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Square Roots: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Nested symbols

Find the value of $\sqrt{\sqrt{625}}$ .
1. Hint 1
  The outer radical acts on the value produced by the inner radical.
2. Hint 2
  Find the zero or positive number whose square is $625$ , then take the root of that result.
Answer

$5$ .

Full solution

The inner radical is the positive square root of $625$ .
$25^2=625$ $\sqrt{625}=25$

Now evaluate the outer radical.
$\sqrt{25}=5$

Answer

$5$ .

Key idea

For nested radicals, the inside value becomes the radicand of the outside radical.
Problem 1: Mark complete
Problem 2 A number under the bar

Fill the box with a positive whole number so that $\sqrt{\frac{245}{\square}}=\frac74$ .
1. Hint 1
  The number under the radical must be the square of the stated root.
2. Hint 2
  Square the fraction on the right, then compare its numerator with $245$ .
3. Hint 3
  The numerator $245$ is a whole-number multiple of $49$ , so scale the denominator by the same factor.
Answer

$80$ .

Full solution

Square the stated positive root to find the required radicand.
$\left(\frac74\right)^2=\frac{49}{16}$

The numerator $245$ is five times $49$ .
$245\div49=5$

The denominator must also be five times as large to preserve the fraction.
$5\times16=80$

The box is $80$ , a positive whole number.
Checking the fraction gives $245/80=49/16$ , whose positive square root is $7/4$ .

Answer

$80$ .

Key idea

Squaring a specified root determines its radicand, which can then be rewritten as an equivalent fraction.
Problem 2: Mark complete
Problem 3 A single symbol

Write $5\sqrt{13}$ as a single square root with a whole-number radicand and no factor outside the radical.
1. Hint 1
  The outside factor must be represented by a square root before the two roots can be combined.
2. Hint 2
  The positive number $5$ is the square root of its square.
Answer

$\sqrt{325}$ .

Full solution

Write the outside factor as a square root.
$5=\sqrt{25}$

Combine the positive radicands as a product.
$5\sqrt{13}=\sqrt{25\times13}$

The whole-number radicand is
$25\times13=325$

The requested expression is $\sqrt{325}$ .

Answer

$\sqrt{325}$ .

Key idea

Moving a positive factor inside a square root puts its square into the radicand.
Problem 3: Mark complete
Problem 4 A poster board

A square poster of area $0.81$ square meters is centered on a square board of area $1.21$ square meters, leaving a border of the same width on all four sides. How wide is that border?
1. Hint 1
  A square has equal side lengths, so its area is the square of one edge length.
2. Hint 2
  Find the edge length of the poster and the edge length of the board from their areas.
3. Hint 3
  Across the board, the extra width is shared by two borders, one on each side of the poster.
Answer

$0.1$ meters.

Full solution

Each edge is the positive square root of its square's area.
$0.9^2=0.81$ $1.1^2=1.21$

So the poster edge is $0.9$ meters and the board edge is $1.1$ meters.
Across the board, the poster edge and two equal borders make up the board edge.
$1.1-0.9=0.2$

That $0.2$ meters is split between the two borders on opposite sides of the poster.
$0.2\div2=0.1$

Each border is $0.1$ meters wide.

Answer

$0.1$ meters.

Key idea

Side lengths found from square areas can then be combined like any other lengths.
Problem 4: Mark complete
Problem 5 Three radical values

Find the exact value of $\frac{\sqrt{14}\times\sqrt{35}}{\sqrt{10}}$ .
1. Hint 1
  The product and quotient rules can combine these roots into one radical.
2. Hint 2
  Simplify the fraction inside the combined radical before finding its root.
3. Hint 3
  Before multiplying out, look for a factor that $14$ or $35$ shares with $10$ .
Answer

$7$ .

Full solution

Combine the product in the numerator and the positive denominator under one radical.
$\frac{\sqrt{14}\sqrt{35}}{\sqrt{10}}=\sqrt{\frac{14\times35}{10}}$

Cancel before multiplying, since $10=2\times5$ , with $2$ a factor of $14$ and $5$ a factor of $35$ .
$\frac{14\times35}{10}=7\times7$
The radicand is $49$ , a perfect square.

Take the positive root.
$\sqrt{49}=7$

Answer

$7$ .

Key idea

Combining products and quotients of roots can reveal a perfect square.
Problem 5: Mark complete
Problem 6 A ruler marker

Two marks lie $8$ cm and $9$ cm from the start of a ruler. A marker lies $\sqrt{74}$ cm from the start. Which mark is closer to the marker? Explain.
1. Hint 1
  First locate the marker between the two whole-number marks.
2. Hint 2
  The point halfway between the marks is $8.5$ cm from the start.
3. Hint 3
  Compare $8.5^2$ with the radicand to decide on which side of the halfway point the marker lies.
Answer

The $9$ cm mark is closer.

Full solution

The surrounding perfect squares locate the marker.
$8^2=64$ $9^2=81$

Since $64<74<81$ , the root lies between $8$ and $9$ .
The halfway point is $8.5$ .
$8.5^2=72.25$

This is less than $74$ , so
$\sqrt{74}>8.5.$
The marker lies in the upper half of the interval and is closer to the $9$ cm mark.

Answer

The $9$ cm mark is closer.

Key idea

To compare distances from a root to neighboring whole numbers, test the square of their halfway point.
Problem 6: Mark complete
Problem 7 A measured cord

A cord is exactly $\sqrt{684}$ cm long. It is cut into three equal pieces. Give the exact length of one piece, with the largest perfect-square factor taken out of the radical.
1. Hint 1
  Divide the full length by the number of equal pieces.
2. Hint 2
  Look for a perfect-square factor in the radicand before simplifying the division.
3. Hint 3
  The perfect square $36$ is a factor of $684$ .
Answer

$2\sqrt{19}$ cm.

