Radicals and Simplification

Problem

Every answer in 3.3 came out whole because every base was a perfect power. Neither

24 5^{1 / 2}

nor

5^{1 / 2}

is a whole number. But power of a product from 3.1 reads in both directions, so two half powers can merge into one. Combine

24 5^{1 / 2} \cdot 5^{1 / 2}

into a single power and enter its value.

Show a hint

The law that splits $(a b)^{1 / 2}$ into $a^{1 / 2} b^{1 / 2}$ also runs right to left. Two half powers of different bases can become one half power.
$245 \cdot 5 = 1225$ . What number squares to $1225$ ?

Show the full solution

Merging the half powers gives

24 5^{1 / 2} \cdot 5^{1 / 2} = (245 \cdot 5)^{1 / 2} = 122 5^{1 / 2}

, and

3 5^{2} = 1225

, so the value is

35

. Two numbers that are not whole on their own multiplied to a number that is. This lesson runs on that move.

The product came out whole, but $24 5^{1 / 2}$ on its own did not go away. It is a genuine number, somewhere between $15$ and $16$ since $1 5^{2} = 225$ and $1 6^{2} = 256$ , and no whole number or fraction squares to exactly $245$ . A number like that cannot be evaluated into anything simpler. It needs a way to be written.

Problem

3.3 settled which roots accept a negative base, and the new symbol changes none of it. Exactly one of

\sqrt{- 1000}

and

\sqrt[3]{- 1000}

names a real number. Enter its value.

Show a hint

An even root needs a nonnegative radicand. An odd root takes any sign.
What number cubed gives $- 1000$ ?

Show the full solution

(- 10)^{3} = (- 10) (- 10) (- 10) = - 1000

, so

\sqrt[3]{- 1000} = - 10

. No real number squares to

- 1000

, since squares are never negative, so

\sqrt{- 1000}

names nothing at all.

$\sqrt{41}$ sits between $6$ and $7$ , since $6^{2} = 36$ and $7^{2} = 49$ , and no fraction squares to $41$ . The symbol is still exact. The square of $\sqrt{41}$ is exactly $41$ , while any decimal you could type squares to nearly $41$ and misses. So $\sqrt{41}$ is not an unfinished computation. It is the answer, and it stays written.

Problem

The rule claims

\sqrt{64 \cdot 49} = \sqrt{64} \cdot \sqrt{49}

. Test it the 1.4 way. The right side is quick. Confirm the left side agrees, then enter the shared value.

Show a hint

Work each side on its own. The claim is true only if the two values match.
$\sqrt{64} \cdot \sqrt{49} = 8 \cdot 7$ . Square that product and compare it with $64 \cdot 49$ .

Show the full solution

\sqrt{64} \cdot \sqrt{49} = 8 \cdot 7 = 56

, and

64 \cdot 49 = 3136

with

5 6^{2} = 3136

, so

\sqrt{3136} = 56

as well. Both sides land on

56

Problem

The factorization is handed to you this time. Use

63 = 9 \cdot 7

to simplify

\sqrt{63}

to the form

a \sqrt{7}

. Enter

a

Show a hint

The product rule splits $\sqrt{63}$ across the two factors you were given.
$\sqrt{9 \cdot 7} = \sqrt{9} \cdot \sqrt{7}$ , and one of those two is a whole number.

Show the full solution

\sqrt{63} = \sqrt{9} \sqrt{7} = 3 \sqrt{7}

. The perfect-square factor

9

leaves the radical as

3

, and the

7

stays inside.

Problem

Simplify

\sqrt{592}

a \sqrt{b}

with

b

as small as possible. A perfect square bigger than

4

divides

592

. Enter

b

Show a hint

Pulling out the first square you spot is not always the last word. Check what is left under the radical.
Splitting off $4$ gives $2 \sqrt{148}$ , and $148$ still has a perfect-square factor.

Show the full solution

592 = 16 \cdot 37

, so

\sqrt{592} = \sqrt{16} \sqrt{37} = 4 \sqrt{37}

and

b = 37

. Stopping at

2 \sqrt{148}

is the common miss.

