What a Square Root Is

Squaring a side gives the area, so $5^{2} = 25$ . This lesson flips the question. Given that a panel uses exactly 36 tiles in a square, how long was one edge? Recovering the side from the area is the square root.

Problem

A square concrete pad covers

196

square feet. Find the length of one edge in feet.

Show a hint

Squaring a side gives the area, so to get the side back you undo the square. You are hunting for the number that, times itself, lands exactly on $196$ .
The two landmarks $13 \times 13 = 169$ and $15 \times 15 = 225$ trap the answer between $13$ and $15$ . Only one whole number lives in that gap, so test it by multiplying it by itself.

Show the full solution

1 3^{2} = 169

falls short and

1 5^{2} = 225

overshoots, so the side is the one whole number in between. Check it,

14 \times 14 = 196

, so one edge is

14

feet. Squaring a side gives the area, so rooting the area gives the side back.

Problem

A video wall is built from

625

identical square panels in a perfect square. Since

2 0^{2} = 400

and

3 0^{2} = 900

, the side length is in the twenties. What is

\sqrt{625}

Show a hint

The side length is a number in the twenties whose square is $625$ . Try squaring a value right in the middle of that range and see how close you land.
Test $25$ . Compute $25 \times 25$ . If it equals $625$ exactly, then $25$ is the side length, since the square root recovers the side of a square from its area.

Show the full solution

The side is in the twenties, so try the middle of that range.

25 \times 25 = 625

exactly, so

\sqrt{625} = 25

panels along one side. Squaring two landmarks first pins a root into a small range, which leaves only a guess or two to test.

Squaring and the square root are opposite trips along one road. Squaring carries a number out to its square, like

7

49

, and the square root brings that square back to where it started,

\sqrt{49} = 7

. On a perfect square the return trip always lands on a whole number.

The radical $\sqrt{36}$ names a single value, $6$ , the nonnegative side of a square with area $36$ . The equation $x^{2} = 36$ asks something different. It wants every number whose square is $36$ , and there are two of those, $6$ and $- 6$ . The two questions look alike on the page, so read carefully to tell which one a problem is handing you.

Problem

Both

10

and

- 10

square to

100

, so two numbers satisfy

x^{2} = 100

. But the radical

\sqrt{100}

names only one of them, the nonnegative one. Evaluate

\sqrt{100}

Show a hint

The radical does not mean "any number that squares to 100". It means the one nonnegative number whose square is 100. Of the two candidates 10 and $- 10$ , only one is zero or positive.
Ask which value is nonnegative. Since $1 0^{2} = 100$ and 10 is positive, the radical names 10, not $- 10$ .

Show the full solution

1 0^{2} = 100

and

10

is nonnegative, so

\sqrt{100} = 10

. Both

10

and

- 10

square to

100

, but the radical is defined to name only the nonnegative one, so it never reports

- 10

Problem

The equation

x^{2} = 49

asks for every number whose square is

49

. Since

7^{2} = 49

and

(- 7)^{2} = 49

, both

7

and

- 7

work. How many values of

x

satisfy

x^{2} = 49

Show a hint

Squaring a negative number gives a positive result, so do not stop at the first number you find. Try both a positive and a negative candidate and square each one.
Check $7^{2}$ and check $(- 7)^{2}$ . If both land on $49$ , then count how many separate values of $x$ you have found.

Show the full solution

7^{2} = 49

, and since a negative times a negative is positive,

(- 7)^{2} = 49

as well. Nothing else squares to

49

, so exactly

2

values work,

7

and

- 7

. The equation keeps both, while

\sqrt{49}

reports only the nonnegative

7

Problem

A depth sensor satisfies

x^{2} = 324

. Two values work:

x = 18

and

x = - 18

. Since depth is negative, which value is

x

Show a hint

Find the two numbers whose square is $324$ . Take the square root of $324$ to get the size, then remember the negative version squares to $324$ as well.
The size of $x$ is $\sqrt{324} = 18$ , so $x$ is $18$ or $- 18$ . Since $x$ measures a depth below the surface, it must be negative, which rules out $18$ .