Full solution

The radicand contains the perfect-square factor $36$ .
$684=36\times19$
Split the radical with the product rule, and take the root of the perfect square, $\sqrt{36}=6$ .
$\sqrt{684}=\sqrt{36}\times\sqrt{19}$ $\sqrt{36}\times\sqrt{19}=6\sqrt{19}$

Divide the length by three.
Since $6\sqrt{19}$ means $6\times\sqrt{19}$ , dividing by $3$ divides the $6$ and leaves $\sqrt{19}$ as it is.
$6\div3=2$ $\frac{6\sqrt{19}}{3}=2\sqrt{19}$

The radicand $19$ has no perfect-square factor greater than $1$ , so the largest one has been taken out.
Each piece is exactly $2\sqrt{19}$ cm long.
Three such lengths give $6\sqrt{19}$ cm, which is the original length.

Answer

$2\sqrt{19}$ cm.

Key idea

Simplifying a radical can make division of an exact length straightforward.
Problem 7: Mark complete
Problem 8 A return trip

Jae says $\sqrt{a^2}=a$ for every integer $a$ . Is the claim correct? If it is not, give an integer for which it fails and the value $\sqrt{a^2}$ takes there.
1. Hint 1
  The radical names the root that is zero or positive.
2. Hint 2
  Test the claim on a few integers of each sign: square each one, then take the root.
3. Hint 3
  Compare the sign of the final result with the sign of the number you started from.
Answer

No. For example, $a=-23$ gives $\sqrt{529}=23$ , not $-23$ ; any negative integer works.

Full solution

Test a negative integer, such as $a=-23$ .
Squaring produces a positive number.
$(-23)^2=529$ $\sqrt{529}=23$

The result is $23$ , not the starting integer $-23$ , so the claim fails there.
The same happens for every negative integer, because the radical never returns a negative number.

For a positive integer, such as $23$ , and for zero, the two operations do return the input.
$\sqrt{23^2}=23$ $\sqrt{0^2}=0$

So the claim is not correct for every integer.
The radical returns the size of the input, with any negative sign removed.

Answer

No. For example, $a=-23$ gives $\sqrt{529}=23$ , not $-23$ ; any negative integer works.

Key idea

The square root of an integer squared gives its size, which equals the integer itself when it is zero or positive.
Problem 8: Mark complete
Problem 9 An unfinished decimal

A student describes $\sqrt{17}$ as a fraction whose decimal has not been finished yet. Is that description possible? Explain.
1. Hint 1
  A decimal may go on forever even when it represents a fraction, but then a pattern repeats.
2. Hint 2
  Decide whether $17$ is a perfect square, then recall what that tells you about the decimal of its root.
3. Hint 3
  A decimal cut off after a few places, such as $4.123=\frac{4123}{1000}$ , is itself a fraction; ask what every fraction's decimal must do, and whether the decimal of $\sqrt{17}$ does it.
Answer

No. $\sqrt{17}$ is irrational: its decimal neither ends nor repeats, and no fraction equals it.

Full solution

A fraction has a decimal that terminates or eventually repeats.

No whole number squares to $17$ , since the perfect squares jump from $16$ to $25$ .
$4^2=16$ $5^2=25$
So $17$ is not a perfect square, and $\sqrt{17}$ is irrational: its decimal neither terminates nor repeats.

No fraction has a decimal like that, so $\sqrt{17}$ is not any fraction, with its decimal finished or not.

A decimal cut off after a few places is only a rounded value, not the exact root.
For example, $4.123$ is the fraction $\frac{4123}{1000}$ , and its square misses $17$ .
$4.123^2=16.999129$
The exact value stays $\sqrt{17}$ .

Answer

No. $\sqrt{17}$ is irrational: its decimal neither ends nor repeats, and no fraction equals it.

Key idea

An irrational root has a decimal that neither terminates nor repeats, so no fraction gives it exactly.
Problem 9: Mark complete
Problem 10 A counting claim

Rosa says exactly $22$ whole numbers $n$ make $\sqrt n$ lie strictly between $11$ and $12$ , and that every one of those roots is irrational. Is she correct about both parts? Explain.
1. Hint 1
  A bigger number has a bigger root, so a condition on $\sqrt n$ can be turned into a condition on $n$ .
2. Hint 2
  Square the two bounds, then count the whole numbers strictly between the results, with neither end included.
3. Hint 3
  Ask whether any perfect square can lie between the squares of two neighboring whole numbers.
Answer

Yes. The numbers are $n=122$ to $n=143$ , twenty-two in all, and each of their roots is irrational.

Full solution

Since a bigger number has a bigger root, $\sqrt n$ lies strictly between $11$ and $12$ exactly when $n$ lies strictly between their squares.
$11^2=121$ $12^2=144$

So $n$ runs from $122$ to $143$ .
Counting from $122$ through $143$ includes both ends.
$143-122+1=22$

The next perfect square after $121$ is $144$ , because $11$ and $12$ are neighboring whole numbers.
So none of $122$ to $143$ is a perfect square.

The square root of a whole number that is not a perfect square is irrational, so all $22$ roots are irrational.
Both parts of Rosa's statement are correct.

Answer

Yes. The numbers are $n=122$ to $n=143$ , twenty-two in all, and each of their roots is irrational.

Key idea

Squaring the bounds turns a condition on a root into a range of radicands, and every whole number strictly between two neighboring perfect squares has an irrational root.
Problem 10: Mark complete

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