148 = 4 \cdot 37

still has a square factor, so the job was only half done.

Factor

3204

into primes and the pairs are visible,

3204 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 89

. Each pair leaves the radical as a single copy, the

2

s as one

2

and the

3

s as one

3

, while the unpaired

89

stays inside. Out front

2 \cdot 3 = 6

, so

\sqrt{3204} = 6 \sqrt{89}

Problem

Two pairs of the same prime this time.

4941 = 3^{4} \cdot 61

. Pair the primes to simplify

\sqrt{4941}

a \sqrt{b}

with

b

as small as possible. Enter

a

Show a hint

$3^{4}$ is two separate pairs of $3$ , and each pair leaves the radical as one copy.
The two exited $3$ s multiply together out front. Only the $61$ stays inside.

Show the full solution

3^{4} = 3^{2} \cdot 3^{2}

is two pairs, and each pair leaves the radical as a

3

, so

\sqrt{4941} = 3 \cdot 3 \cdot \sqrt{61} = 9 \sqrt{61}

. One pair at a time also works,

\sqrt{4941} = 3 \sqrt{549}

and

549 = 9 \cdot 61

, it just takes two steps.

Problem

Not every radical can be simplified. Of

\sqrt{66}

\sqrt{85}

\sqrt{110}

, and

\sqrt{348}

, how many can be? Enter the count.

Show a hint

A radical simplifies exactly when some prime repeats in the radicand's factorization.
Factor each radicand and look for a repeated prime. Three of the four have none.

Show the full solution

66 = 2 \cdot 3 \cdot 11

85 = 5 \cdot 17

, and

110 = 2 \cdot 5 \cdot 11

have no repeated prime, so those three are already simplified. Only

348 = 2^{2} \cdot 3 \cdot 29

has a pair, giving

2 \sqrt{87}

, so the count is

1

. A squarefree radicand is not stuck, it is finished.

Problem

Evaluate

\sqrt{\frac{490}{810}}

exactly, reducing the fraction under the radical first. Enter the value as a fraction in lowest terms.

Show a hint

Simplify the fraction before touching the radical.
Divide the top and the bottom by $10$ . Both results are perfect squares.

Show the full solution

\frac{490}{810} = \frac{49}{81}

, so

\sqrt{\frac{490}{810}} = \sqrt{\frac{49}{81}} = \frac{\sqrt{49}}{\sqrt{81}} = \frac{7}{9} .

Rooting first leaves

\sqrt{490}

over

\sqrt{810}

, and neither is a whole number. Reducing first made both perfect squares.

Problem

Combine

\sqrt{53} \cdot \sqrt{212}

under one radical. Neither factor alone is a whole number, but the product is. Enter the whole-number value.

Show a hint

One radical first, then look for pairs among the prime factors.
$212 = 4 \cdot 53$ , so the combined radicand is $4 \cdot 5 3^{2}$ .

Show the full solution

\sqrt{53} \cdot \sqrt{212} = \sqrt{53 \cdot 212} = \sqrt{4 \cdot 5 3^{2}} = 2 \cdot 53 = 106

. Two numbers with no exact decimal form multiplied to a plain whole number, which is why exact forms are worth keeping until the end.

Problem

\sqrt{204} + \sqrt{459}

looks like a sum of unlike radicals. Simplify each term, then enter the coefficient of

\sqrt{51}

in the sum.

Show a hint

Neither term is simplified yet, and likeness only shows after both are.
$204 = 4 \cdot 51$ and $459 = 9 \cdot 51$ .

Show the full solution

\sqrt{204} = 2 \sqrt{51}

and

\sqrt{459} = 3 \sqrt{51}

, so the sum is

2 \sqrt{51} + 3 \sqrt{51} = 5 \sqrt{51}

. The two radicals were like all along. Simplifying is what made it visible.

Problem

Compute

\sqrt{2303} - \sqrt{752} - \sqrt{423}

exactly. No decimals and no estimating. The exact value is a number you can type. Enter it.

Show a hint

Three ugly radicals can all be multiples of one small radical. Simplify each term first.
Each radicand is a perfect square times $47$ .

