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$(a^{m})^{n} = a^{m \cdot n}$

a^{n \cdot \frac{1}{n}}

$a \cdot \frac{1}{a} = 1$

a^{1}

$a^{1} = a$

a

Therefore,

n a^{n}

is equal to

a

a

is non-negative.

$a < 0$

In case of a negative value of $a,$ there are also two cases two consider.

$n$ is even
$n$ is odd

Even $n$

Recall that a root with an even index

n_{e}

is defined only for non-negative numbers. Although

a

is negative,

a^{n_{e}}

is positive. Also, a power with a negative base and an even exponent can be rewritten as a power with a positive base.

a^{n_{e}} = (- a)^{n_{e}}

Now, since

- a

is positive, the Power of a Power Property for Rational Exponents can be applied again to simplify

n_{e} a^{n_{e}} .

n_{e} a^{n_{e}}

▼

Simplify

NegBaseToPosPow

$(- a)^{n_{e}} = a^{n_{e}}$

n_{e} (- a)^{n_{e}}

RootToPowD

$n a = a^{\frac{1}{n}}$

((- a)^{n_{e}})^{\frac{1}{n _{e}}}

$(a^{m})^{n} = a^{m \cdot n}$

(- a)^{n_{e} \cdot \frac{1}{n _{e}}}

$a \cdot \frac{1}{a} = 1$

(- a)^{1}

$a^{1} = a$

- a

Odd $n$

A root with an odd index

n_{o}

is defined for all real numbers. By the definition of the

n^{th}

root, the expression

n_{o} a^{n_{o}}

is the number

y

that, when multiplied by itself

n_{o}

times, will result in

a^{n_{o}} .

n_{o} times y \cdot y \cdot \dots \cdot y = a^{n_{o}}

Because

n_{o}

is odd and

a

is negative,

a^{n_{o}}

is also negative. This means that the best candidate for

y

is simply

a .

n a^{n} = a

Summary

If $a$ is non-negative, $n a^{n}$ is always equal to $a .$ However, in case of negative $a,$ the value of $n a^{n}$ depends on the parity of $n .$

	$a \geq 0$	$a < 0$
Even $n$	$n a^{n} = a$	$n a^{n} = - a$
Odd $n$	$n a^{n} = a$	$n a^{n} = a$

To conclude, for odd values of $n,$ the expression $n a^{n}$ is equal to $a .$ On the other hand, if $n$ is even, $n a^{n}$ can be written as $∣ a ∣ .$

n a^{n} = {∣ a, ∣ a ∣, if n is odd if n is even

Extra

Simplifying

n a^{m}

Depending on the index of the root and the power in the radicand, simplifying $n a^{m}$ may be problematic. Because real $n^{th}$ roots with an even index are defined only for non-negative numbers, the absolute value is sometimes needed.

If $n$ is even, $n a^{m}$ is defined only for non-negative $a^{m} .$

The following applet presents a decision tree to simplify

n a^{m} .

In this applet,

k

is the greatest common factor of

m

and

n .

Consider example

n^{th}

roots

n a^{m}

that can be simplified by using the decision tree.

$a$	$m$	$n$	$n a^{m}$	Simplify
$2$	$8$	$4$	$4 2^{8}$	$2^{\frac{8}{4}} = 2^{2} = 4$
$- 3$	$6$	$2$	$(- 3)^{6}$	$∣ ∣ ∣ ∣ (- 3)^{\frac{6}{2}} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ (- 3)^{3} ∣ ∣ ∣ = ∣ - 27 ∣ = 27$
$x$	$6$	$3$	$3 x^{6}$	$x^{\frac{6}{3}} = x^{2}$
$- 3$	$2$	$8$	$8 (- 3)^{2}$	$4 ∣ - 3 ∣ = 43$
$2$	$3$	$6$	$6 2^{3}$	$2$
$- 2$	$3$	$9$	$9 (- 2)^{3}$	$3 - 2$
$2$	$6$	$8$	$8 2^{6}$	$4 2^{3} = 48$
$- 2$	$6$	$8$	$8 (- 2)^{6}$	$4 ∣ - 2 ∣^{3} = 4 2^{3} = 48$

Example

Homework on Radical Expressions

In addition to physics and astronomy, Ignacio is also interested in algebra. Unfortunately, he missed one class and he cannot solve his homework on radical expressions. Help Ignacio pair each expression to its simplified form.

