Solving equations that contain radicals. Includes squaring both sides to remove a radical, isolating the radical before squaring, extraneous solutions and why squaring produces them, equations with no solution because a radical cannot be negative, and applying the method to a model relating pressure and flow rate.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 12 — Radicals and More Connections to Geometry
Solving Radical Equations
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 704-709 — the lesson these objectives are drawn from
Warm-up
Lesson 11.7 warned that clearing denominators can produce a value the original equation forbids. Squaring both sides does the same thing, and this lesson is where it happens.
Discussion prompt
Is the statement one equals negative one true? Now square both sides. What happened?
Hint: Compare the two statements.
Answer:
\[ 1 = -1 \text{ is false, but } 1^2 = (-1)^2 \text{ is true} \]
Squaring turned a false statement into a true one, because it cannot tell a quantity from its negative. That single fact is the entire reason every answer in this lesson has to be checked.
Concept
If two expressions are equal then their squares are equal, so squaring both sides of an equation removes a square root. The step is not reversible, so every candidate must be checked.
extraneous solution — A value that satisfies an equation obtained by squaring both sides but does not satisfy the original equation.
Squaring can turn a false statement into a true one.
Figure (svg): The rule for squaring both sides of an equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 704-705
Section
Section 1
Concept
If a equals b then a squared equals b squared. Applying this to an equation containing a square root removes the radical and leaves an ordinary equation.
\[ \sqrt{x - 1} = 5 \;\Longrightarrow\; x - 1 = 25 \]
Squaring a square root leaves the radicand.
Figure (svg): The rule for squaring both sides of an equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 704-704 — the Squaring Both Sides of an Equation property and Example 1
Picture it
Squaring is the inverse.
Figure (svg): The rule for squaring both sides of an equation
Squaring and taking a square root are inverse operations, so applying one undoes the other. The complication is that the undoing is not perfect, which the later sections address.
Worked example
This is Example 1 from the textbook.
\[ \text{Solve } \sqrt{x} - 7 = 0 \text{ and } 3\sqrt{x + 4} = 15. \]
Isolate the first radical
Why: Add seven.
\[ \sqrt{x} = 7 \]
Square both sides
Why: The radical vanishes.
\[ x = 49 \]
Isolate the second radical
Why: Divide by three.
\[ \sqrt{x + 4} = 5 \]
Square and solve
Why: Then subtract four.
\[ x = 21 \]
Figure (svg): The rule for squaring both sides of an equation
\[ x = 49, \qquad x = 21 \]
Verify: check both in the originals
Why: The root of forty-nine is seven, and seven minus seven is nought. For the second, the root of twenty-five is five and three times five is fifteen. Both candidates satisfy their original equations, so both are genuine.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 704-704
Faded example
The radical becomes its radicand.
Fill in the blanks
\sqrt25 = 5 \;\Longrightarrow\; x + 4 = 21 \;\Longrightarrow\; x = ___
Why: Squaring the left side leaves the radicand and squaring the right gives twenty-five. Both whole sides were squared, not individual terms.
Worked example
Guided Practice 1 to 4.
\[ \text{Solve } \sqrt{x} = 3, \; \sqrt{m} - 4 = 0, \; \sqrt{x + 6} = 4 \text{ and } \sqrt{n - 1} = 1. \]
Take the first two
Why: Square directly, or isolate first.
\[ x = 9, \; m = 16 \]
Take the third
Why: Square, then subtract six.
\[ x + 6 = 16 \]
Finish the third
Why: Ten.
\[ x = 10 \]
Take the fourth
Why: Square, then add one.
\[ n = 2 \]
Figure (svg): The rule for squaring both sides of an equation
\[ x = 9, \; m = 16, \; x = 10, \; n = 2 \]
Verify: check one with a shifted radicand
Why: For the third, x equal to ten gives a radicand of sixteen and a root of four, matching the equation. Substituting into the original rather than into a later line is what makes the check meaningful.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 705-705
Trap
\[ \sqrt{x} - 7 = 0 \;\Longrightarrow\; x - 7 = 0 \;\Longrightarrow\; x = 7 \]
Square the radical and leave the rest
Why: The aim was to remove the radical, so only it was squared.