Show the full solution

Since

1 8^{2} = 324

, the two numbers that square to

324

are

18

and

- 18

. A depth below the surface is negative, so

x = - 18

. Squaring erases the sign, so the equation alone leaves both candidates and the story decides which one you keep.

Problem

Squaring any real number gives a result that is zero or positive. So how many real values of

x

satisfy

x^{2} = - 9

Show a hint

Try a few numbers. $3^{2} = 9$ , $(- 3)^{2} = 9$ , $0^{2} = 0$ . Every square you get is $0$ or positive. Can any of them ever equal $- 9$ ?
Since squaring any real number gives a result of $0$ or more, no real number can square to a negative like $- 9$ . Count how many real $x$ work, and that count is the answer.

Show the full solution

A positive squared is positive, a negative squared is positive, and

0^{2} = 0

. That covers every real number, so a square is always

0

or greater and nothing can reach

- 9

. The count of real solutions is

0

The same number line, read two ways. The equation

x^{2} = 36

lights up both

- 6

and

+ 6

, since squaring either one gives

36

. The radical

\sqrt{36}

is defined to name only the nonnegative root, so it points at the single value

6

Problem

Evaluate

\sqrt{(- 12)^{2}}

. Square first, then take the root. What whole number do you get?

Show a hint

Follow the order of operations and square first. What is $(- 12)^{2}$ ? Remember that a negative times a negative gives a positive result, so the inside becomes a positive number before you ever touch the radical.
Once the inside is $144$ , you just need $\sqrt{144}$ . Find the nonnegative number whose square is $144$ . The radical always reports the nonnegative root, so the sign you started with does not survive.

Show the full solution

Square first.

(- 12)^{2} = 144

, and

\sqrt{144} = 12

, so the value is

12

. Squaring drops the sign and the radical returns the nonnegative root, which is why the answer comes out positive even though the number inside started negative.

Problem

A square garden covers

1, 296

sq ft. Notice

1, 296 = 16 \times 81

. Use

\sqrt{1, 296} = \sqrt{16} \times \sqrt{81}

to find the side length.

Show a hint

Split the work using $1, 296 = 16 \times 81$ . You can take the root of each piece on its own, since $\sqrt{16 \times 81} = \sqrt{16} \times \sqrt{81}$ .
Find each small root, then multiply them. $\sqrt{16} = 4$ because $4^{2} = 16$ , and $\sqrt{81} = 9$ because $9^{2} = 81$ . Now compute $4 \times 9$ .

Show the full solution

Since

1, 296 = 16 \times 81

, the root splits, so

\sqrt{1, 296} = \sqrt{16} \times \sqrt{81} = 4 \times 9 = 36

. Checking,

36 \times 36 = 1, 296

, so the side is

36

feet. Splitting into perfect-square factors turns one hard root into two easy ones.

A big perfect square does not have to be rooted all at once. Split

1296

into the perfect-square factors

16

and

81

, root each piece on its own to get

4

and

9

, then multiply those roots to land on

\sqrt{1296} = 4 \times 9 = 36

. The two little grids show why, a side of

4

builds an area of

16

and a side of

9

builds an area of

81

, so a side of

4 \times 9 = 36

builds the full area

1296

. And the check holds,

36 \times 36 = 1296

Problem

A classmate claims

\sqrt{1024} = 34

. But

3 4^{2} = 1156 > 1024

, and a number ending in 4 squares to something ending in 6, not 4. What is the correct value of

\sqrt{1024}

Show a hint

The size test says the root sits below 34, and the last-digit test says it cannot end in 4. So try a slightly smaller whole number and square it to see if you hit $1, 024$ on the nose.
You want a number whose square ends in 4. A number ending in 2 works, since $2 \times 2 = 4$ . Check $3 2^{2}$ .

