Raising Equations to Powers

Learning goals

Distinguish a reversible step from one that can add solutions
Raise to an odd power safely, unlike an even one
Write $\sqrt{x^2}$ as $\lvert x \rvert$ , not $x$
Isolate the radical, raise, solve, then check every candidate
Clear two radicals by squaring twice
Attach $\pm$ when taking an even root of both sides

Operating on both sides of an equation

Solving is a chain of reversible steps. When you add $3$ to both sides you can always subtract it again. That means the new equation has exactly the same solutions as the old one: nothing is gained, nothing is lost. The two equations are equivalent. Every basic move you have used to solve linear and quadratic equations was reversible in this way, which is why you never had to look back and check.

Raising both sides to a power breaks the pattern. Half of the guarantee still holds, and it is worth stating on its own before we find the crack.

Raising both sides to a power never loses a solution#

Suppose some value of the variable makes the original equation true, so that its two sides, call them $a$ and $b$ , are equal:

a = b.

Equal quantities are the very same number written two ways. Raising that one number to the $n$ th power produces one definite result, so $a^n$ and $b^n$ are that same result:

a^n = b^n.

The value therefore still satisfies the new equation. Nothing in the argument depended on which power $n$ we chose, so it holds for squaring, cubing, and every power beyond. Whatever solutions the original equation had, the raised equation still has all of them.

What the argument does not promise is the reverse, that the raised equation has no others. Going from $a^n = b^n$ back to $a = b$ is a separate claim, and it is exactly where new solutions slip in.

The gap is easy to see with squaring. Knowing $a = b$ certainly gives $a^2 = b^2$ . But knowing $a^2 = b^2$ does not give back $a = b$ , because a square forgets a sign. Both $3$ and $-3$ square to $9$ , so $a^2 = b^2$ leaves open whether $a = b$ or $a = -b$ . The raised equation quietly carries an extra possibility that the original never contained.

Squaring is two-to-one: the inputs 3 and negative 3 both land on 9. That collapse is why undoing a square can hand back a value the original problem never contained, so squaring both sides of an equation can add a solution you must check.

Why even powers add solutions and odd powers do not

The sign that squaring forgets is the whole story. Cubing does not forget it, and that single difference splits every power into a safe kind and a risky kind. The proof turns on two factoring patterns you already know, the difference of squares and the difference of cubes.

Why an even power can add solutions but an odd power cannot#

Start from the raised equation and ask when it forces the original. Take squaring first. Suppose

a^2 = b^2.

Bring everything to one side and factor with the difference of squares:

a^2 - b^2 = 0 \quad\Longrightarrow\quad (a - b)(a + b) = 0.

A product is zero only when one of its factors is zero, so $a = b$ or $a = -b$ . The squared equation is satisfied by both branches. The branch $a = b$ is the one you wanted, and the branch $a = -b$ is the extra. Any value that fits the extra branch without fitting $a = b$ is a new solution the original equation never had.

Now cube instead. Suppose

a^3 = b^3.

Factor with the difference of cubes:

a^3 - b^3 = 0 \quad\Longrightarrow\quad (a - b)(a^2 + ab + b^2) = 0.

The first factor gives the wanted $a = b$ . The second factor cannot open a new branch over the real numbers, and completing the square shows why:

a^2 + ab + b^2 = \left(a + \tfrac{b}{2}\right)^2 + \tfrac{3}{4}b^2.

This is a sum of two squares, and a sum of squares is zero only when each square is zero. That needs $b = 0$ and then $a = 0$ , which is already the case $a = b$ . So over the reals $a^3 = b^3$ forces $a = b$ with nothing extra. Cubing adds no solutions.

The pattern continues up the powers. An even power, like a square, keeps the sign-forgetting factor and splits into $a = b$ or $a = -b$ , so it can inflate the solution set. An odd power, like a cube, keeps $a^n = b^n$ equivalent to $a = b$ , so it is safe. The dividing line is simply whether the power hides the sign of its base.

This is the practical headline. Raising both sides to an odd power gives an equivalent equation, with the same solutions. Raising to an even power can hand you extra candidates that solve the new equation but not the one you started with. Because even roots (square roots, fourth roots) are cleared by even powers, radical equations are precisely where the extra candidates show up.