Show the full solution

2303 = 49 \cdot 47

752 = 16 \cdot 47

, and

423 = 9 \cdot 47

, so the expression is

7 \sqrt{47} - 4 \sqrt{47} - 3 \sqrt{47} = (7 - 4 - 3) \sqrt{47} = 0

. The surprise is the point. Simplifying can collapse a whole expression to nothing.

Problem

A cube root to close.

344 = 2^{3} \cdot 43

. Simplify

\sqrt[3]{344}

a \sqrt[3]{b}

with

b

as small as possible. Enter

b

Show a hint

Under a cube root a prime needs three copies to leave, not two.
The triple of $2$ s leaves as a single $2$ . Count what has no triple.

Show the full solution

The triple of

2

s leaves the radical as one

2

, so

\sqrt[3]{344} = 2 \sqrt[3]{43}

and

b = 43

. Pulling out a mere pair of

2

s is the square-root habit misfiring, since

4

is not a perfect cube.

Look at what this lesson actually did. Radicals were multiplied, split apart, and added when like. Every division came out clean, because every fraction under a radical reduced to perfect squares. 3.5 faces the division that does not come out clean. Its first answers are correct but awkward to work with, and they get a cleanup rule of their own.

Practice these ideas

Practice

Evaluate

\sqrt{4900}

Show the solution

4900 = 49 \cdot 100

, so

\sqrt{4900} = \sqrt{49} \sqrt{100} = 7 \cdot 10 = 70

. The product rule turns one big square root into two easy ones.

Practice

Enter the exact value of

(\sqrt{83})^{2}

Show the solution

\sqrt{83}

is the nonnegative number whose square is

83

, so

(\sqrt{83})^{2} = 83

. That is the symbol's whole job. In exponent form,

(8 3^{1 / 2})^{2} = 8 3^{1}

says the same thing.

Practice

Exactly one of

\sqrt{- 512}

and

\sqrt[3]{- 512}

names a real number. Enter its value.

Show the solution

(- 8)^{3} = - 512

, so

\sqrt[3]{- 512} = - 8

. No real number squares to

- 512

, so

\sqrt{- 512}

names nothing, while an odd root passes a negative sign straight through.

Practice

A student answers

- 34

for

\sqrt{1156}

, and the check

(- 34)^{2} = 1156

comes out right. The answer is still wrong. Enter the value of

\sqrt{1156}

Show the solution

3 4^{2} = 1156

, so

\sqrt{1156} = 34

. Both

34

and

- 34

square to

1156

, and the radical sign names the nonnegative one, the same convention

115 6^{1 / 2}

carried in 3.3. The other candidate is written

- \sqrt{1156}

Practice

A student claims that

\sqrt{a} + \sqrt{b} = \sqrt{a + b}

. Test the claim the 1.4 way at

a = 9

and

b = 16

, then enter the true value of

\sqrt{9} + \sqrt{16}

Show the solution

\sqrt{9} + \sqrt{16} = 3 + 4 = 7

. The claim would need

\sqrt{9 + 16} = \sqrt{25} = 5

to equal

7

, and it does not. The product rule splits multiplication across a radical, but it has no addition twin.

Practice

Simplify

\sqrt{152}

to the form

a \sqrt{38}

. Enter

a

Show the solution

152 = 4 \cdot 38

, so

\sqrt{152} = \sqrt{4} \sqrt{38} = 2 \sqrt{38}

Practice

Simplify

\sqrt{142}

a \sqrt{b}

with

b

as small as possible. Enter

b

Show the solution

142 = 2 \cdot 71

has no repeated prime, so nothing comes out.

\sqrt{142}

is already

1 \cdot \sqrt{142}

, and

b = 142

. A squarefree radicand means the radical is finished, not that you missed a trick.

Practice

Simplify

\sqrt{585}

a \sqrt{b}

with

b

as small as possible. Enter the product

a \cdot b

, a times b, as a single number.

Show the solution

585 = 9 \cdot 65

, so

\sqrt{585} = 3 \sqrt{65}

, and

a \cdot b = 3 \cdot 65 = 195

. Since

65 = 5 \cdot 13

has no repeated prime,

b

is as small as it can get.