Hint

Write the radicand as a power. Then, use the $n^{th}$ Roots of $n^{th}$ Powers Property to simplify the radical expression.

Solution

To simplify the given expressions, the

n^{th}

Roots of

n^{th}

Powers Property will be used.

n a^{n} = {∣ a, ∣ a ∣, if n is odd if n is even

Unfortunately, only the second expression is written in the form that allows to use the property right away. Because the power and the index of the root are both

5

and

5

is an odd number, the second expression simplifies to

x - 2 .

Radical Expression	Simplified Form
$x^{2} - 4 x + 4$	$?$
$5 (x - 2)^{5}$	$x - 2$
$4 16 x^{8}$	$?$
$4 x^{2}$	$?$

Each of the remaining expressions should be rewritten as a power with the exponent equal to the index of the corresponding root. Notice that the radicand in the first expression is a perfect square trinomial.

x^{2} - 4 x + 4 = x^{2} - 2 (2) x + 2^{2}

Now, the radical expression can be simplified by writing the radicand as the square of a binomial. Because the index of the root is an even number, the result will require the absolute value.

x^{2} - 2 (2) x + 2^{2}

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

(x - 2)^{2}

$a^{2} = ∣ a ∣$

∣ x - 2 ∣

Therefore, the first expression simplifies to

∣ x - 2 ∣ .

Radical Expression	Simplified Form
$x^{2} - 4 x + 4$	$∣ x - 2 ∣$
$5 (x - 2)^{5}$	$x - 2$
$4 16 x^{8}$	$?$
$4 x^{2}$	$?$

Next, properties of exponents will be used to write the radicand

16 x^{8}

as a power with the exponent of

4 .

4 16 x^{8}

▼

Rewrite

Write as a power

4 2^{4} x^{8}

Split into factors

4 2^{4} x^{2 \cdot 4}

ProdInExponent

$a^{m \cdot n} = (a^{m})^{n}$

4 2^{4} (x^{2})^{4}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

4 (2 x^{2})^{4}

$4 a^{4} = ∣ a ∣$

∣ ∣ ∣ 2 x^{2} ∣ ∣ ∣

Since

x^{2}

is non-negative,

2 x^{2}

is also non-negative. This means that

∣ ∣ ∣ 2 x^{2} ∣ ∣ ∣

is equal to

2 x^{2} .

Radical Expression	Simplified Form
$x^{2} - 4 x + 4$	$∣ x - 2 ∣$
$5 (x - 2)^{5}$	$x - 2$
$4 16 x^{8}$	$2 x^{2}$
$4 x^{2}$	$?$

Now the final expression

4 x^{2}

will be examined. Similar to the previous expression, properties of exponents will be used to rewrite the radicand as a square.

4 x^{2}

▼

Simplify

Write as a power

2^{2} x^{2}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

(2 x)^{2}

$a^{2} = ∣ a ∣$

∣ 2 x ∣

This time, it is not known whether

2 x

is non-negative. Therefore, the absolute value is needed in this expression. Finally, all simplified forms may be written in the table.

Radical Expression	Simplified Form
$x^{2} - 4 x + 4$	$∣ x - 2 ∣$
$5 (x - 2)^{5}$	$x - 2$
$4 16 x^{8}$	$2 x^{2}$
$4 x^{2}$	$∣ 2 x ∣$

Discussion

Product and Quotient Properties of Radicals

Usually, the $n^{th}$ Roots of $n^{th}$ Powers Property is not enough to simplify radical expressions. Therefore, more properties will be presented and proven in this lesson. The first one refers to the root of a product.

Rule

Product Property of Radicals

Given two non-negative numbers $a$ and $b,$ the $n^{th}$ root of their product equals the product of the $n^{th}$ root of each number.