Squaring must be applied to whole sides, not to individual terms. Checking exposes it at once: the root of seven is about 2.65, so root seven minus seven is about negative 4.35 rather than nought.
\[ \sqrt{x} = 7 \;\Longrightarrow\; x = 49 \]
Isolate the radical, then square both entire sides
Why: That is what the property permits.
The property is stated as a squared equals b squared, with a and b whole sides.
Matching
Isolate, square, solve.
Match the pairs
Why: Three of these needed something moved before the squaring and one did not. In every case the squaring itself was the shortest step.
Elimination
For the equation root (x + 4) = 5.
Eliminate the wrong options
What is the squared equation?
Survives elimination: A
Why: Squaring the radical leaves its entire radicand and squaring the five gives twenty-five. Option C is the trap of squaring individual terms rather than whole sides.
Socratic
It is used without comment.
Discussion prompt
Explain why squaring a square root leaves the radicand. Then say what condition on the radicand makes this work.
Hint: What does the square root mean?
Answer:
The square root of a quantity is the number whose square is that quantity, so squaring it must give the quantity back. That is the definition of a square root rather than a separate rule, which is why the step needs no justification beyond Lesson 9.1.
It works whenever the radicand is non-negative, which it must be for the radical to exist at all. So within the domain of the equation the step is always valid — the difficulty is not that squaring fails but that the squared equation permits values the original never did.
Section
Section 2
Concept
Squaring a side containing a radical plus something else does not remove the radical, because squaring a sum produces a cross term. The radical must be alone on its side first.
Squaring a sum is Lesson 10.3's pattern.
Figure (svg): A radical isolated before both sides are squared
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 705-705 — Example 2, where the radical must be isolated first
Picture it
Isolate, square, solve.
Figure (svg): A radical isolated before both sides are squared
The subtraction of four came first for a reason. Squaring root of two x minus three plus four would have produced a middle term containing the radical, leaving the problem no simpler.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } \sqrt{2x - 3} + 4 = 5. \]
Isolate the radical
Why: Subtract four from each side.
\[ \sqrt{2x - 3} = 1 \]
Square both sides
Why: The radicand survives.
\[ 2 x - 3 = 1 \]
Add three
Why: Move the constant.
\[ 2 x = 4 \]
Divide by two
Why: And check.
\[ x = 2 \]
Figure (svg): A radical isolated before both sides are squared
\[ x = 2 \]
Verify: check in the original
Why: At x equal to two the radicand is one, its root is one, and one plus four is five. The candidate satisfies the equation as first written, so it is genuine.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 705-705
Faded example
Move everything else away.
Fill in the blanks
\sqrt1 + 4 = 5 \;\Longrightarrow\; \sqrt1 = ___ \;\Longrightarrow\; 2x - 3 = ___
Why: Subtracting four leaves the radical alone, and only then does squaring remove it cleanly. Squaring before the subtraction would leave a radical in the cross term.
Worked example
Seeing why the isolation matters.
\[ \text{What happens if } \sqrt{2x - 3} + 4 = 5 \text{ is squared immediately?} \]
Square the left side
Why: It is a sum, so use the pattern.
\[ (a + b) ^{2} = a ^{2} + 2 a b + b ^{2} \]
Expand it
Why: The middle term keeps a radical.
\[ 2x - 3 + 8\sqrt{2x-3} + 16 \]
Compare with the aim
Why: The radical is still there.
Conclude
Why: Isolation was necessary.
Figure (svg): A radical isolated before both sides are squared
\[ 2x - 3 + 8\sqrt{2x - 3} + 16 = 25 \]
Verify: notice what would be needed next
Why: The new equation still contains a radical, so it would have to be isolated and squared all over again — more work than the original problem. Isolating first avoids creating that cross term at all.