Practice

Evaluate

\sqrt{\frac{96}{150}}

exactly. Enter a fraction in lowest terms.

Show the solution

\frac{96}{150} = \frac{16}{25}

, so

\sqrt{\frac{96}{150}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5}

4 / 5

. Rooting

96

and

150

separately leaves a mess that the one reduction avoids.

Practice

\sqrt{624} - \sqrt{351}

simplifies all the way down to a single radical

\sqrt{b}

. Enter

b

Show the solution

\sqrt{624} = 4 \sqrt{39}

and

\sqrt{351} = 3 \sqrt{39}

, so the difference is

4 \sqrt{39} - 3 \sqrt{39} = 1 \sqrt{39}

and

b = 39

. Subtracting radicands would give

\sqrt{273}

, which is wrong, since no rule pushes subtraction under a radical.

Practice

For positive

y

\sqrt{y^{194}} = y^{k}

. Enter

k

Show the solution

\sqrt{y^{194}} = (y^{194})^{1 / 2} = y^{194 \cdot \frac{1}{2}} = y^{97}

, so

k = 97

. Halving the exponent works whenever the base is positive and the exponent is even.

Practice

Evaluate

(\sqrt{46})^{4}

Show the solution

(\sqrt{46})^{4} = (4 6^{1 / 2})^{4} = 4 6^{2} = 2116

. Grouping as

((\sqrt{46})^{2})^{2} = 4 6^{2}

says the same thing without exponent notation.

Practice

Combine

\sqrt{657} \cdot \sqrt{146}

under one radical, then simplify to

a \sqrt{b}

with

b

as small as possible. Enter

a

Show the solution

657 \cdot 146 = 3^{2} \cdot 7 3^{2} \cdot 2

, so

\sqrt{657} \sqrt{146} = 3 \cdot 73 \cdot \sqrt{2} = 219 \sqrt{2}

. Multiplying out to

95922

first also works, but the pairs are easier to see before the multiplication buries them.

Practice

Evaluate

\sqrt{\frac{126}{686}}

exactly. Enter a fraction in lowest terms.

Show the solution

\frac{126}{686} = \frac{9}{49}

, so the root is

\frac{\sqrt{9}}{\sqrt{49}} = \frac{3}{7}

3 / 7

. Neither

126

nor

686

is a perfect square on its own, and the single reduction is what makes both layers come out clean.

Practice

Simplify

\sqrt[3]{472}

a \sqrt[3]{b}

with

b

as small as possible. Enter

b

Show the solution

472 = 2^{3} \cdot 59

, and the triple of

2

s exits as one

2

, so

\sqrt[3]{472} = 2 \sqrt[3]{59}

and

b = 59

. Writing

2 \sqrt[3]{118}

from

472 = 4 \cdot 118

is the square-root habit misfiring, since a pair of

2

s does nothing under a cube root.

Practice

Factor

5056

into primes to simplify

\sqrt{5056}

a \sqrt{b}

with

b

as small as possible. Enter the product

a \cdot b

, a times b, as a single number.

Show the solution

5056 = 2^{6} \cdot 79

. The three pairs of

2

s leave as

2 \cdot 2 \cdot 2 = 8

and the unpaired

79

stays, so

\sqrt{5056} = 8 \sqrt{79}

and

a \cdot b = 8 \cdot 79 = 632

. Stopping early at

4 \sqrt{316}

2 \sqrt{1264}

leaves a square inside, and

b

is only as small as possible once no prime repeats.

Practice

Simplify

\sqrt{335} \cdot \sqrt{737}

a \sqrt{b}

with

b

as small as possible. Enter the sum

a + b

, a plus b, as a single number.

Show the solution

335 \cdot 737 = 5 \cdot 67 \cdot 11 \cdot 67 = 6 7^{2} \cdot 55

, so

\sqrt{335} \sqrt{737} = 67 \sqrt{55}

, and

a + b = 67 + 55 = 122

. The shared

67

is invisible until both radicands sit under one radical, which is why combining comes first.