$n a b = n a \cdot n b,$ for $a \geq 0$ and $b \geq 0$

If $n$ is an odd number, the $n^{th}$ root of a negative number is defined. In this case, the Product Property of Radicals for negative $a$ and $b$ is also true.

Proof

First, a special case of the property will be proven. Assume that

a

b

is equal to

0 .

By the Zero Property of Multiplication, the radicand in left-hand side of the formula is equal to

0 .

\underline{Left - Hand Side} n 0 = ? \underline{Right - Hand Side} n a \cdot n b

Because

0^{n} = 0,

n^{th}

root of

0

is equal to

0 .

This means that both the left-hand side and one of the factors on the right-hand side equal

0 .

Again, by the Zero Property of Multiplication, the right-hand side is also

0 .

\underline{Left - Hand Side} 0 = \underline{Right - Hand Side} 0

Therefore, the property is true when

a

b

is equal to

0 .

Next, assume that both

a

and

b

are nonzero. Let

x,

y,

and

z

be real numbers such that

x = n a,

y = n b,

and

z = n a b .

By the definition of an

n^{th}

root, each of these numbers raised to the

n^{th}

power is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{n} = a y^{n} = b z^{n} = a b (I) (II) (III)

Next, because

b

is not equal to

0,

Equation (I) can be multiplied by

y^{n},

which is equal to

b .

x^{n} = a ⇓ x^{n} y^{n} = a y^{n}

Now, substitute Equations (II) and (III) into the obtained equation.

x^{n} y^{n} = a y^{n}

▼

Substitute values and simplify

$y^{n} = b$

x^{n} y^{n} = a b

$a b = z^{n}$

x^{n} y^{n} = z^{n}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

(x y)^{n} = z^{n}

Rearrange equation

z^{n} = (x y)^{n}

Since

x y

and

z

are of the same sign, the final equation implies that

z = x y .

z^{n} = (x y)^{n} \Rightarrow z = x y

The last step is substituting

z = n a b,

x = n a,

and

y = n b

into this equation.

z = x y \Leftrightarrow n a b = n a \cdot n b

The following property indicates how to work with roots of a quotient.

Rule

Quotient Property of Radicals

Let $a$ be a non-negative number and $b$ be a positive number. The $n^{th}$ root of the quotient $\frac{a}{b}$ equals the quotient of the $n^{th}$ roots of $a$ and $b .$

$n \frac{a}{b} = \frac{n a}{n b},$ for $a \geq 0,$ $b > 0$

If $n$ is an odd number, the $n^{th}$ root of a negative number is defined. In this case, the Quotient Property of Radicals for negative $a$ and $b$ is also true.

Proof

First, a special case of the property will be proven. Assume that

a

is equal to

0 .

Because

\frac{0}{b} = 0,

the radicand in left-hand side of the formula is equal to

0 .

\underline{Left - Hand Side} n 0 = ? \underline{Right - Hand Side} \frac{n 0}{n b}

Because

0^{n} = 0,

n^{th}

root of

0

is equal to

0 .

This means that both the left-hand side and the numerator on the right-hand side equal

0 .

Again, since dividing

0

by a nonzero number results in

0,

the right-hand side is also

0 .

\underline{Left - Hand Side} 0 = ✓ \underline{Right - Hand Side} 0

Therefore, the property is true when

a

is equal to

0 .

Next, assume that both

a

and

b

are nonzero. Let

x,

y,

and

z

be real numbers such that

x = n a,

y = n b,

and

z = n \frac{a}{b} .

By the definition of an

n^{th}

root, each of these numbers raised to the

n^{th}

power is equal to its corresponding radicand.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x^{n} = a y^{n} = b z^{n} = \frac{a}{b} (I) (II) (III)

Next, because

b

is not equal to

0,

Equation (I) can be divided by

y^{n},

which is equal to

b .

x^{n} = a ⇓ \frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

Now, substitute Equations (II) and (III) into the obtained equation.

