Why do extraneous solutions show up at all?

Because squaring destroys sign information. The false statement -3=3 becomes the true statement 9=9 once you square. So the squared equation can be satisfied by values that made the original two sides opposites rather than equals. Substituting each candidate into the original equation is the only reliable way to catch them.

Do I have to check every answer, even when the equation looks simple?

Checking is fast and it is the step that makes your answer complete, so do it every time. You are safest checking in the original equation, since the squared version will accept extraneous values. Equations like sqrt(x-4)=7, where the right side is a positive constant, essentially never produce extraneous answers, but the habit of checking protects you on the harder ones where a variable sits on the other side.

How do I know which side is the hypotenuse?

The hypotenuse is always opposite the right angle and is always the longest side. In a picture, find the little square marking the right angle and look straight across from it. In a word problem, the hypotenuse is usually the slanted or direct-line distance: the ladder, the wire, the diagonal, the straight path home.

Should I give the answer as a simplified radical or a decimal?

Follow the instructions in the problem. If nothing is specified, the exact simplified radical such as 2sqrt(13) is the most precise answer, and a rounded decimal like 7.2 is useful for real-world measurements. If your result continues into another calculation, keep the exact radical until the very last step so rounding error does not build up.

ALG1-9.2

Radical Equations & Pythagorean Applications

Learn to solve radical equations by isolating and squaring, spot extraneous solutions, and use the Pythagorean theorem to find a missing leg or hypotenuse.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Radical Equations & Pythagorean Applications, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Squaring and taking a square root undo each other — that single fact powers this whole lesson. When a variable is trapped under a radical sign, you can free it by isolating the radical and squaring both sides. But squaring is a slightly dangerous move: it can create answers that look fine on paper and fail when you plug them back in. Those fakes are called extraneous solutions, and checking for them is part of solving, not an optional extra.

The same square-root machinery shows up in geometry. The Pythagorean theorem,

a^2+b^2=c^2

, almost always leaves you with a squared length that you must undo with a radical. By the end of this lesson you should be able to solve an equation like

\sqrt{x+7}=x-5

, explain why one of its two algebraic answers must be thrown out, and find the missing side of a right triangle in both exact and decimal form.

Isolating the Radical and Squaring Both Sides

A radical equation is an equation in which a variable sits under a radical sign, such as

\sqrt{2x-1}=5

. The strategy is to undo the radical with its inverse operation, squaring — but only after the radical stands alone on one side.

Why the order matters: squaring a sum does not square the pieces. If you square

\sqrt{x}+3

you get

x+6\sqrt{x}+9

, which still contains a radical and is worse than what you started with. Isolate first, and the left side collapses cleanly:

(\sqrt{A})^2=A

Step	What you do	Example: $\sqrt{x-4}+6=10$
1	Get the radical alone	$\sqrt{x-4}=4$
2	Square both sides	$x-4=16$
3	Solve the result	$x=20$
4	Check in the original	$\sqrt{16}+6=10$ ✓

If a coefficient multiplies the radical, divide it out before squaring: from

3\sqrt{x}=12

go to

\sqrt{x}=4

, then

x=16

. A frequent mistake is squaring immediately to get

3x=144

, which produces

x=48

— wrong, because

(3\sqrt{x})^2=9x

, not

3x

.

Watch for equations with no solution before you do any work. Since the principal square root is never negative,

\sqrt{x+1}=-7

cannot be true for any real

x

. Squaring anyway gives

x=48

, a purely fake answer. Recognizing an impossible setup early saves you from trusting that fake.

Extraneous Solutions and Why They Appear

Squaring both sides is not a reversible move. The equation

t=-4

is false, but squaring it gives

t^2=16

, which is true for

t=4

as well. Squaring erases the sign information, so the squared equation can have solutions the original never had. Those extra answers are extraneous solutions: they solve the squared equation but not the original one.

This is exactly why checking is required. Substitute each candidate back into the original equation — not into the squared version, which will accept the impostor every time. Keep the values that make a true statement and discard the rest.

A quick way to predict trouble: whenever the equation has the form

\sqrt{\text{something}}=\text{expression with a variable}

, that expression must come out non-negative, because a principal square root is never negative. In

\sqrt{x+7}=x-5

, any candidate less than 5 is doomed.