The square root of a square

The same sign-forgetting appears from the other direction, in an identity you will use constantly. The symbol $\sqrt{\phantom{x}}$ returns the principal root, the one that is not negative. So $\sqrt{x^2}$ is not simply $x$ : it is the nonnegative number whose square is $x^2$ . If $x$ is already zero or positive, that number is $x$ itself. If $x$ is negative, then $x$ cannot be the answer (the answer must be nonnegative), and the nonnegative number that squares to $x^2$ is $-x$ . Both cases are captured at once by the absolute value:

\sqrt{x^2} = \lvert x \rvert.

Test it on a negative input. With $x = -4$ ,

\sqrt{(-4)^2} = \sqrt{16} = 4 = \lvert -4 \rvert,

which is $-x$ , not $x$ . Contrast this with squaring after rooting a nonnegative number: there $\left(\sqrt{x}\right)^2 = x$ exactly, because $\sqrt{x}$ is already nonnegative and squaring it just undoes the root. The order is what matters. Root then square returns $x$ ; square then root passes through $\lvert x \rvert$ and can change the sign.

Check your understanding

Simplify $\sqrt{(-6)^2}$ .

-6

6

\pm 6

36

Solving radical equations

A radical equation hides the variable under a root, as in $\sqrt{2x - 1} = 5$ . The plan uses everything above. Raising both sides to the power that matches the index clears the root. For a square root, that matching power is even, so you finish by checking, because the even power may have added a candidate. The reliable method is four steps:

Isolate the radical on one side by itself.
Raise both sides to the power that matches the index (square for a square root, cube for a cube root).
Solve the equation that remains.
Check every candidate in the original equation, and discard any that fail.

A candidate that solves the raised equation but fails the original is called an extraneous solution. The name comes from the Latin for “foreign”: it is an outsider the squaring let in. Step 4 is not politeness. It is the step that repairs the one-way logic, because raising to an even power only promised to keep the real solutions, not to reject impostors.

Worked example 1 Solve $4 + \sqrt{3x - 2} = 9$

Isolate the radical first. Subtract $4$ from both sides so the root stands alone:

\sqrt{3x - 2} = 5.

Both sides are now single quantities, so square them to clear the root:

3x - 2 = 25 \quad\Longrightarrow\quad 3x = 27 \quad\Longrightarrow\quad x = 9.

Check the candidate in the original equation, not the squared one:

4 + \sqrt{3(9) - 2} = 4 + \sqrt{25} = 4 + 5 = 9. \checkmark

The check holds, so $x = 9$ is the solution. Isolating first was essential: squaring while the $4$ was still attached would have left a radical behind.

Worked example 2 Solve $\sqrt{x + 2} = x$ and catch the impostor

The radical is already alone, so square both sides. The right side becomes $x^2$ :

x + 2 = x^2.

Rearrange into a quadratic and factor:

x^2 - x - 2 = 0 \quad\Longrightarrow\quad (x - 2)(x + 1) = 0,

so the candidates are $x = 2$ and $x = -1$ . Squaring is an even power, so test both in the original equation $\sqrt{x + 2} = x$ .

For $x = 2$ : the left side is $\sqrt{4} = 2$ and the right side is $2$ . They agree, so $x = 2$ is a genuine solution.

For $x = -1$ : the left side is $\sqrt{1} = 1$ and the right side is $-1$ . Since $1 \ne -1$ , the candidate $x = -1$ fails, so it is extraneous. It is exactly the intruder the theory predicts. It solves the other branch $\sqrt{x + 2} = -x$ , since $-(-1) = 1 = \sqrt{1}$ . But the original equation needs the root to equal $x$ , and a root is never negative, so $x$ cannot be $-1$ . The only solution is $x = 2$ .

Check your understanding

Squaring $\sqrt{x + 6} = x$ gives $x + 6 = x^2$ , whose candidates are $x = 3$ and $x = -2$ . Which of these are actual solutions of the original equation?

A only

x = 3

B only

x = -2

C both

x = 3

and

x = -2

D neither

Odd roots behave more gently, because clearing a cube root uses cubing, an odd power that adds nothing.

Worked example 3 Solve $\sqrt[3]{5x + 2} = -2$

Here the root is a cube root, so raise both sides to the third power to clear it. Cubing is the matching odd power:

5x + 2 = (-2)^3 = -8.