Error analysis
The student squared both sides of a radical equation without isolating.
Annotate
On: \( \begin{aligned} \sqrt{2x - 3} + 4 &= 5 \\ 2x - 3 + 16 &= 25 \\ 2x &= 12, \quad x = 6 \end{aligned} \)
This is Lesson 10.3's error — squaring a sum term by term — appearing inside an equation. Isolating the radical first means the sum never has to be squared, which sidesteps the whole difficulty.
Sorting
The radical must be alone on its side.
Sort into buckets
Sort each equation by whether it can be squared as it stands.
Half of these need one preliminary step. A coefficient must be divided out and an added constant subtracted, and neither can be dealt with by the squaring itself.
Hypothesis
It seems to remove the radical.
Predict first
What does squaring root a plus four actually give?
Correct: a plus eight root a plus sixteen, which still contains a radical.
\[ (\sqrt{a} + 4)^2 = a + 8\sqrt{a} + 16 \]
Why: Squaring a sum gives the square of the first, twice the product of the two, and the square of the second — and the middle term keeps one factor of the radical. So squaring a sum containing a radical does not remove it, which defeats the purpose entirely. This is precisely Lesson 10.3's warning that squaring does not distribute over addition.
Socratic
Both steps have to happen.
Discussion prompt
Explain why the order matters. Then say what to do if an equation contains two separate radicals.
Hint: What does squaring a sum produce?
Answer:
Squaring only removes a radical if the radical is the entire side being squared. If anything is added to it, the square of the sum contains a cross term with the radical still in it, so the equation is no simpler and often longer.
With two radicals the usual approach is to isolate one of them, square, and then isolate and square again — each squaring removes one radical provided it is alone at that moment. That takes two rounds and the same checking discipline at the end, since each squaring can introduce false candidates.
Section
Section 3
Concept
Squaring both sides can produce a value that satisfies the squared equation but not the original. Such a value is called an extraneous solution and must be discarded.
Squaring cannot tell a quantity from its negative.
Figure (svg): Two candidates, one of which fails the original equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 705-705 — the Extraneous Solutions note and Example 3
Picture it
Both came from correct algebra.
Figure (svg): Two candidates, one of which fails the original equation
The rejected candidate is not the result of a mistake. It is a genuine solution of the squared equation and simply not of the original, which is why substitution is the only way to tell.
Worked example
This is Example 3 from the textbook.
\[ \text{Solve } \sqrt{x + 2} = x \text{ and check.} \]
Square both sides
Why: The radical goes.
\[ x + 2 = x ^{2} \]
Write in standard form
Why: Move everything across.
\[ x ^{2} - x - 2 = 0 \]
Factor and solve
Why: Two and negative one.
\[ x = 2 \text{ or } x = -1 \]
Check both
Why: Only one works.
\[ x = 2 \]
Figure (svg): Two candidates, one of which fails the original equation
\[ x = 2 \]
Verify: substitute each candidate
Why: At two the root of four is two, matching the right side. At negative one the root of one is one, but the right side is negative one — and a radical is never negative, so it can never equal a negative number.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 705-705
Faded example
Substitute into the original.
Fill in the blanks
x = 2: \; \sqrt2 = -1 \;\checkmark \qquad x = -1: \; \sqrt___ = 1 \ne ___
Why: The first candidate makes both sides equal and the second does not. The failure is a sign mismatch, which is exactly what squaring is blind to.
Worked example
Tracing the mechanism.
\[ \text{Why did } x = -1 \text{ survive the squaring?} \]
Substitute into the original
Why: The two sides differ in sign.
Note it is false
Why: One is not negative one.
Square both sides of that
Why: The sign difference vanishes.
\[ 1 = 1 \]
Conclude
Why: The squared version is true.
