Algebra 2 · A.REI.A.2

Solving radical equations

The unknown is trapped under a root sign. How do you get it out, and why can the answer you get be false?

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Multi-step equationsMath 8 Simplifying a square rootAlgebra 1 Simplifying radicalsGrade 9 Rational exponentsAlgebra 2 +1 more

Where you are 5 in Radicals and rational expressions

Algebra9

Radicals and rational expressions5

Simplifying radicals3 Simplifying rational expressions3

Solving radical equations

Solving a rational equation Inverse, joint and combined variation

When you reach for it

A radical equation is an equation with the variable inside a root, such as \displaystyle \sqrt{2x + 7} = x + 2 or \displaystyle \sqrt[3]{2x - 5} + 4 = 1. They turn up whenever a quantity is a root of something unknown: the length of a slanted wire is a square root of a sum of squares, by Pythagoras, and a formula such as \displaystyle T = 2\pi\sqrt{L/g} for a pendulum, read backwards, asks for the L inside the root.

None of the moves for multi-step equations can reach a variable under a root. One move can: a root is undone by the matching power, so raising both sides to that power frees the variable. The price is that squaring is not reversible, so the equation you end with can have solutions the original does not. Those are called extraneous solutions, and checking for them is part of the method, not an afterthought.

This lesson runs over two days. Part 1 solves equations with one root: isolate it, raise both sides to a power, solve, check. Part 2 uses the same move on equations with two roots and on equations written with rational exponents.

The procedure

Part 1: one radical

Take the equation

\sqrt{2x + 7} = x + 2

Isolate the radical. Get the root alone on one side, using the ordinary moves of adding, subtracting, multiplying and dividing both sides. Here it is already alone. The reason for doing this first is that squaring has to remove the root completely. Squaring a sum such as \displaystyle \sqrt{x + 5} + 1 does not do that: (a + b)^2 = a^2 + 2ab + b^2, and the middle term 2ab still carries the root (Example 2).

Raise both sides to the power that matches the index. A square root is undone by squaring, because \displaystyle \left(\sqrt{A}\right)^2 = A whenever \displaystyle \sqrt{A} is real, and a cube root by cubing, as in the lesson on roots. If two numbers are equal their squares are equal, so this step keeps every solution:

\begin{aligned} \left(\sqrt{2x + 7}\right)^2 &= (x + 2)^2 \\ 2x + 7 &= x^2 + 4x + 4 \end{aligned}

The right side is the square of a binomial and has to be multiplied out in full; squaring x + 2 does not give x^2 + 4.

Solve what is left. It is now a polynomial equation. Gather everything on one side and factor:

\begin{aligned} 0 &= x^2 + 2x - 3 \\ 0 &= (x + 3)(x - 1) \end{aligned}

So x = -3 or x = 1. These are candidates, not yet solutions.

Check every candidate in the original equation. For x = 1: the left side is \displaystyle \sqrt{9} = 3 and the right side is 1 + 2 = 3. It is a solution. For x = -3: the left side is \displaystyle \sqrt{1} = 1 and the right side is -3 + 2 = -1. They differ, so x = -3 is extraneous. The equation has exactly one solution, x = 1.

Why squaring adds solutions. The step “equal numbers have equal squares” runs only one way. The reverse is false: 1^2 = (-1)^2, but 1 \ne -1. Two numbers with the same square are either equal or opposite. So the squared equation 2x + 7 = (x + 2)^2 collects the solutions of two equations at once:

\begin{aligned} \sqrt{2x + 7} &= x + 2 \\ -\sqrt{2x + 7} &= x + 2 \end{aligned}

The value x = -3 solves the second one: \displaystyle -\sqrt{1} = -1. The figure shows this. The graph of \displaystyle y = \sqrt{2x + 7} is the top half of a sideways parabola, and the graph of the squared equation is the whole parabola, top and bottom halves together. The line y = x + 2 meets the top half once and the bottom half once, and squaring cannot tell the two meetings apart.