\frac{x ^{n}}{y ^{n}} = \frac{a}{y ^{n}}

▼

Substitute values and simplify

$y^{n} = b$

\frac{x ^{n}}{y ^{n}} = \frac{a}{b}

$\frac{a}{b} = z^{n}$

\frac{x ^{n}}{y ^{n}} = z^{n}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

(\frac{x}{y})^{n} = z^{n}

Rearrange equation

z^{n} = (\frac{x}{y})^{n}

Since

\frac{x}{y}

and

z

are of the same sign, the final equation implies that

z = \frac{x}{y} .

z^{n} = (\frac{x}{y})^{n} \Rightarrow z = \frac{x}{y}

The last step is substituting

z = n \frac{a}{b},

x = n a,

and

y = n b

into this equation.

z = \frac{x}{y} \Leftrightarrow n \frac{a}{b} = \frac{n a}{n b}

Example

The Voltage in an Electric Circuit

To work on physics experiments in his astronomical observatory, Ignacio needs the right lighting for the new workstation. He has already designed a simple electric circuit for a $48 -$ watt light bulb.

A simple electric circuit representing the situation

The voltage $V$ required for a circuit is given by $V = P \cdot R .$ In this formula, $P$ is the power in watts and $R$ is the resistance in ohms.

a What voltage of battery is needed to light the light bulb if its resistance is

12

ohms?

b Does the battery need to have a higher, lower, or the same voltage in order to light a

20 -

watt light bulb with a

32 -

ohm resistance?

Hint

a Substitute the given values into the formula for the voltage

V .

Then, evaluate the expression using the Product Property of Radicals.

b Similar to Part A, evaluate the voltage needed to light the other light bulb. How is the lower bound of the obtained root calculated?

Solution

a The formula for the voltage

V

required for a circuit is given. In this equation,

P

represents the power in watts and

R

represents the resistance in ohms.

V = P \cdot R

The circuit connects the battery to a

48 -

watt light bulb with a resistance of

12

ohms. The voltage

V_{1}

needed to light the bulb is calculated by substituting

P = 48

and

R = 12

into the given formula.

V_{1} = P \cdot R

$P = 48$ , $R = 12$

V_{1} = 48 \cdot 12

▼

Evaluate right-hand side

ProdSqrt

$a \cdot b = a \cdot b$

V_{1} = 48 \cdot 12

V_{1} = 576

Calculate root

V_{1} = 24

Therefore, a

24 -

volt battery is needed to power the light bulb.

b In Part A, it was found that the voltage needed to light a

48 -

watt light bulb is

24

volts. Now, the number of volts

V_{2}

needed to light a

20 -

watt light bulb with a resistance of

32

ohms will be calculated. To do so, substitute

P = 20

and

R = 32

into the given formula and simplify.

V_{2} = P \cdot R

$P = 20$ , $R = 32$

V_{2} = 20 \cdot 32

▼

Evaluate right-hand side

ProdSqrt

$a \cdot b = a \cdot b$

V_{2} = 20 \cdot 32

V_{2} = 640

UseCalc

Use a calculator

V_{2} = 25.298221 \dots

RoundDec

Round to $1$ decimal place(s)

V_{2} \approx 25.3

This means that a battery with a higher voltage is needed to light the

20 -

watt light bulb.

Example

Calculating the Length of the Diagonal in TV Screen

Ignacio wants to organize a movie night to celebrate the grand opening of his astronomical observatory. He plans to buy a brand new TV for the occasion, but he does not know what size of TV screen will fit on his wall.

The shape of a TV screen is represented by its aspect ratio, which is the ratio of the width of a screen to its height. The most common aspect ratio for TV screens is $16 : 9,$ which means that the width of the screen is $\frac{1 6}{9}$ times its height.

a Write an expression for the width of a TV screen with this aspect ratio in terms of the length of the diagonal

d .

b Find the area of the screen if the length of the diagonal is

3033

inches.

Hint

a The length of the diagonal of a rectangle can by found by using the Pythagorean Theorem.

b Use the expression found in Part A. The area of a rectangle is calculated by multiplying its length times its width.