Figure (svg): Why squaring can make a false statement true
\[ 1 \ne -1 \quad \text{but} \quad 1^2 = (-1)^2 \]
Verify: see that no error was made
Why: Every algebraic step was valid, and the candidate is a genuine solution of x plus two equals x squared. The squared equation is simply a larger problem than the original, and its extra solution has to be filtered out by hand.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 705-705
Trap
\[ \sqrt{x + 2} = x \;\Longrightarrow\; x = 2 \text{ and } x = -1 \]
Report both roots of the quadratic
Why: Both came out of correct factoring.
At negative one the left side is one and the right side is negative one, so the equation is false there. A quadratic has two roots and this radical equation has one.
\[ x = 2 \quad \text{only} \]
Substitute every candidate into the original before reporting it
Why: Correct algebra is not enough here.
The number of candidates and the number of solutions are different quantities in this topic.
Prediction
In an equation where a radical equals an expression.
Predict first
Which kind of candidate is most likely to be extraneous?
Correct: One that makes the other side negative.
\[ \sqrt{x+2} \ge 0 \text{ always, so } x \text{ must be } \ge 0 \]
Why: A square root is never negative, so if a candidate makes the other side of the equation negative it cannot possibly be a solution. That single observation identifies most extraneous candidates before any substitution is done. Fractions, large values and large radicands are all perfectly ordinary and cause no difficulty on their own.
Elimination
Earlier chapters rarely required it.
Eliminate the wrong options
What makes checking essential in this lesson?
Survives elimination: A
Why: Squaring is not reversible, so the squared equation can hold where the original does not. That is a property of the operation rather than of radicals or of quadratics in general.
Socratic
Squaring is not unique.
Discussion prompt
Name another operation met in this course that can produce extraneous solutions, and say why. Then say what the two have in common.
Hint: Think about Chapter 11.
Answer:
Multiplying both sides by an expression containing the variable, as when clearing denominators in Lesson 11.7, can do it. At a value making that expression nought, the multiplication is by nought and turns any equation into a true statement — so the cleared equation can hold where the original was not even defined.
What they have in common is that both are irreversible: you cannot undo a squaring without a plus-or-minus, and you cannot undo a multiplication by nought at all. Any step that loses information can enlarge the solution set, and the only remedy is to test every candidate against the original.
Section
Section 4
Concept
If isolating the radical leaves it equal to a negative number, the equation has no solution. Squaring will still produce a candidate, and that candidate will always fail.
\[ \sqrt{x} = -13 \;\Longrightarrow\; \text{no solution} \]
The square root of any number is non-negative.
Figure (svg): An equation asking a square root to be negative
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 706-706 — Example 4, Check for Extraneous Solutions, where the equation has no solution
Picture it
No square root is negative.
Figure (svg): An equation asking a square root to be negative
The impossibility is visible the moment the radical is isolated, before any squaring. Noticing it then saves the whole calculation and the check afterwards.
Worked example
This is Example 4 from the textbook.
\[ \text{Solve } \sqrt{x} + 13 = 0. \]
Isolate the radical
Why: Subtract thirteen.
\[ \sqrt{x} = -13 \]
Square both sides
Why: A candidate appears.
\[ x = 169 \]
Check it
Why: The root of 169 is 13.
\[ 13 + 13 = 26 \]
Conclude
Why: Twenty-six is not nought.
Figure (svg): An equation asking a square root to be negative
\[ \text{no solution, since } \sqrt{x} \ge 0 \text{ always} \]
Verify: say why no value could work
Why: The square root of any permitted x is at least nought, so adding thirteen gives at least thirteen — never nought. No amount of searching would find a solution, and the reasoning covers every value at once rather than testing them one by one.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 706-706
Sorting
Isolate the radical and look at the other side.
Sort into buckets
Sort each equation by whether a solution exists.
Three of the six are impossible and each becomes obvious after one line of isolating. Doing that line first is what turns a full calculation into a one-sentence answer.