Equation	Candidates after squaring	Check	Solution set
$\sqrt{x+7}=x-5$	$x=9$ , $x=2$	$x=2$ gives $3=-3$ , false	$x=9$
$\sqrt{2x+3}=x$	$x=3$ , $x=-1$	$x=-1$ gives $1=-1$ , false	$x=3$
$\sqrt{x-1}=-2$	$x=5$	$\sqrt{4}=2\neq-2$	no solution

Where students go wrong: they solve the quadratic correctly, list both roots, and stop. Reporting an extraneous root as a solution means the answer is incomplete, and writing "no solution" when a valid root exists is just as much a problem. Always check both, and say plainly which one you rejected and why.

The Pythagorean Theorem: Legs, Hypotenuse, and Square Roots

In a right triangle, the two sides that form the right angle are the legs (

a

and

b

) and the side opposite the right angle is the hypotenuse (

c

), always the longest side. The Pythagorean theorem states

a^2+b^2=c^2

.

The theorem is only about right triangles, and

c

is only the hypotenuse. The single most common error in this topic is plugging the hypotenuse in as a leg. If the given sides are 8 and 17 and 17 is the hypotenuse, then

8^2+b^2=17^2

, so

b^2=289-64=225

and

b=15

. Adding instead —

8^2+17^2

— gives about 18.8, a "leg" longer than the hypotenuse, which is impossible.

Missing side	Setup	Formula
Hypotenuse	both legs known	$c=\sqrt{a^2+b^2}$
Leg	hypotenuse and one leg known	$b=\sqrt{c^2-a^2}$

After you get

c^2=52

, take the square root:

c=\sqrt{52}=2\sqrt{13}\approx7.2

. Both the simplified radical (exact) and the rounded decimal (approximate) are legitimate answers; give whichever your problem asks for, and use the exact form if a later step continues the computation.

Notice that

c^2=52

technically has two square roots,

\pm 2\sqrt{13}

. We keep only the positive one because a side length cannot be negative — the same reasoning that kills extraneous solutions in radical equations. Also remember

\sqrt{a^2+b^2}\neq a+b

: for legs 3 and 4 the hypotenuse is 5, not 7.

Applying Right Triangles to Real Situations

Most Pythagorean word problems hide a right triangle inside a picture. Ladders lean against walls (the wall and ground are the legs, the ladder is the hypotenuse). Someone walks 3 blocks east then 4 blocks north, and the straight-line distance home is the hypotenuse. A television's advertised size is the diagonal of the screen rectangle, which splits that rectangle into two right triangles.

Sketch first and label. Ask one question before computing: is the unknown side across from the right angle? If yes, add the squares; if no, subtract. For a 25 foot ladder whose base sits 7 feet from a wall, the ladder is the hypotenuse, so the height reached is

\sqrt{25^2-7^2}=\sqrt{625-49}=\sqrt{576}=24

feet.

Units and rounding matter in applications. Keep every length in the same unit before squaring — mixing feet and inches produces nonsense. Round only at the end, and round in the direction the situation demands: if you need a rope to reach at least 18.4 feet, buy 19 feet, not 18.

The converse of the theorem is also useful: if

a^2+b^2=c^2

holds for the three side lengths, the triangle is right. A carpenter checking a corner measures 3 feet along one wall and 4 feet along the other; if the diagonal is exactly 5 feet, the corner is square. Sides of 6, 8, and 11 fail the test, since

36+64=100\neq121

, so that triangle is not a right triangle.

Key terms

Radical equation.: An equation in which a variable appears under a radical sign, such as $\sqrt{3x-2}=5$ .
Radicand.: The expression underneath the radical symbol; in $\sqrt{x+7}$ the radicand is $x+7$ .
Principal square root.: The non-negative square root of a number. $\sqrt{25}=5$ only, which is why $\sqrt{\;}$ of anything can never equal a negative number.
Extraneous solution.: A value that satisfies the equation obtained after squaring but makes the original equation false; it must be discarded.
Hypotenuse.: The side of a right triangle opposite the right angle; always the longest side and always $c$ in $a^2+b^2=c^2$ .
Leg.: Either of the two sides of a right triangle that meet at the right angle, labeled $a$ and $b$ .
Pythagorean theorem.: For any right triangle, $a^2+b^2=c^2$ , where $a$ and $b$ are legs and $c$ is the hypotenuse.
Converse of the Pythagorean theorem.: If three side lengths satisfy $a^2+b^2=c^2$ , then the triangle they form is a right triangle.

Worked example

Solve

\sqrt{x+7}=x-5

for all real solutions, and state which candidates (if any) are extraneous. Then use the same square-root skill to find the missing leg of a right triangle with hypotenuse 26 cm and one leg 10 cm.

