Solving Radical Equations: Free Response

Move the radical by itself first. Adding $\sqrt{2x+1}$ to both sides and subtracting $2$ gives

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

Now the radical is the whole left side, so squaring clears it with nothing left over:

$2x+1=9$

Solving the linear equation gives $2x=8$, so $x=4$.

Check the candidate in the original equation, not in the isolated line above it:

$5-\sqrt{2(4)+1}=5-\sqrt9=5-3=2$

The two sides agree, so $x=4$ is a genuine solution.

The radical already stands alone, so square both sides:

$x+14=(x-6)^2=x^2-12x+36$

Collect everything on one side:

$x^2-13x+22=0$

This factors as $(x-11)(x-2)=0$, so the candidates are $x=11$ and $x=2$.

Check $x=11$ in the original equation. The left side is $\sqrt{11+14}=\sqrt{25}=5$, and the right side is $11-6=5$. The two agree, so $x=11$ is a genuine solution.

Check $x=2$. The left side is $\sqrt{2+14}=\sqrt{16}=4$, and the right side is $2-6=-4$:

$4\ne-4$

So $x=2$ is extraneous, even though its radicand, $16$, is perfectly positive. The solution set is $\{11\}$.

Let $A=\sqrt{x+14}$ and $B=x-6$. The original equation says $A=B$. Squaring both sides produces $A^2=B^2$, and $A^2-B^2$ factors as a difference of squares:

$(A-B)(A+B)=0$

A product is zero only when one factor is, so $A^2=B^2$ says exactly "$A=B$ or $A=-B$", nothing more and nothing less. The second branch is the twin equation

$\sqrt{x+14}=-(x-6)=6-x$

Check $x=2$ there: the left side is $\sqrt{16}=4$ and the right side is $6-2=4$. They agree, so $x=2$ is an honest solution of the twin, even though it solves nothing in the original.

The reason this cannot be avoided is that squaring is built from multiplying $A$ by itself, and $(-A)\cdot(-A)$ gives the same product as $A\cdot A$. The operation has no way to record which sign $A$ started with, so once it is applied, the equation it produces is symmetric in $A$ and $-A$ by construction. Any candidate belonging to either branch survives; sorting out which branch a candidate actually solves is exactly what the final check against the original equation is for.

Isolate one radical by moving the other to the right side:

$\sqrt{x+13}=2+\sqrt{x+1}$

Square both sides. The right side is a binomial, so it expands with a middle term:

$x+13=4+4\sqrt{x+1}+(x+1)=x+5+4\sqrt{x+1}$

Subtracting $x+5$ from both sides isolates what is left of the radical:

$8=4\sqrt{x+1}$

which is the same equation as $\sqrt{x+1}=2$. Exactly one radical remains, exactly as expected: squaring a binomial that contains one radical always leaves that radical's cross term behind.

Starting from $\sqrt{x+1}=2$, the radical is already alone, so square once more:

$x+1=4$

which gives $x=3$.

Check in the original equation:

$\sqrt{3+13}-\sqrt{3+1}=\sqrt{16}-\sqrt4=4-2=2$

The two sides agree, so $x=3$ is the only solution, and it is genuine.

The classmate is treating any radical still present after squaring as a sign of failure, but that mixes up two different things: a radical that never gets cleared, and a radical that has not been cleared yet.

Squaring is guaranteed to clear an isolated radical, since then $(\sqrt u)^2=u$ with nothing left over. Here the squared side was a binomial instead, so its nonzero cross term leaves a radical behind:

$\left(2+\sqrt{x+1}\right)^2=4+4\sqrt{x+1}+(x+1)$

Squaring a binomial always leaves its cross term behind, so a radical was always going to survive this particular squaring. That is expected, not a failure of the method.

The method has an explicit next step for exactly this situation: isolate whatever radical remains and square again. Doing so here turns $\sqrt{x+1}=2$ into $x+1=4$, a radical-free equation with the solution $x=3$, which checks in the original equation. So the equation does have a solution; the classmate stopped one round of the method too early.

Rewrite the left side using $a^{1/2}=\sqrt a$:

$\sqrt{2x-3}=x-3$

The radical already stands alone, so square both sides:

$2x-3=(x-3)^2=x^2-6x+9$

Collecting terms gives

$x^2-8x+12=0$

which factors as $(x-6)(x-2)=0$, so the candidates are $x=6$ and $x=2$.

Check $x=6$: the left side is $(2(6)-3)^{1/2}=9^{1/2}=3$, and the right side is $6-3=3$. They agree, so $x=6$ is genuine.

Check $x=2$: the left side is $(2(2)-3)^{1/2}=1^{1/2}=1$, and the right side is $2-3=-1$:

$1\ne-1$

So $x=2$ is extraneous, exactly the same failure a plain radical equation can produce, since raising to the power 2 is still squaring underneath the fractional notation.

Write the left side as a cube of a square root: $(x-2)^{3/2}=\left[(x-2)^{1/2}\right]^3$. Naming $u=(x-2)^{1/2}$, note that u is a principal square root, so $u\ge0$, and the equation becomes

$u^3=64$

Raising to an odd power is reversible, so $u=4$ is the only real cube root of $64$.

Undoing the substitution, $(x-2)^{1/2}=4$, so $x-2=16$ and $x=18$.