Show the full solution

The root sits below

34

, and it has to end in

2

, since only a number ending in

2

squares to something ending in

4

. That points at

32

, and

32 \times 32 = 1, 024

exactly, so

\sqrt{1024} = 32

. The size check and the last-digit check together usually leave a single candidate to test.

Problem

A square herb bed has a side of

15

feet, so

\sqrt{x} = 15

where

x

is the area. Square both sides to free

x

. What is

x

Show a hint

The variable is stuck under the radical sign. Whatever you do to one side of an equation you must do to the other, so apply the same operation to both sides that will cancel the square root.
Squaring undoes a square root. The left side $(\sqrt{x})^{2}$ collapses back to just $x$ , and the right side becomes $1 5^{2}$ . Now just compute $1 5^{2}$ .

Show the full solution

Square both sides.

{(\sqrt{x})}^{2} = 1 5^{2}

The left side collapses to

x

, and

1 5^{2} = 225

, so

x = 225

square feet. Squaring is the move that frees a variable from under a radical, and the check holds, since

\sqrt{225} = 15

Problem

Solve

\sqrt{3 x + 4} = 11

. Square both sides, then solve the resulting linear equation. What is

x

Show a hint

Squaring is the move that undoes a square root. If $\sqrt{3 x + 4} = 11$ , then squaring both sides gives $3 x + 4 = 1 1^{2}$ . Work out $1 1^{2}$ first.
Now $3 x + 4 = 121$ is just a linear equation like the ones from Chapter 6. Subtract $4$ from both sides, then divide by $3$ .

Show the full solution

Squaring both sides gives

3 x + 4 = 1 1^{2} = 121

. Subtract

4

to get

3 x = 117

, then divide by

3

, so

x = 39

. Checking,

3 \times 39 + 4 = 121

and

\sqrt{121} = 11

. Squaring clears the radical and leaves an ordinary linear equation.

Problem

How many solutions does

\sqrt{x - 2} = - 4

have?

Show a hint

The radical $\sqrt{}$ always means the nonnegative root. So whatever $x - 2$ turns out to be, can $\sqrt{x - 2}$ ever come out negative?
Try the tempting answer anyway. Squaring gives $x = 18$ , and then $\sqrt{18 - 2} = \sqrt{16} = 4$ , not $- 4$ . So even that candidate fails the check. If the one value squaring hands you does not work, count how many values are left.

Show the full solution

The radical always returns a value that is zero or positive, and the right side is

- 4

, so nothing can match. Squaring hands you

x - 2 = 16

and

x = 18

, but

\sqrt{18 - 2} = \sqrt{16} = 4

, not

- 4

, so that candidate fails too. The number of solutions is

0

. Squaring can produce answers that do not satisfy the original equation, which is why you check every candidate back.

Practice these ideas

Practice

A square garden bed covers

121

sq ft. What is the side length, in feet? Evaluate

\sqrt{121}

Show the solution

1 0^{2} = 100

is too small and

1 2^{2} = 144

is too big, so try

11

. Since

11 \times 11 = 121

, the bed is

11

feet on a side.

Practice

A square rug covers

256

sq ft. What is its side length in feet?

Show the solution

The side length is

\sqrt{256}

. Since

1 5^{2} = 225

is too small and

1 7^{2} = 289

is too big, test

16

, and

16 \times 16 = 256

. The rug is

16

feet on a side.

Practice

A square patio covers

169

sq ft. Since

1 0^{2} = 100

and

2 0^{2} = 400

, the side length is between

10

and

20

. Find

\sqrt{169}

Show the solution

The root is between

10

and

20

, and since

169

ends in

9

the root must end in

3

7

. Testing the smaller option,

13 \times 13 = 169

, so each side of the patio is

13

feet.

Practice

When you write

x^{2} = 9

, both

3

and

- 3

work. But

\sqrt{9}

names only the nonnegative one. Evaluate

\sqrt{9}

Show the solution

3 \times 3 = 9

and

3

is nonnegative, so

\sqrt{9} = 3

. The equation

x^{2} = 9

keeps both

3

and

- 3

, but the radical names only the nonnegative one.