Solve the linear equation:

5x = -10 \quad\Longrightarrow\quad x = -2.

A quick check confirms it: $\sqrt[3]{5(-2) + 2} = \sqrt[3]{-8} = -2$ . Notice two things. The right side was negative, which is fine, because a cube root happily returns a negative value. And because cubing is an odd power, no extraneous candidate could have appeared, so this equation cannot spring a surprise the way a squared one can. Checking is still cheap insurance, but here it was guaranteed to pass.

Equations with two radicals

When two separate roots appear, one squaring will not clear them both, because a root left next to other terms survives the square. The fix is to square twice. Isolate one radical and square to remove it, which leaves a simpler equation with a single remaining root; isolate that one and square again.

Worked example 4 Solve $\sqrt{3x + 1} - \sqrt{x + 4} = 1$

Isolate one radical by moving the other across:

\sqrt{3x + 1} = 1 + \sqrt{x + 4}.

Square both sides. The right side is a binomial, so expand it in full, and a single root survives:

3x + 1 = 1 + 2\sqrt{x + 4} + (x + 4) = x + 5 + 2\sqrt{x + 4}.

Isolate the surviving root, then divide the whole equation by $2$ :

2x - 4 = 2\sqrt{x + 4} \quad\Longrightarrow\quad x - 2 = \sqrt{x + 4}.

Square a second time and collect into a quadratic:

(x - 2)^2 = x + 4 \quad\Longrightarrow\quad x^2 - 4x + 4 = x + 4 \quad\Longrightarrow\quad x^2 - 5x = 0.

Factor out $x$ :

x(x - 5) = 0 \quad\Longrightarrow\quad x = 0 \ \text{ or } \ x = 5.

Two squarings mean two chances for an impostor, so test both in the original equation.

For $x = 5$ : $\sqrt{16} - \sqrt{9} = 4 - 3 = 1$ . Correct.

For $x = 0$ : $\sqrt{1} - \sqrt{4} = 1 - 2 = -1 \ne 1$ . This one fails, so it is extraneous. The only solution is $x = 5$ . The intermediate line $x - 2 = \sqrt{x + 4}$ already warned us, since its right side cannot be negative, so $x$ must be at least $2$ , which $x = 0$ violates.

Even roots, the plus-or-minus sign, and rational powers

Raising to a power undoes a root; taking a root undoes a power. When you undo an even power by rooting both sides, the sign-forgetting comes back and you must write $\pm$ . Starting from $(x - h)^2 = k$ with $k \ge 0$ , take the square root of both sides using $\sqrt{(x-h)^2} = \lvert x - h \rvert$ :

\lvert x - h \rvert = \sqrt{k} \quad\Longrightarrow\quad x - h = \pm\sqrt{k}.

The absolute value is what forces the two signs, and forgetting it is the same mistake as forgetting the negative root. Both signs give genuine solutions here, because taking a root is undoing a square, not performing one, so nothing extraneous is created. The $\pm$ sign simply records the two honest answers.

Check your understanding

Solve $(x + 2)^2 = 9$ for all real $x$ .

x = 1

x = -5

x = 1

only C

x = \pm 3

x = -5

only

The same idea reaches rational exponents. To undo $x^{2/3}$ you raise to the reciprocal power $\tfrac{3}{2}$ , and because the exponent $\tfrac{2}{3}$ has an even numerator, a $\pm$ appears.

Worked example 5 Solve $x^{2/3} = 4$

Read $x^{2/3}$ as $\left(x^{1/3}\right)^2$ , the cube root of $x$ then squared. Let $u = x^{1/3}$ , which is defined for every real number and one-to-one. The equation becomes

u^2 = 4 \quad\Longrightarrow\quad u = \pm 2,

where the $\pm$ is the even-power sign-forgetting once more. Now undo the cube root by cubing, $x = u^3$ :

u = 2 \Rightarrow x = 2^3 = 8, \qquad u = -2 \Rightarrow x = (-2)^3 = -8.

Both check in the original: $8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$ , and $(-8)^{2/3} = \left(\sqrt[3]{-8}\right)^2 = (-2)^2 = 4$ . So $x = \pm 8$ . The even numerator is what produced two answers; an equation like $x^{3/2} = 8$ , with an odd numerator, has the single solution $x = 4$ .