Solution

a A TV screen can be modeled by a rectangle. Let

w

and

h

denote the width and height of the rectangle, respectively. Since it is given that the aspect ratio of the screen is

16 : 9,

the height can be written in terms of the width.

w = \frac{1 6}{9} h ⇕ h = \frac{9}{1 6} w

Next, the height and the width of the rectangle will be labeled on a diagram. The length of the diagonal

d

will be marked in the rectangle as well.

Because two sides and the diagonal of the rectangle form a right triangle, the length of the diagonal can be related to the width by applying the Pythagorean Theorem.

d^{2} = w^{2} + (\frac{9}{1 6} w)^{2}

Now, the equation will be solved for

w .

Keep in mind that the absolute value of a positive number is equal to itself.

d^{2} = w^{2} + (\frac{9}{1 6} w)^{2}

▼

Solve for

w

PowProdII

$(a b)^{m} = a^{m} b^{m}$

d^{2} = w^{2} + (\frac{9}{1 6})^{2} w^{2}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

d^{2} = w^{2} + \frac{8 1}{2 5 6} w^{2}

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

d^{2} = w^{2} + \frac{8 1 w ^{2}}{2 5 6}

NumberToFrac

$a = \frac{2 5 6 \cdot a}{2 5 6}$

d^{2} = \frac{2 5 6 w ^{2}}{2 5 6} + \frac{8 1 w ^{2}}{2 5 6}

AddFrac

Add fractions

d^{2} = \frac{2 5 6 w ^{2} + 8 1 w ^{2}}{2 5 6}

Add terms

d^{2} = \frac{3 3 7 w ^{2}}{2 5 6}

MovePartNumRight

$\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

d^{2} = \frac{3 3 7}{2 5 6} w^{2}

MultEqn

$LHS \cdot \frac{2 5 6}{3 3 7} = RHS \cdot \frac{2 5 6}{3 3 7}$

\frac{2 5 6}{3 3 7} d^{2} = w^{2}

Rearrange equation

w^{2} = \frac{2 5 6}{3 3 7} d^{2}

$LHS = RHS$

w^{2} = \frac{2 5 6}{3 3 7} d^{2}

$a^{2} = ∣ a ∣$

∣ w ∣ = \frac{2 5 6}{3 3 7} d^{2}

ProdSqrt

$a \cdot b = a \cdot b$

∣ w ∣ = \frac{2 5 6}{3 3 7} d^{2}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

∣ w ∣ = \frac{2 5 6}{3 3 7} d^{2}

Calculate root

∣ w ∣ = \frac{1 6}{3 3 7} d^{2}

$a^{2} = ∣ a ∣$

∣ w ∣ = \frac{1 6}{3 3 7} ∣ d ∣

$∣ w ∣ = w$ $&$ $∣ d ∣ = d$

w = \frac{1 6}{3 3 7} d

b It is given that the length of the diagonal is

3033

inches. Therefore,

d = 3033

will be substituted into the equation written in Part A to calculate the width of the rectangle.

w = \frac{1 6}{3 3 7} d

$d = 3033$

w = \frac{1 6}{3 3 7} 3033

▼

Evaluate right-hand side

$\frac{a}{c} \cdot b = a \cdot \frac{b}{c}$

w = 16 \cdot \frac{3 0 3 3}{3 3 7}

DivSqrt

$\frac{a}{b} = \frac{a}{b}$

w = 16 \cdot \frac{3 0 3 3}{3 3 7}

CalcQuot

Calculate quotient

w = 16 \cdot 9

Calculate root

w = 16 \cdot 3

w = 48

The width of the rectangle is

48

inches. The area

A

of the rectangle is calculated by multiplying the rectangle's width by its height. The height will now be found by using the aspect ratio of the TV screen. Substitute

w = 48

into the formula

h = \frac{9}{1 6} w .

h = \frac{9}{1 6} w

$w = 48$

h = \frac{9}{1 6} (48)

▼

Evaluate right-hand side

$\frac{a}{c} \cdot b = a \cdot \frac{b}{c}$

h = 9 \cdot \frac{4 8}{1 6}

CalcQuot

Calculate quotient

h = 9 \cdot 3

h = 27

Finally, the area of the TV screen can be found.