Figure 1Squaring \displaystyle \sqrt{2x + 7} = x + 2 gives 2x + 7 = (x + 2)^2, whose graph is the whole sideways parabola, both halves. The line meets the solid half at x = 1, a true solution. It meets the dashed half at x = -3, which solves \displaystyle -\sqrt{2x + 7} = x + 2 instead.

Squaring never loses a solution, only adds some, so the check catches every extraneous one. It is also a check on the arithmetic, which is reason enough to do it even when no candidate could be extraneous.

A quicker test for square roots. A square root is never negative. That gives an exact description of what squaring keeps:

\begin{aligned} &\sqrt{A} = B \\ &\text{exactly when } A = B^2 \text{ and } B \ge 0 \end{aligned}

If \displaystyle \sqrt{A} = B, then B is a square root, so B \ge 0, and squaring gives A = B^2. Conversely, if A = B^2 with B \ge 0, then A \ge 0, and B is the non-negative number whose square is A, which is what \displaystyle \sqrt{A} means. So a candidate of the squared equation is a true solution exactly when it makes the side opposite the root non-negative. Here x + 2 is 3 at x = 1 and -1 at x = -3, which sorts the two candidates without working out a single root. This test is a real saving when the candidates are irrational (Example 3). Notice also that the condition A \ge 0 never needs checking separately: A = B^2 already forces it.

Odd roots need no such test. Cubing is reversible: if a^3 = b^3 then a = b, because an odd power keeps the sign and no two real numbers have the same cube. So cubing both sides of an equation with a cube root gives an equation with exactly the same solutions, and the same holds for any odd index. An even index, 4 or 6, behaves like the square root: raising to an even power can add solutions, because a and -a have the same even power.

Part 2: two radicals and rational exponents

A radical on each side. For \displaystyle \sqrt{A} = \sqrt{B}, squaring gives A = B. Each side of the original must be a real number, so a candidate must make the radicands non-negative. Since A = B at a candidate, it is enough to check one of them (Example 5).

Two radicals on one side. For an equation such as \displaystyle \sqrt{2x + 3} - \sqrt{x + 1} = 1, squaring as it stands leaves a product of the two roots in the middle term. Isolate one radical first, so that one side is a single root:

\sqrt{2x + 3} = 1 + \sqrt{x + 1}

Squaring now removes the root on the left, and the right side becomes \displaystyle 1 + 2\sqrt{x + 1} + (x + 1), which still has one root in it. That is progress: there is one radical where there were two. Isolate it and square a second time. Each squaring can add candidates, so the check at the end is against the very first equation (Example 6).

Rational exponents. A power with a fractional exponent is a root and a power, as rational exponents defines it: for \tfrac mn in lowest terms,

A^{m/n} = \left(\sqrt[n]{A}\right)^m

So an equation such as (x - 4)^{2/3} = 9 is the radical equation \displaystyle \left(\sqrt[3]{x - 4}\right)^2 = 9 in other clothes, and it is solved by undoing the outer power, then the root. Undoing an even power gives two signs, \pm, and undoing an odd one gives one. Taking every value into account is the whole difficulty; a shortcut that raises both sides to the power \tfrac32 uses the law \left(A^{2/3}\right)^{3/2} = A, which holds only for A \ge 0, and loses solutions (Example 7).

In a formula or a story. The same moves solve a formula for a variable under a root, or a problem whose unknown is a length under a square root. The story adds its own check: a length or a time cannot be negative.

Worked examples

Examples 1 to 4 go with Part 1 and Examples 5 to 8 with Part 2.

Example 1

Isolate first

Solve (a) \displaystyle \sqrt{x - 3} + 5 = 9, (b) \displaystyle 3\sqrt{2x} = 12 and (c) \displaystyle \sqrt{4x + 1} + 6 = 2.