The same substitution rescues an equation that is quadratic in $\sqrt{x}$ . Setting $s = \sqrt{x}$ turns $x - 5\sqrt{x} + 6 = 0$ into $s^2 - 5s + 6 = 0$ , since $x = s^2$ , and that is an ordinary quadratic to factor. The one extra care is that $s = \sqrt{x}$ is a principal square root, so $s$ cannot be negative. Discard any negative value of $s$ before squaring back to $x = s^2$ . That nonnegativity check is the same watch-the-branch habit the rest of the lesson is built on.

Changing both sides in other ways

Raising to a power is the headline case, but the same logic governs any operation that is not reversible. Two more come up often enough to name.

Multiplying both sides by an expression that can be zero adds solutions. To solve $\frac{x^2}{x - 2} = \frac{4}{x - 2}$ you would multiply both sides by $x - 2$ to get $x^2 = 4$ , whose roots are $x = 2$ and $x = -2$ . But $x = 2$ makes the original denominators zero, so the original equation is undefined there; it is extraneous, introduced by multiplying by $x - 2$ , which is itself zero at $x = 2$ . Only $x = -2$ survives. This is the very extraneous-root check you met with rational equations, and it has the same cause as squaring.

Dividing both sides by an expression that can be zero loses solutions. Given $x^2 = 5x$ , dividing both sides by $x$ gives $x = 5$ and silently discards $x = 0$ , which is a perfectly good solution since $0^2 = 5 \cdot 0$ . The safe move is never to divide by something that might vanish. Bring everything to one side and factor instead:

x^2 - 5x = 0 \quad\Longrightarrow\quad x(x - 5) = 0 \quad\Longrightarrow\quad x = 0 \ \text{ or } \ x = 5.

Multiplying by a vanishing expression inflates the solution set, so you check and discard; dividing by one deflates it, so you avoid the division entirely.

Common mistakes

Skipping the check. Raising both sides to an even power can create extraneous solutions, so every candidate must be tested in the original equation. A root you have not verified is not yet an answer.
Writing $\sqrt{x^2} = x$ . The principal root is never negative, so $\sqrt{x^2} = \lvert x \rvert$ . This is the same collapse that produces extraneous roots, seen from the other side.
Squaring term by term. Squaring $\sqrt{x} = x - 2$ gives $x = (x-2)^2 = x^2 - 4x + 4$ , not $x^2 + 4$ . A binomial must be expanded in full, and $\sqrt{a + b}$ is not $\sqrt{a} + \sqrt{b}$ .
Dropping the $\pm$ . Taking an even root of both sides of $(x - h)^2 = k$ gives $x - h = \pm\sqrt{k}$ , two solutions, because $\sqrt{(x-h)^2} = \lvert x - h \rvert$ . Keeping only the positive sign loses one.
Assuming the smaller candidate is always the extraneous one. Sometimes both candidates are valid, sometimes the larger fails. Only substitution into the original equation decides.
Dividing by a variable expression. Dividing $x^2 = 5x$ by $x$ throws away $x = 0$ . Move everything to one side and factor rather than dividing by something that might be zero.

Key takeaways

Doing the same thing to both sides keeps every solution the equation already had. A step that is not reversible, like squaring, can also add solutions that were never there.
Raising both sides to an odd power gives an equivalent equation, with the same solutions. Raising both sides to an even power can inflate the solution set, because an even power forgets the sign of its base.
The square root of a square is the absolute value: $\sqrt{x^2} = \lvert x \rvert$ , not $x$ .
To solve a radical equation, isolate the radical, raise both sides to the matching power, then solve. Finally, check every candidate in the original equation and discard the extraneous ones.
Two radicals are cleared by squaring twice: isolate one root and square, then isolate the survivor and square again.
Taking an even root of both sides introduces $\pm$ ; multiplying both sides by a vanishing expression adds solutions (check them), while dividing by one loses solutions (avoid it).

Raising Equations to Powers

Learning goals

Operating on both sides of an equation

Why even powers add solutions and odd powers do not

The square root of a square

Solving radical equations

Equations with two radicals

Even roots, the plus-or-minus sign, and rational powers

Changing both sides in other ways

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)