A = w \cdot h

$w = 48$ , $h = 27$

A = 48 \cdot 27

A = 1296

Therefore, the area of the TV screen is

1296

square inches.

Example

The Surface Area of the Earth

Ignacio wants to decorate his observatory by hanging a model of the solar system on the ceiling. He has already bought some of the planets, which are modeled by gleaming spheres. The volume of the miniature Earth is $\frac{4 π}{3}$ cubic inches.

The volume of a sphere is given by the formula $V = \frac{4}{3} π r^{3} .$ In this formula, $r$ is the radius of the sphere. Ignacio wants to find the surface area of the model to approximate the surface area of the Earth by using the model scale.

a Using the formula for the surface area of a sphere,

S = 4 π r^{2},

represent

S

in terms of the volume

V .

Choose the correct answer written as a product of radicals.

b Calculate the surface area of the model of the Earth.

Hint

a Solve the formula for the volume of a sphere for

r .

Then, substitute the expression for

r

into the formula for the surface area of a sphere.

b Substitute the given volume into the formula written in Part A. Use the properties of radicals to calculate the surface area.

Solution

a The formulas for both the volume and the surface area of a sphere are given. Note that both quantities depend on the radius of the sphere.

Volume : Surface Area : V = \frac{4}{3} π r^{3} S = 4 π r^{2}

Therefore, the formula for the volume can be solved for

r,

which can then be substituted into the formula for the surface area. This will give the formula for the surface area in terms of the volume.

V = \frac{4}{3} π r^{3}

▼

Solve for

r

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

V = \frac{4 π}{3} r^{3}

MultEqn

$LHS \cdot \frac{3}{4 π} = RHS \cdot \frac{3}{4 π}$

\frac{3}{4 π} V = r^{3}

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

\frac{3 V}{4 π} = r^{3}

Rearrange equation

r^{3} = \frac{3 V}{4 π}

RadicalEqn

$3 LHS = 3 RHS$

3 r^{3} = 3 \frac{3 V}{4 π}

$3 a^{3} = a$

r = 3 \frac{3 V}{4 π}

Now, the rewritten expression in terms of

r

will be substituted into the formula for the surface area

S = 4 π r^{2} .

Use the properties of exponents and radicals to simplify the formula.

S = 4 π r^{2}

$r = 3 \frac{3 V}{4 π}$

S = 4 π (3 \frac{3 V}{4 π})^{2}

▼

Simplify right-hand side

RootToPowD

$n a = a^{\frac{1}{n}}$

S = 4 π ((\frac{3 V}{4 π})^{\frac{1}{3}})^{2}

$(a^{m})^{n} = a^{m \cdot n}$

S = 4 π (\frac{3 V}{4 π})^{\frac{1}{3} \cdot 2}

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

S = 4 π (\frac{3 V}{4 π})^{\frac{2}{3}}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

S = 4 π \frac{( 3 V ) ^{\frac{2}{3}}}{( 4 π ) ^{\frac{2}{3}}}

$a \cdot \frac{b}{c} = \frac{a}{c} \cdot b$

S = \frac{4 π}{( 4 π ) ^{\frac{2}{3}}} (3 V)^{\frac{2}{3}}

$a = a^{1}$

S = \frac{( 4 π ) ^{1}}{( 4 π ) ^{\frac{2}{3}}} (3 V)^{\frac{2}{3}}

DivPow

$\frac{a ^{m}}{a ^{n}} = a^{m - n}$

S = (4 π)^{1 - \frac{2}{3}} (3 V)^{\frac{2}{3}}

OneToFrac

Rewrite $1$ as $\frac{3}{3}$

S = (4 π)^{\frac{3}{3} - \frac{2}{3}} (3 V)^{\frac{2}{3}}

SubFrac

Subtract fractions

S = (4 π)^{\frac{1}{3}} (3 V)^{\frac{2}{3}}

MoveNumLeft

$\frac{a}{b} = a \cdot \frac{1}{b}$

S = (4 π)^{\frac{1}{3}} (3 V)^{2 \cdot \frac{1}{3}}

ProdInExponent

$a^{m \cdot n} = (a^{m})^{n}$

S = (4 π)^{\frac{1}{3}} ((3 V)^{2})^{\frac{1}{3}}

PowToRootD

$a^{\frac{1}{n}} = n a$

S = 3 4 π \cdot 3 (3 V)^{2}

This corresponds to option C.