Solution

(a) Subtract 5 from both sides to isolate the root: \displaystyle \sqrt{x - 3} = 4. Both sides are now ready to square, and the right side, 4, is non-negative, so nothing extraneous can appear:

x - 3 = 16 \qquad\Longrightarrow\qquad x = 19

Check. \displaystyle \sqrt{19 - 3} + 5 = \sqrt{16} + 5 = 4 + 5 = 9.

(b) Divide both sides by 3: \displaystyle \sqrt{2x} = 4. Squaring gives 2x = 16, so x = 8.

Check. \displaystyle 3\sqrt{16} = 3 \cdot 4 = 12.

(c) Subtract 6: \displaystyle \sqrt{4x + 1} = -4. Stop here. A square root is never negative, so no value of x works, and the equation has no solution. Squaring anyway would give 4x + 1 = 16 and the candidate x = \tfrac{15}{4}, which the check rejects: \displaystyle \sqrt{16} + 6 = 10, not 2. Isolating the root first showed the answer before any squaring.

Example 2

When both candidates are real, and when one is not

Solve (a) \displaystyle \sqrt{6x - 5} = x and (b) \displaystyle \sqrt{x + 5} + 1 = x.

Solution

(a) The root is alone. Square both sides and gather terms:

\begin{aligned} 6x - 5 &= x^2 \\ 0 &= x^2 - 6x + 5 \\ 0 &= (x - 1)(x - 5) \end{aligned}

The candidates are x = 1 and x = 5. Check: \displaystyle \sqrt{1} = 1, and \displaystyle \sqrt{25} = 5. Both are solutions. Squaring can add a solution, but it does not have to; here each candidate makes the right side, x, positive, which is the test from the procedure.

(b) First isolate the root by subtracting 1:

\sqrt{x + 5} = x - 1

Squaring the original equation as it was written would have turned the left side into \displaystyle (x + 5) + 2\sqrt{x + 5} + 1, still with a root in it. With the root alone, squaring clears it:

\begin{aligned} x + 5 &= x^2 - 2x + 1 \\ 0 &= x^2 - 3x - 4 \\ 0 &= (x - 4)(x + 1) \end{aligned}

Check in the original. For x = 4: \displaystyle \sqrt{9} + 1 = 4, which matches. For x = -1: \displaystyle \sqrt{4} + 1 = 3, not -1. The only solution is x = 4, and x = -1 is extraneous: it makes x - 1 negative, so it solves \displaystyle -\sqrt{x + 5} = x - 1 instead.

Example 3

Irrational candidates and the sign test

Solve \displaystyle \sqrt{x + 3} = x - 1.

Solution

Square both sides:

\begin{aligned} x + 3 &= x^2 - 2x + 1 \\ 0 &= x^2 - 3x - 2 \end{aligned}

This does not factor over the integers, so use the quadratic formula:

x = \frac{3 \pm \sqrt{9 + 8}}{2} = \frac{3 \pm \sqrt{17}}{2}

Substituting these into the original means working out the square root of an irrational number. The sign test avoids that. A candidate is a solution exactly when it makes x - 1 \ge 0. Since \displaystyle \sqrt{17} is between 4 and 5,

\begin{aligned} \frac{3 + \sqrt{17}}{2} - 1 &= \frac{1 + \sqrt{17}}{2} > 0 \\ \frac{3 - \sqrt{17}}{2} - 1 &= \frac{1 - \sqrt{17}}{2} < 0 \end{aligned}

So the only solution is \displaystyle x = \dfrac{3 + \sqrt{17}}{2} \approx 3.56, and the other candidate is extraneous.

Check exactly. With \displaystyle x = \dfrac{3 + \sqrt{17}}{2}, the right side is \displaystyle \dfrac{1 + \sqrt{17}}{2}, and its square is \displaystyle \dfrac{1 + 2\sqrt{17} + 17}{4} = \dfrac{9 + \sqrt{17}}{2}. The left side’s radicand is \displaystyle x + 3 = \dfrac{9 + \sqrt{17}}{2}, the same number, and the right side is positive, so it is that number’s square root.