Worked example
Saving the calculation entirely.
\[ \text{How can you tell } \sqrt{x} + 13 = 0 \text{ has no solution without squaring?} \]
Isolate the radical
Why: One subtraction.
\[ \sqrt{x} = -13 \]
Recall the range
Why: Square roots are never negative.
\[ \sqrt{x} \ge 0 \]
Compare
Why: Negative thirteen is below nought.
Stop
Why: No need to square.
Figure (svg): An equation asking a square root to be negative
\[ \sqrt{x} \ge 0 > -13 \]
Verify: compare the two routes
Why: Squaring gives a candidate of a hundred and sixty-nine which then has to be tested and rejected. Noticing the impossibility takes one line and reaches the same conclusion with more certainty.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 706-706
Trap
\[ \sqrt{x} = -13 \;\Longrightarrow\; x = 169 \]
Square both sides and report the result
Why: The algebra produced a number, so the number was reported.
The root of a hundred and sixty-nine is positive thirteen, not negative thirteen, so the candidate fails the original equation. Every equation of this shape produces a candidate and every one of them is extraneous.
The equation has no solution, since a square root cannot be negative.
Compare the isolated radical with nought before squaring
Why: A negative right side ends the problem.
This is the counterpart of Lesson 9.2's rule that x squared equals a negative has no solution.
Faded example
After isolating.
Fill in the blanks
Since a square root is never negative, the equation root x equals -13 has no solution.
Why: The range of the square root function is the non-negative numbers, so it can never equal a negative value. That fact from Lesson 12.1 settles the whole equation.
Elimination
After isolating a radical.
Eliminate the wrong options
What is worth looking at before squaring?
Survives elimination: A
Why: A negative right side means no solution exists and the squaring is a waste of effort. It is a one-second check that occasionally saves the entire problem.
Socratic
The equation has no solution at all.
Discussion prompt
Explain why squaring an impossible equation still produces a number. Then say what that number actually solves.
Hint: What equation does 169 satisfy?
Answer:
Squaring turns the statement that root x equals negative thirteen into the statement that x equals a hundred and sixty-nine, and the second is a perfectly ordinary equation with a solution. The squaring destroyed the sign requirement, so the resulting equation asks something weaker than the original.
The number a hundred and sixty-nine solves root x equals positive thirteen, not negative thirteen — squaring cannot distinguish those two, so it merges them into one problem. Every candidate from an impossible radical equation is really a solution of the sign-flipped version, which is why it always fails the check.
Section
Section 5
Concept
When a model gives one quantity as a square root of another, finding the second from the first means solving a radical equation. The physical situation usually rules out the extraneous candidate.
Negative quantities are usually rejected twice over.
Figure (svg): Nozzle pressure related to flow rate through a radical equation
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 704-709 — the lesson opener on nozzle pressure and flow rate
Picture it
Squaring finds the pressure.
Figure (svg): Nozzle pressure related to flow rate through a radical equation
The model runs from pressure to rate, and the question runs the other way. Solving a radical equation is what reverses the direction.
Worked example
The lesson opener's situation, with a model supplied.
\[ \text{With } r = 1.2\sqrt{p}, \text{ find the pressure } p \text{ giving a rate of } 12. \]
Substitute the rate
Why: Twelve equals the model.
\[ 12 = 1.2\sqrt{p} \]
Isolate the radical
Why: Divide by 1.2.
\[ \sqrt{p} = 10 \]
Square both sides
Why: The radical goes.
\[ p = 100 \]
Check
Why: 1.2 times ten is twelve.
\[ \;\checkmark \]
Figure (svg): Nozzle pressure related to flow rate through a radical equation
\[ p = 100 \]
Verify: check against the situation
Why: A pressure of a hundred is positive, which it must be, and substituting gives 1.2 times the root of a hundred, which is twelve. Both the algebra and the situation accept the answer.
Faded example
Isolate, then square.
Fill in the blanks
12 = 1.2\sqrt10 \;\to\; \sqrt100 = ___ \;\to\; p = ___
Why: Dividing by the coefficient isolates the radical and squaring then gives the pressure. Both steps are the ordinary procedure, applied inside a formula.