b The volume of the model of the Earth is given as

\frac{4 π}{3}

cubic inches. The surface area of the model can be calculated by substituting

V = \frac{4 π}{3}

into the formula found in Part A.

S = 3 4 π \cdot 3 (3 V)^{2}

$V = \frac{4 π}{3}$

S = 3 4 π \cdot 3 (3 (\frac{4 π}{3}))^{2}

▼

Evaluate right-hand side

DenomMultFracToNumber

$3 \cdot \frac{a}{3} = a$

S = 3 4 π \cdot 3 (4 π)^{2}

$3 a \cdot 3 b = 3 a \cdot b$

S = 3 4 π (4 π)^{2}

S = 3 (4 π)^{3}

$3 a^{3} = a$

S = 4 π

Therefore, the surface area of the model Earth is

4 π

square inches.

Discussion

Irrational Conjugate of an $n^{th}$ Root

Some irrational numbers are written as expressions involving

n^{t h}

roots of powers. Previously, the Product Property of Radicals and the

n^{th}

Roots of

n^{th}

Powers Property were used to eliminate the root. Consider the following example.

3516 = 3 5 4^{2}

Given an irrational number in this form, it is possible to find another irrational number by changing the power in the radicand.

Concept

Irrational Conjugate of an $n^{th}$ Root

Let

a,

b,

and

x

be rational numbers and

n

a natural number so that the

n^{th}

root

n b^{x}

is irrational. The irrational conjugate of an

n^{th}

root

a n b^{x}

is obtained by raising

b

to the power of

n - x

in the radicand.

\underline{Number} a n b^{x} Change Power \underline{Conjugate} a n b^{n - x}

For example, the conjugate of

4 5 7^{2}

4 5 7^{3} .

4 5 7^{2} Change Power 4 5 7^{5 - 2} = 4 5 7^{3}

The irrational conjugate can be used to rationalize a denominator containing an

n^{th}

root.

In the next part of this lesson, the conjugate will be used to rationalize the denominator of a numeric expression involving an

n^{th}

root.

Pop Quiz

Determining the Irrational Conjugate of a Radical Expression

Find the irrational conjugate of $a + b c$ or the irrational conjugate of an $n^{th}$ root.

Discussion

Rationalizing a Denominator Involving an $n^{th}$ Root

When a numeric expression has a denominator that involves an irrational

n^{th}

root, it is convenient to rewrite the expression so that the denominator is a rational number. This process is known as rationalization. Consider the following example.

\frac{4}{5 2 7}

Since the denominator involves an

n^{th}

root, the fraction can be rewritten by multiplying both the numerator and denominator by the irrational conjugate of the denominator. The resulting expression can then be simplified.

Multiply by the Conjugate of the Denominator

First, the radicand should be rewritten as a power.

\frac{4}{5 2 7} = \frac{4}{5 3 ^{3}}

To get rid of the radical expression in the denominator, the numerator and the denominator are multiplied by its

irrational

conjugate .

Since the radicand involves the

third

power and the index of the root is

5,

multiply by

5 3^{5 - 3} = 5 3^{2} .

\frac{4 \cdot 5 3 ^{2}}{5 3 ^{3} \cdot 5 3 ^{2}}

Simplify the Expression

The obtained expression can now be simplified.