Example 4

Cube roots and fourth roots

Solve (a) \displaystyle \sqrt[3]{2x - 5} + 4 = 1, (b) \displaystyle 2\sqrt[4]{x - 1} = 6 and (c) \displaystyle \sqrt[4]{x + 3} = -2.

Solution

(a) Isolate the root: \displaystyle \sqrt[3]{2x - 5} = -3. A cube root can be negative, so this is no cause to stop. Cube both sides, using (-3)^3 = -27:

\begin{aligned} 2x - 5 &= -27 \\ 2x &= -22 \\ x &= -11 \end{aligned}

Cubing is reversible, so no extraneous solution can have appeared. The check confirms the arithmetic: 2(-11) - 5 = -27, \displaystyle \sqrt[3]{-27} = -3, and -3 + 4 = 1.

(b) Divide by 2: \displaystyle \sqrt[4]{x - 1} = 3. Raise both sides to the fourth power: x - 1 = 81, so x = 82. Check: \displaystyle \sqrt[4]{81} = 3, and 2 \cdot 3 = 6.

(c) A fourth root, like a square root, is never negative, so there is no solution. Raising to the fourth power anyway gives x + 3 = 16 and x = 13, but \displaystyle \sqrt[4]{16} = 2, not -2. The fourth power is even, so it cannot tell 2 from -2.

Example 5

A radical on each side

Solve (a) \displaystyle \sqrt{3x - 2} = \sqrt{x + 6} and (b) \displaystyle \sqrt{x - 5} = \sqrt{2x - 3}.

Solution

(a) Each side is a single root, so squaring removes both:

\begin{aligned} 3x - 2 &= x + 6 \\ 2x &= 8 \\ x &= 4 \end{aligned}

Check: both radicands are 10, so both sides are \displaystyle \sqrt{10}. The solution is x = 4.

(b) Squaring gives x - 5 = 2x - 3, so x = -2. But then both radicands are -7, and \displaystyle \sqrt{-7} is not a real number. Neither side of the original has a value at x = -2, so this candidate is extraneous, and the equation has no solution. The two radicands are equal at x = -2, but equal radicands give equal roots only when the radicands are not negative.

Example 6

Squaring twice

Solve (a) \displaystyle \sqrt{2x + 3} - \sqrt{x + 1} = 1 and (b) \displaystyle \sqrt{x} + \sqrt{x - 5} = 1.

Solution

(a) Isolate one radical by adding \displaystyle \sqrt{x + 1} to both sides, then square. The right side is the square of a sum, \displaystyle 1 + 2\sqrt{x + 1} + (x + 1):

\begin{aligned} \sqrt{2x + 3} &= 1 + \sqrt{x + 1} \\ 2x + 3 &= x + 2 + 2\sqrt{x + 1} \\ x + 1 &= 2\sqrt{x + 1} \end{aligned}

One radical is left, and it is isolated. Square again:

\begin{aligned} x^2 + 2x + 1 &= 4(x + 1) \\ x^2 - 2x - 3 &= 0 \\ (x - 3)(x + 1) &= 0 \end{aligned}

Check both candidates in the original equation, since either squaring could have added one. For x = 3: \displaystyle \sqrt{9} - \sqrt{4} = 3 - 2 = 1. For x = -1: \displaystyle \sqrt{1} - \sqrt{0} = 1 - 0 = 1. Both are solutions, x = 3 and x = -1.

The line \displaystyle x + 1 = 2\sqrt{x + 1} can also be read with \displaystyle u = \sqrt{x + 1}, where it says u^2 = 2u, so u = 0 or u = 2. That gives x + 1 = 0 or x + 1 = 4, the same two answers.