Worked example
Extraneous candidates in a physical setting.
\[ \text{Could a negative pressure ever be a solution here?} \]
Look at the radicand
Why: It must be non-negative.
\[ p \ge 0 \]
Look at the situation
Why: Pressure is a positive quantity.
\[ p > 0 \]
Compare the two restrictions
Why: The situation is stricter.
Conclude
Why: Negatives fail twice over.
Figure (svg): Nozzle pressure related to flow rate through a radical equation
\[ p > 0 \]
Verify: note the double protection
Why: The algebra already forbids a negative radicand, and the situation forbids a negative pressure independently. When a model has such restrictions, extraneous candidates are usually rejected by the situation before the algebra even needs to.
Trap
Square both sides, report the number, and move on.
Trust the algebra, since the model is a formula
Why: Formulas are reliable, so the answer must be.
The squaring step is just as irreversible inside a model as outside one. A candidate that makes the modelled quantity negative, or that lies outside the range the model was built for, has to be rejected regardless of where it came from.
Check the candidate in the original equation and against the situation.
Apply both tests, algebraic and physical
Why: They can reject different things.
A model usually supplies extra restrictions that the equation alone does not.
Prediction
In a model where rate depends on the root of pressure.
Predict first
To double the flow rate, how must the pressure change?
Correct: It must be four times as large.
\[ r = 6: \; p = 25 \qquad r = 12: \; p = 100 \]
Why: The rate is proportional to the square root of the pressure, so doubling the rate requires the root to double, which requires the pressure to quadruple. Going from a rate of six to a rate of twelve takes the pressure from twenty-five to a hundred. That is the same square root flattening met in Lessons 12.1 and 12.2, seen from the other direction — the cost of extra performance rises sharply.
Sorting
For a pressure model.
Sort into buckets
Sort each candidate by whether it could be a valid pressure.
Nought is rejected on physical grounds rather than algebraic ones — the radicand may be nought, but a hose at zero pressure delivers no flow. Situations are often stricter than the algebra.
Socratic
The algebra is the same.
Discussion prompt
Say why a physical model often filters out the false candidate automatically. Then say why the check is still worth doing.
Hint: What values does the situation permit?
Answer:
Physical quantities such as pressure, length and time are usually required to be positive, and extraneous candidates from squaring are typically negative or otherwise outside the sensible range. So the situation rejects them for reasons of its own, before the algebraic check is reached.
The check is still worth doing because not every extraneous candidate is negative, and because a model can have restrictions that are not obvious — an upper limit on pressure, say, or a range over which the formula was fitted. Relying on the situation to catch errors works until the one time it does not, and substitution costs a few seconds.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Operation | Why it can invent solutions | Where it appeared |
|---|---|---|
| squaring both sides | it cannot tell a quantity from its negative | this lesson |
| clearing denominators | it may multiply by zero at an excluded value | Lesson 11.7 |
| adding to both sides | it cannot; it is reversible | everywhere, needing no such check |
Only irreversible operations can enlarge the solution set. The bottom row is there as a reminder that most of algebra does not need this precaution.
Pattern
To solve any equation containing a square root, these five moves cover it.
Step two costs one glance and occasionally ends the problem immediately, and step five is the one that separates genuine solutions from the ones squaring invented.
OpenStax Elementary Algebra 2e, §9.6 Solve Equations with Square Roots §9.6
Check
Isolate, then square.
Check your understanding
Solve root (2x - 3) + 4 = 5.
Answer: A
Why: Subtracting four gives a radical equal to one, and squaring gives two x minus three equals one, so x is two.
Check
Check both candidates.
Check your understanding
Solve root (x + 2) = x.
Answer: A
Why: Squaring gives x squared minus x minus two equals nought, whose roots are two and negative one, but at negative one the radical would have to equal a negative number.
Check
Look at the isolated radical.
Check your understanding
Solve root x + 13 = 0.