Practice

A square garden covers

400

sq ft. What is

\sqrt{400}

Show the solution

Split into perfect-square factors, since

400 = 4 \times 100

. Then

\sqrt{400} = \sqrt{4} \times \sqrt{100} = 2 \times 10 = 20

, and squaring back gives

2 0^{2} = 400

. The garden is

20

feet on a side.

Practice

Both

9

and

- 9

square to

81

, since

9^{2} = 81

and

(- 9)^{2} = 81

. How many values of

x

satisfy

x^{2} = 81

Show the solution

9^{2} = 81

, and a negative times a negative is positive, so

(- 9)^{2} = 81

as well. Nothing else squares to

81

, so exactly

2

values work,

9

and

- 9

. Solving

x^{2} = 81

gives a pair, while evaluating

\sqrt{81}

gives just

9

Practice

The equation

x^{2} = 225

has two solutions,

15

and

- 15

. Engineers know

x

is negative. What is

x

Show the solution

Since

1 5^{2} = 225

, both

15

and

- 15

satisfy

x^{2} = 225

. The engineers know

x

is negative, so

x = - 15

. The equation keeps both roots, and the context tells you which one to report.

Practice

Squaring any real number gives zero or a positive result, never negative. How many real values of

x

satisfy

x^{2} = - 16

Show the solution

A positive squared is positive, a negative squared is positive, and

0^{2} = 0

, so

x^{2}

is never negative for a real

x

. Nothing can square to

- 16

, so the count of real solutions is

0

Practice

Squaring undoes a square root and vice versa. Evaluate

{(\sqrt{47})}^{2}

without finding the decimal value of

\sqrt{47}

Show the solution

\sqrt{47}

is the nonnegative number whose square is

47

, so squaring it hands

47

straight back, giving

{(\sqrt{47})}^{2} = 47

. No decimal is needed, since squaring and the square root undo each other for any nonnegative number.

Practice

A recorded side length of

- 8

has the wrong sign. Evaluate

\sqrt{(- 8)^{2}}

to recover the nonnegative side length.

Show the solution

Square first,

(- 8)^{2} = 64

, then take the root,

\sqrt{64} = 8

. Squaring drops the sign and the radical returns the nonnegative value, so

\sqrt{x^{2}}

gives the size rather than the original

- 8

Practice

A square hall uses

4, 900

tiles. Since

4, 900 = 49 \times 100

, use

\sqrt{4, 900} = \sqrt{49} \times \sqrt{100}

to find the number of tiles along one edge.

Show the solution

Since

4, 900 = 49 \times 100

, the root splits, so

\sqrt{4, 900} = \sqrt{49} \times \sqrt{100} = 7 \times 10 = 70

. Checking,

7 0^{2} = 4, 900

, so one edge of the hall holds

70

tiles.

Practice

A friend claims

\sqrt{289} = 17

. Does

1 7^{2} = 289

? What is

\sqrt{289}

Show the solution

17 \times 17 = 289

exactly, so the friend is right and

\sqrt{289} = 17

. The last-digit clue agrees, since a square ending in

9

can only come from a root ending in

3

7

Practice

You know

\sqrt{x} = 19

. Square both sides to find

x

. What is

x

Show the solution

Square both sides. The left side gives back

x

, and the right gives

1 9^{2} = 361

, so

x = 361

. Checking the original,

\sqrt{361} = 19

since

19 \times 19 = 361

Practice

A game requires solving

\sqrt{x + 7} = - 3

. The radical is always nonnegative. How many solutions exist?

Show the solution

The radical is always zero or positive, and

- 3

is negative, so no value of

x

can make the two sides match. The count is

0

. Squaring would give

x + 7 = 9

and

x = 2

, but

\sqrt{2 + 7} = 3

, not

- 3

, so even that candidate fails the check.

Squares and Their Roots

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