(b) Isolate \displaystyle \sqrt{x} and square:

\begin{aligned} \sqrt{x} &= 1 - \sqrt{x - 5} \\ x &= 1 - 2\sqrt{x - 5} + (x - 5) \\ 2\sqrt{x - 5} &= -4 \\ \sqrt{x - 5} &= -2 \end{aligned}

A square root cannot be -2, so there is no solution. Squaring once more would give x - 5 = 4 and the candidate x = 9, which the original rejects: \displaystyle \sqrt{9} + \sqrt{4} = 5, not 1. A look at the original confirms it. Both roots are real only when x \ge 5, and then \displaystyle \sqrt{x} \ge \sqrt{5} > 2, so the left side is always more than 2. It can never equal 1.

Example 7

Rational exponents

Solve (a) (x - 4)^{2/3} = 9, (b) (2x + 1)^{3/2} = 27 and (c) (x + 1)^{-1/2} = \tfrac15.

Solution

(a) Write the power as a root and a power: \displaystyle \left(\sqrt[3]{x - 4}\right)^2 = 9. A number whose square is 9 is 3 or -3, so

\sqrt[3]{x - 4} = 3 \quad\text{or}\quad \sqrt[3]{x - 4} = -3

Cube each, which is reversible: x - 4 = 27 or x - 4 = -27, so x = 31 or x = -23.

Check. 27^{2/3} = 3^2 = 9 and (-27)^{2/3} = (-3)^2 = 9. Both are solutions. Raising both sides to the power \tfrac32 would have given only x - 4 = 9^{3/2} = 27, losing x = -23, because the law it relies on needs a base that is not negative.

(b) \displaystyle (2x + 1)^{3/2} = \left(\sqrt{2x + 1}\right)^3, so the equation says \displaystyle \left(\sqrt{2x + 1}\right)^3 = 27. Only one real number has cube 27, so \displaystyle \sqrt{2x + 1} = 3. Squaring, 2x + 1 = 9 and x = 4.

Check. \displaystyle 9^{3/2} = \left(\sqrt{9}\right)^3 = 3^3 = 27.

(c) A negative exponent means one over the positive power, so the equation says \displaystyle \dfrac{1}{\sqrt{x + 1}} = \dfrac15. Two positive numbers with equal reciprocals are equal, so \displaystyle \sqrt{x + 1} = 5, and squaring gives x + 1 = 25, so x = 24.

Check. \displaystyle 25^{-1/2} = \dfrac{1}{\sqrt{25}} = \dfrac15.

Example 8

A guy wire

A wire runs from the top of a 12 m pole to a stake in the ground. The wire is 3 m longer than the distance from the foot of the pole to the stake. How far is the stake from the pole, and how long is the wire?

Solution

Call the distance from the foot of the pole to the stake d meters. The pole, the ground and the wire form a right triangle with legs 12 and d, so by the Pythagorean theorem the wire is \displaystyle \sqrt{d^2 + 144} meters long. The statement gives a second expression for the same length:

\sqrt{d^2 + 144} = d + 3

The root is isolated. Square both sides; the d^2 terms cancel, which leaves a linear equation:

\begin{aligned} d^2 + 144 &= d^2 + 6d + 9 \\ 135 &= 6d \\ d &= 22.5 \end{aligned}

The sign test passes, since d + 3 = 25.5 > 0, and a distance of 22.5 m fits the story. The stake is 22.5 m from the pole and the wire is 25.5 m long.

Check. 22.5^2 + 12^2 = 506.25 + 144 = 650.25, and 25.5^2 = 650.25.

Boundaries

The radical must be isolated before it is squared. Squaring a side that is a sum, such as \displaystyle \sqrt{x + 5} + 1, produces a middle term that still holds the root. The move works only on a side that is a single root, perhaps times a number.

Raising both sides to an even power can add solutions. It keeps every solution of the original, and may add solutions of the equation with the root’s sign flipped. Every candidate has to be checked in the original equation, or, for a single square root \displaystyle \sqrt{A} = B, tested for B \ge 0. When an equation has been squared twice, the check is against the first equation, not the one in the middle.