Answer: A
Why: Isolating gives a square root equal to negative thirteen, which is impossible since square roots are never negative.
Real world
This is the de-icing hose question from the lesson opener. The flow rate of a hose depends on the square root of the nozzle pressure, so finding the pressure needed for a given rate means solving a radical equation.
Discussion prompt
With r equal to 1.2 times the square root of p, find the pressure that gives a flow rate of 12. Then say how the pressure must change to double the rate.
Hint: Isolate the radical, then square.
Answer:
\[ 12 = 1.2\sqrt{p} \;\Longrightarrow\; \sqrt{p} = 10 \;\Longrightarrow\; p = 100 \]
Checking confirms it: 1.2 times the root of a hundred is 1.2 times ten, which is twelve, and a pressure of a hundred is positive as it must be.
To double the rate the square root must double, so the pressure must quadruple — from a hundred to four hundred. That is the square root relationship working against you: each further increase in flow costs disproportionately more pressure, which is a real constraint on what a pump can achieve and why high-flow equipment is so much more demanding than the numbers first suggest.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Solve the square root of (x + 2) equals x.
Correct: x = 2 only.
\[ x = -1: \; \sqrt{1} = 1 \ne -1 \]
Why: Squaring gives x plus two equals x squared, which rearranges to x squared minus x minus two equals nought and factors as x minus two times x plus one — so the candidates are two and negative one. At two the root of four is two and both sides agree. At negative one the root of one is one while the right side is negative one, so the equation is false there: a square root is never negative and can therefore never equal a negative number. The algebra was entirely correct and still produced a candidate that is not a solution, because squaring cannot distinguish one from negative one. That is what makes the checking step compulsory here rather than merely prudent.
Explain it
They reported both roots of the quadratic as solutions of a radical equation.
Discussion prompt
In no more than four sentences, explain why one of them fails. Then give them the quick way to spot a suspicious candidate.
Hint: What can a square root never be?
Answer:
A usable answer: squaring both sides cannot tell a number from its negative, so it lets through values where the two sides had opposite signs. At negative one the left side is one and the right is negative one, which is false — a square root is never negative.
The quick check is to look at the side the radical is set equal to. If a candidate makes that side negative, it cannot possibly work and you can reject it without any substitution at all.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Isolating is fixed by making it the first thing you write, before any squaring. Squaring whole sides is fixed by remembering that squaring a sum produces a cross term. Extraneous solutions are fixed by substituting every candidate into the original. Impossible equations are fixed by comparing the isolated radical with nought before doing anything else. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write the squaring property, and beneath it write the statement one equals negative one, square both sides, and note that a false statement became true — this is the mechanism behind everything else on the page. Underneath, solve two straightforward radical equations, showing the isolating step separately from the squaring in each. In the middle, take an equation where the radical is not alone, work it correctly, and beside it show what squaring the sum would have produced, ringing the cross term that still contains a radical. Beneath that, solve an equation with two candidates, substitute both into the original, and strike through the extraneous one with a one-line reason. In the lower half, write an equation whose isolated radical equals a negative number, note the impossibility before squaring, then square anyway to see the candidate appear and fail. Finally, in the margin, list the two operations in this course that can invent solutions and what they have in common.
Every candidate on your page should be substituted into the original equation, not into a later line. Checking against a squared version would pass exactly the candidates the squaring invented.
Recap
Five things, and the last is what makes the first four safe.
| If the question says | Your first move is |
|---|---|
| Solve a radical equation | Isolate the radical |
| The radical is not alone | Move the other terms first |
| The isolated radical equals a negative | There is no solution |
| You have two candidates | Substitute both into the original |
| A model gives a candidate | Check it against the situation too |
Lesson 12.4 introduces a new way of writing radicals. A fractional exponent means the same thing as a root, which lets all of Chapter 8's exponent rules apply to radicals as well.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.3 Solving Radical Equations §12.3, pp. 704-709 — everything on these slides traces back here
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