collapses them together. This is the parity distinction that has run through Chapter 6, appearing here as the difference between a step that needs auditing and one that does not.
Explain it
They have just solved a radical equation and been told one of their answers is wrong.
Discussion prompt
In four sentences or fewer, explain what an extraneous solution is and why squaring produces them.
Hint: Talk about losing the sign.
Answer:
Squaring both sides turns two different equations into the same one, because a number and its negative have the same square. So the squared equation is really solving your equation and its sign-flipped twin at once, and some of the answers you get belong to the twin.
Those answers are called extraneous: they satisfy the squared equation but not the one you started with. The only way to tell which is which is to substitute each answer back into the original equation, and a quick shortcut is that a square root can never equal a negative number.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For isolating, divide out any coefficient before touching the radical. For reciprocals, check that the two exponents multiply to 1. For checking, make substitution the last written line of every solution that used an even power. For two radicals, expect a cross term after the first squaring and plan for a second. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the equation x plus 1 equals the square root of 7x plus 15 and work it four ways on one page. Top left: solve it algebraically, showing the squaring, the quadratic and both roots. Top right: check both roots in the original equation, writing out the two sides separately for each so the failure is visible. Bottom left: sketch y equals x plus 1 and y equals the square root of 7x plus 15 on the same axes, mark the single intersection, and add the reflected line y equals negative x minus 1 as a dashed curve through the false solution. Bottom right: write one sentence explaining what squaring did to the two lines, and one sentence giving the quick sign test that would have rejected the false root without substituting. In a margin, write which powers are safe to raise both sides to and which are not.
If your dashed reflected line does not pass through the false solution, recheck it: the point where the original line and the curve have equal magnitudes but opposite signs is exactly where it should cross.
Recap
Five things, and the fourth is the one that makes the others trustworthy.
| If you see | Then |
|---|---|
| A radical with a coefficient | Divide it out before raising a power |
| An index of n | Raise both sides to the nth power |
| A rational exponent | Raise to its reciprocal |
| An even power used | Check every answer |
| An odd power used | No extraneous solution can appear |
| Two radicals | Square twice, isolating in between |
| A negative side opposite a root | That candidate is extraneous |
That closes Chapter 6. Chapter 7 changes families again, moving the variable from the base into the exponent and taking up exponential growth and decay.
McDougal Littell Algebra 2 (Texas Edition), Ch. 6 Rational Exponents and Radical Functions — Lesson 6.6 Solve Radical Equations §6.6, pp. 452-459 — everything on these slides traces back here
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