Raising to an odd power adds nothing. Cubing, or any odd power, is reversible, so an equation with one cube root has exactly the solutions of the cubed equation. Checking then only guards the arithmetic.

An even root is never negative. An equation that, once the root is isolated, sets a square root or fourth root equal to a negative number has no solution. A cube root can be negative.

Each root must be a real number. A candidate that makes the radicand of an even root negative is not a solution, even if the squared equation holds there, as in Example 5(b).

A rational exponent with a negative base needs care. A^{m/n} is read as \displaystyle \left(\sqrt[n]{A}\right)^m with \tfrac mn in lowest terms. When the numerator m is even (so n is odd), the equation A^{m/n} = c with c > 0 has a solution with A negative as well as one with A positive, and raising both sides to the power \tfrac nm finds only the positive one.

In context, the story has the last word. A candidate can solve the equation and still be a negative length or time, which the problem rules out.

Where this goes next

Raising both sides to a power is the first of several moves that are true in one direction only. The same thing happens with rational equations, where multiplying by an expression that can be zero adds candidates, and with logarithmic equations, where an answer must be dropped if it makes a logarithm take a value that is not positive. In each case the method is the same: know which step is not reversible, and check what it could have added.

The equations of this lesson are also the backward questions of radical functions in context: given the period of a pendulum or the speed of a car, find the length or the braking distance that produced it. And the picture of a square root as half of a sideways parabola comes from graphing a square root function, where a solution of \displaystyle \sqrt{A} = B is a point where two graphs meet.

Exercises

Part 1: isolate and solve

\displaystyle \sqrt{x + 6} = 7
\displaystyle \sqrt{5x - 1} - 4 = 3
\displaystyle 4 + 2\sqrt{3x} = 16
\displaystyle \sqrt{2x + 9} + 7 = 4

Part 1: candidates to check

Solve, and name any extraneous candidate.

\displaystyle \sqrt{x + 10} = x - 2
\displaystyle \sqrt{8x - 7} = x
\displaystyle \sqrt{x + 1} + 5 = x
\displaystyle 3\sqrt{x - 2} = x
\displaystyle \sqrt{2 - x} = x + 4
\displaystyle \sqrt{x + 7} = x - 3

Part 1: other roots

\displaystyle \sqrt[3]{x + 2} = -4
\displaystyle \sqrt[3]{4x - 1} + 2 = 5
\displaystyle \sqrt[4]{2x + 6} = 2
\displaystyle \sqrt[3]{x^2 + 7} = 2

Part 2: two radicals

\displaystyle \sqrt{5x - 4} = \sqrt{2x + 5}
\displaystyle \sqrt{x^2 + 3x} = \sqrt{x + 8}
\displaystyle \sqrt{x + 12} - \sqrt{x} = 2
\displaystyle \sqrt{x + 4} + \sqrt{x - 1} = 5
\displaystyle \sqrt{3x + 1} - \sqrt{x + 4} = 1

Part 2: rational exponents

(x - 1)^{3/2} = 8
(3x + 1)^{2/3} = 16
2(x + 3)^{1/4} = 4

Part 2: problems

A point P = (x, 0) with x > 0 is 4 units farther from (0, 6) than it is from the origin. Find x.
Find every x for which the point (x, 2) is 5 units from the point (5, -1).

Reasoning

Squaring \displaystyle \sqrt{2x + 7} = x + 2 produced the candidate x = -3. Name an equation that x = -3 does solve, and explain why squaring cannot tell the two equations apart.
Explain why cubing both sides of an equation never produces an extraneous solution.
Prove that \displaystyle \sqrt{A} = B holds exactly when A = B^2 and B \ge 0.
To solve \displaystyle \sqrt{x + 5} + 1 = x, Ravi squares both sides straight away. Show what he gets, and explain why isolating the root first is the better move.
Without solving, explain why \displaystyle \sqrt{x - 3} + \sqrt{x + 1} = 1 has no solution.