Part 1. The radical is already isolated, so square both sides:

(\sqrt{x+7})^2=(x-5)^2

, which gives

x+7=x^2-10x+25

. Expand carefully —

(x-5)^2

x^2-10x+25

, not

x^2-25

.

Move everything to one side:

0=x^2-11x+18

. Factor:

0=(x-9)(x-2)

, so the candidates are

x=9

and

x=2

.

Check both in the ORIGINAL equation. For

x=9

: left side

\sqrt{9+7}=\sqrt{16}=4

, right side

9-5=4

. True, so

x=9

is a solution. For

x=2

: left side

\sqrt{2+7}=\sqrt{9}=3

, right side

2-5=-3

. Since

3\neq-3

x=2

is extraneous and is discarded. It appeared because squaring turned

3=-3

into the true statement

9=9

.

Solution:

x=9

only.

Part 2. The 26 cm side is opposite the right angle, so it is the hypotenuse

c

, and 10 cm is a leg. Use

a^2+b^2=c^2

10^2+b^2=26^2

, so

100+b^2=676

. Subtract:

b^2=576

. Take the positive square root because a length cannot be negative:

b=\sqrt{576}=24

.

The missing leg is 24 cm. Sanity check: 24 is less than the hypotenuse 26, as it must be, and

100+576=676

✓.

Practice questions

What is the complete solution set of

\sqrt{2x+3}=x

$x=3$ and $x=-1$
$x=3$ only
$x=-1$ only
No real solution

Answer: $x=3$ only

Squaring both sides gives

2x+3=x^2

, so

x^2-2x-3=0

and

(x-3)(x+1)=0

, producing candidates

x=3

and

x=-1

. Check each in the original. For

x=3

\sqrt{9}=3

✓. For

x=-1

\sqrt{1}=1

, but the right side is

-1

, and

1\neq-1

. A principal square root is never negative, so

x=-1

is extraneous. Listing both roots is the most common wrong answer here — it skips the check.

A support wire runs from the top of a 12 meter pole to a stake in the ground 9 meters from the base of the pole. How long is the wire? Then explain why you keep only the positive square root.

Answer: 15 meters. The wire is the hypotenuse: $12^2+9^2=c^2$ , so $144+81=225$ and $c=\sqrt{225}=15$ . Only the positive root is kept because $-15$ cannot describe a physical length.

The pole and the ground meet at a right angle, so they are the legs and the wire is the hypotenuse. Because the unknown is opposite the right angle, you add the squares rather than subtract. The equation

c^2=225

has two algebraic roots,

15

and

-15

, but the negative one is rejected for the same reason extraneous solutions are rejected in radical equations: it does not fit the original situation. A check confirms the answer, since 15 is longer than either leg, as a hypotenuse must be.

Solve

4+\sqrt{x-1}=2

, or explain why no solution exists.

Answer: No real solution.

Isolate the radical first: subtract 4 from both sides to get

\sqrt{x-1}=-2

. A principal square root is never negative, so no real value of

x

can make this true, and you can stop here. If you square anyway you get

x-1=4

and

x=5

, but checking gives

4+\sqrt{4}=6

, not 2 — so

x=5

is extraneous and the correct response is that there is no solution. This is why the check is part of the solving process, not an afterthought.

FAQ

Why do extraneous solutions show up at all?: Because squaring destroys sign information. The false statement $-3=3$ becomes the true statement $9=9$ once you square. So the squared equation can be satisfied by values that made the original two sides opposites rather than equals. Substituting each candidate into the original equation is the only reliable way to catch them.
Do I have to check every answer, even when the equation looks simple?: Checking is fast and it is the step that makes your answer complete, so do it every time. You are safest checking in the original equation, since the squared version will accept extraneous values. Equations like $\sqrt{x-4}=7$ , where the right side is a positive constant, essentially never produce extraneous answers, but the habit of checking protects you on the harder ones where a variable sits on the other side.
How do I know which side is the hypotenuse?: The hypotenuse is always opposite the right angle and is always the longest side. In a picture, find the little square marking the right angle and look straight across from it. In a word problem, the hypotenuse is usually the slanted or direct-line distance: the ladder, the wire, the diagonal, the straight path home.
Should I give the answer as a simplified radical or a decimal?: Follow the instructions in the problem. If nothing is specified, the exact simplified radical such as $2\sqrt{13}$ is the most precise answer, and a rounded decimal like 7.2 is useful for real-world measurements. If your result continues into another calculation, keep the exact radical until the very last step so rounding error does not build up.

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