Checking, $(18-2)^{3/2}=16^{3/2}=(\sqrt{16})^3=4^3=64$, which confirms the solution.

The same substitution as part B applies: $(x-2)^{3/2}=\left[(x-2)^{1/2}\right]^3=u^3$ where $u=(x-2)^{1/2}$.

Because u is a principal square root, it can never be negative: $u\ge0$ for every x in the domain. A nonnegative number raised to the power 3 is still nonnegative, so

$u\ge0 \;\Longrightarrow\; u^3\ge0$

The equation $u^3=-64$ asks $u^3$ to equal a negative number, which that inequality rules out immediately. No value of u, and therefore no value of x, can satisfy it. Recognizing this before computing anything is the whole point: no domain check, no squaring, and no candidate list are needed to reach the answer.

Substitute each candidate into the equation as it was originally written, not into the domain condition.

At x=13: the left side is $\sqrt{6(13)-14}=\sqrt{64}=8$, and the right side is $13-5=8$. The two sides agree, so x=13 is a genuine solution.

At x=3: the left side is $\sqrt{6(3)-14}=\sqrt4=2$, and the right side is $3-5=-2$:

$2\ne-2$

The two sides disagree, so x=3 is extraneous. Yet the domain condition $6x-14\ge0$ is satisfied at x=3, since $6(3)-14=4\ge0$: the domain check gave both candidates a pass, and one of them is still wrong. Passing the domain check and being a genuine solution are not the same thing here.

From part A, x=13 checks and x=3 does not, so the solution set of the original equation is $\{13\}$.

The domain condition

$6x-14\ge0$

was derived from requiring the radicand to be nonnegative, which is exactly what it is good for: it guarantees that $\sqrt{6x-14}$ names a real number at all. It says nothing about the right side, $x-5$, and nothing about whether the two sides agree once both are evaluated. A candidate can make the left side a perfectly good real number and still disagree with the right side in size or in sign, and x=3 is exactly that case: $\sqrt4=2$ is a real number, but 2 is not -2.

In a rational equation, the step that can go wrong is multiplying both sides by the LCD, and it can only go wrong where the LCD equals zero, since that is the one place the multiplication cannot be undone by dividing back. The excluded values are defined to be exactly those zeros, so checking a candidate against the excluded values is checking the exact condition under which the step might have failed. Nothing else about that step is capable of losing or inventing information.

Squaring is a different operation with a different weak point. It is not reversible in general, because squaring $A=B$ factors as

$A^2-B^2=(A-B)(A+B)=0$

which is satisfied by both $A=B$ and $A=-B$, and there is no way to see from $A^2$ alone which of the two signs A actually had. The information squaring discards is the sign of each side, not the location of a zero denominator.

Those are two different kinds of information, so a check built for one cannot substitute for the other. The domain check for a radical equation, $6x-14\ge0$, only certifies that a radicand is nonnegative; it says nothing about which sign the two sides end up with once evaluated, which is exactly the piece squaring erased. Only substituting into the original equation reads off the sign directly, which is why it is the one check that works for both kinds of equation.

Cube both sides:

$2x+7=(-3)^3=-27$

Solving the linear equation gives $2x=-34$, so $x=-17$.

No separate check for extraneous roots is needed because cubing both sides of $A=B$ is a reversible step: for real numbers, $a^3=b^3$ holds exactly when $a=b$, with no second branch the way squaring produces $A=-B$ as well. So the cubed equation $2x+7=-27$ says exactly what the original equation said, no more and no less, and whatever solves one solves the other.

Confirming anyway: $\sqrt[3]{2(-17)+7}=\sqrt[3]{-27}=-3$, since a cube root reports the sign of its radicand. The negative right side caused no trouble, because an odd-index root is allowed to be negative.

The left side, $\sqrt{2x+7}$, is a principal square root. By definition, a principal square root names the nonnegative root, whatever nonnegative number it turns out to be, for every x that keeps the radicand nonnegative:

$\sqrt{2x+7}\ge0$

The right side is the fixed number $-3$, which is negative. A nonnegative quantity can never equal a negative one, so no value of x, whatever it does to the radicand, can make the two sides agree. The equation has no solution, and no squaring, no candidate list, and no check is needed to see that: the impossibility is visible from the equation as written.

Isolate one cube root:

$\sqrt[3]{x+2}=-\sqrt[3]{x-6}$

Cube both sides once. On the right, $\left(-\sqrt[3]{x-6}\right)^3=-(x-6)$, since cubing a negative keeps the sign:

$x+2=-(x-6)=-x+6$

Solving gives $2x=4$, so $x=2$.

Check in the original equation:

$\sqrt[3]{2+2}+\sqrt[3]{2-6}=\sqrt[3]4+\sqrt[3]{-4}=\sqrt[3]4-\sqrt[3]4=0$

using $\sqrt[3]{-4}=-\sqrt[3]4$, which holds because an odd-index root carries the sign of its radicand. The two terms cancel exactly, so x=2 checks.

This cubing step needed no extraneous-root check, for the same reason part A's did not: cube both sides of $A=B$, whatever A and B happen to be built from, and the reversible fact $a^3=b^3\iff a=b$ still applies to the whole expressions standing on each side. Nothing about A being $\sqrt[3]{x+2}$ and B being $-\sqrt[3]{x-6}$ changes that. The verification above confirms the arithmetic rather than filtering out an impostor, because this equation never had one to filter.