Answers

x = 43. Square: x + 6 = 49. Check: \displaystyle \sqrt{49} = 7.
x = 10. Add 4: \displaystyle \sqrt{5x - 1} = 7, so 5x - 1 = 49. Check: \displaystyle \sqrt{49} - 4 = 3.
x = 12. Subtract 4 and divide by 2: \displaystyle \sqrt{3x} = 6, so 3x = 36. Check: 4 + 2 \cdot 6 = 16.
No solution. Subtract 7: \displaystyle \sqrt{2x + 9} = -3, and a square root is never negative.
x = 6; x = -1 is extraneous. Square: x + 10 = x^2 - 4x + 4, so x^2 - 5x - 6 = 0 and (x - 6)(x + 1) = 0. At x = 6, \displaystyle \sqrt{16} = 4 = 6 - 2. At x = -1 the right side is -3.
x = 1 and x = 7, both solutions. Square: x^2 - 8x + 7 = 0, so (x - 1)(x - 7) = 0. Check: \displaystyle \sqrt{1} = 1 and \displaystyle \sqrt{49} = 7.
x = 8; x = 3 is extraneous. Isolate: \displaystyle \sqrt{x + 1} = x - 5. Square: x^2 - 11x + 24 = 0, so (x - 3)(x - 8) = 0. At x = 8, \displaystyle \sqrt{9} + 5 = 8. At x = 3, \displaystyle \sqrt{4} + 5 = 7 \ne 3.
x = 3 and x = 6, both solutions. Square: 9(x - 2) = x^2, so x^2 - 9x + 18 = 0 and (x - 3)(x - 6) = 0. Check: \displaystyle 3\sqrt{1} = 3 and \displaystyle 3\sqrt{4} = 6.
x = -2; x = -7 is extraneous. Square: 2 - x = x^2 + 8x + 16, so x^2 + 9x + 14 = 0 and (x + 2)(x + 7) = 0. At x = -2, \displaystyle \sqrt{4} = 2 = -2 + 4. At x = -7 the right side is -3.
\displaystyle x = \dfrac{7 + \sqrt{41}}{2} \approx 6.70; \displaystyle x = \dfrac{7 - \sqrt{41}}{2} is extraneous. Square: x + 7 = x^2 - 6x + 9, so x^2 - 7x + 2 = 0 and \displaystyle x = \dfrac{7 \pm \sqrt{41}}{2}. The sign test needs x - 3 \ge 0: the first gives \displaystyle \dfrac{1 + \sqrt{41}}{2} > 0, the second \displaystyle \dfrac{1 - \sqrt{41}}{2} < 0.
x = -66. Cube: x + 2 = -64. Check: \displaystyle \sqrt[3]{-64} = -4.
x = 7. Isolate: \displaystyle \sqrt[3]{4x - 1} = 3. Cube: 4x - 1 = 27. Check: \displaystyle \sqrt[3]{27} + 2 = 5.
x = 5. Raise to the fourth power: 2x + 6 = 16. Check: \displaystyle \sqrt[4]{16} = 2.
x = 1 or x = -1. Cube: x^2 + 7 = 8, so x^2 = 1. Cubing is reversible, and both give \displaystyle \sqrt[3]{8} = 2.
x = 3. Square: 5x - 4 = 2x + 5, so 3x = 9. Both radicands are 11, which is not negative.
x = 2 and x = -4, both solutions. Square: x^2 + 3x = x + 8, so x^2 + 2x - 8 = 0 and (x - 2)(x + 4) = 0. At x = 2 both radicands are 10; at x = -4 both are 4.
x = 4. Isolate: \displaystyle \sqrt{x + 12} = 2 + \sqrt{x}. Square: \displaystyle x + 12 = 4 + 4\sqrt{x} + x, so \displaystyle \sqrt{x} = 2. Check: \displaystyle \sqrt{16} - \sqrt{4} = 2.
x = 5. Isolate: \displaystyle \sqrt{x + 4} = 5 - \sqrt{x - 1}. Square: \displaystyle x + 4 = 25 - 10\sqrt{x - 1} + x - 1, so \displaystyle 10\sqrt{x - 1} = 20 and \displaystyle \sqrt{x - 1} = 2. Check: \displaystyle \sqrt{9} + \sqrt{4} = 5.
x = 5; x = 0 is extraneous. Isolate: \displaystyle \sqrt{3x + 1} = 1 + \sqrt{x + 4}. Square: \displaystyle 3x + 1 = x + 5 + 2\sqrt{x + 4}, so \displaystyle x - 2 = \sqrt{x + 4}. Square again: x^2 - 5x = 0, so x = 0 or x = 5. At x = 5, \displaystyle \sqrt{16} - \sqrt{9} = 1. At x = 0, \displaystyle \sqrt{1} - \sqrt{4} = -1.
x = 5. \displaystyle \left(\sqrt{x - 1}\right)^3 = 8, and only 2 has cube 8, so \displaystyle \sqrt{x - 1} = 2 and x - 1 = 4. Check: 4^{3/2} = 2^3 = 8.
x = 21 or x = -\tfrac{65}{3}. \displaystyle \left(\sqrt[3]{3x + 1}\right)^2 = 16, so \displaystyle \sqrt[3]{3x + 1} = \pm 4 and 3x + 1 = \pm 64. Check: 64^{2/3} = 16 and (-64)^{2/3} = (-4)^2 = 16.
x = 13. Divide by 2: \displaystyle \sqrt[4]{x + 3} = 2, so x + 3 = 16. Check: \displaystyle 2\sqrt[4]{16} = 4.
x = \tfrac52. The distance from the origin is x and from (0, 6) is \displaystyle \sqrt{x^2 + 36}, so \displaystyle \sqrt{x^2 + 36} = x + 4. Square: x^2 + 36 = x^2 + 8x + 16, so 8x = 20. Check: \displaystyle \sqrt{\tfrac{25}{4} + 36} = \sqrt{\tfrac{169}{4}} = \tfrac{13}{2} = \tfrac52 + 4.
x = 9 or x = 1. By the distance formula, \displaystyle \sqrt{(x - 5)^2 + 9} = 5. Square: (x - 5)^2 = 16, so x - 5 = \pm 4. Both check, since the right side, 5, is positive.
x = -3 solves \displaystyle -\sqrt{2x + 7} = x + 2: both sides are -1. Squaring that equation gives 2x + 7 = (x + 2)^2, the same squared equation, because (-a)^2 = a^2. Squaring throws away the sign of the side with the root, so the squared equation holds wherever either equation does.
If a = b then a^3 = b^3, so no solution is lost. Conversely, if a^3 = b^3 then a = b: an odd power keeps the sign, and for numbers of the same sign a larger size gives a larger cube, so two different real numbers never have the same cube. The cubed equation therefore holds for exactly the same values as the original.
If \displaystyle \sqrt{A} = B, then B is a square root, so B \ge 0, and squaring gives A = B^2. Conversely, if A = B^2 and B \ge 0, then A \ge 0, so \displaystyle \sqrt{A} is defined, and it is the non-negative number whose square is A. B is such a number, so \displaystyle \sqrt{A} = B.
He gets \displaystyle (x + 5) + 2\sqrt{x + 5} + 1 = x^2, which still has a root in it, because the square of a sum has a middle term 2ab. Isolating first, \displaystyle \sqrt{x + 5} = x - 1, makes one side a single root, and squaring it gives x + 5 with no root left.
Both roots are real only when x \ge 3. Then x + 1 \ge 4, so \displaystyle \sqrt{x + 1} \ge 2, and \displaystyle \sqrt{x - 3} \ge 0. The left side is therefore at least 2 for every allowed x, and never equals 1.