\frac{4 \cdot 5 3 ^{2}}{5 3 ^{3} \cdot 5 3 ^{2}}

▼

Simplify

$5 a \cdot 5 b = 5 a \cdot b$

\frac{4 \cdot 5 3 ^{2}}{5 3 ^{3} \cdot 3 ^{2}}

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

\frac{4 \cdot 5 3 ^{2}}{5 3 ^{3 + 2}}

Add terms

\frac{4 \cdot 5 3 ^{2}}{5 3 ^{5}}

$5 a^{5} = a$

\frac{4 \cdot 5 3 ^{2}}{3}

Calculate power and product

\frac{4 5 9}{3}

The denominator has now been rationalized because it contains an integer number and no roots.

Notice that this method also works when the denominator is the product of two roots with different indexes. In these cases, the method should be applied twice. However, if the denominator involves a sum of two $n^{th}$ roots with different indexes, rationalizing is a more complicated task.

Example

Calculating the Circular Velocity of a Satellite

While reading some research papers on astronomy, Ignacio came across the fact that the circular velocity

v

of a satellite circling the Earth is given by the following formula.

v = \frac{K}{r}

In this formula,

v

is measured in miles per hour,

r

is the distance in miles from the satellite to the center of the Earth, and

K

is a constant depending on the gravitational constant and the mass of the Earth.

a Rationalize the denominator of the formula for the velocity

v

of a satellite when

r

is equal to

3 26252 8^{2} .

Choose the correct answer.

b A second satellite is orbiting the Earth at a distance

(3 - 2)^{2}

times greater than the distance of the satellite from Part A. Write the velocity

v_{2}

of this second satellite in terms of

v .

Rationalize the denominator if needed.

Hint

a Rewrite the formula using the Quotient Property of Square Roots. Then, multiply both the numerator and denominator by the irrational conjugate of the

n^{th}

root.

b Write the distance from the second satellite in terms of

r .

Consider the quotient of the velocities to cancel out common factors.

Solution

a First, notice that the formula for the circular velocity

v

can be rewritten using the Quotient Property of Square Roots.

v = \frac{K}{r} = \frac{K}{r}

It is given that the distance from the Earth to the satellite is

3 26252 8^{2}

miles. Therefore,

r = 3 26252 8^{2}

will be substituted into the formula and the denominator rationalized.

v = \frac{K}{r}

$r = 3 26252 8^{2}$

v = \frac{K}{3 2 6 2 5 2 8 ^{2}}

▼

Rewrite

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 3 2 6 2 5 2 8 ^{2}}{b \cdot 3 2 6 2 5 2 8 ^{2}}$

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2}}{3 2 6 2 5 2 8 ^{2} \cdot 3 2 6 2 5 2 8 ^{2}}

MultSqrt

$a \cdot a = a$

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2}}{3 2 6 2 5 2 8 ^{2}}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 3 2 6 2 5 2 8 ^{1}}{b \cdot 3 2 6 2 5 2 8 ^{1}}$

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2} \cdot 3 2 6 2 5 2 8 ^{1}}{3 2 6 2 5 2 8 ^{2} \cdot 3 2 6 2 5 2 8 ^{1}}

$3 a \cdot 3 b = 3 a \cdot b$

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2} \cdot 3 2 6 2 5 2 8 ^{1}}{3 2 6 2 5 2 8 ^{2} \cdot 2 6 2 5 2 8 ^{1}}

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2} \cdot 3 2 6 2 5 2 8 ^{1}}{3 2 6 2 5 2 8 ^{3}}

$3 a^{3} = a$

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2} \cdot 3 2 6 2 5 2 8 ^{1}}{2 6 2 5 2 8}

Now that the denominator has been rationalized, the numerator of the obtained expression will be further simplified.

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2} \cdot 3 2 6 2 5 2 8 ^{1}}{2 6 2 5 2 8}

▼

Simplify right-hand side

SqrtToPowD

$a = a^{\frac{1}{2}}$

v = \frac{K \cdot ( 3 2 6 2 5 2 8 ^{2} ) ^{\frac{1}{2}} \cdot 3 2 6 2 5 2 8 ^{1}}{2 6 2 5 2 8}

RootToPowD

$n a = a^{\frac{1}{n}}$

v = \frac{K \cdot ( ( 2 6 2 5 2 8 ^{2} ) ^{\frac{1}{3}} ) ^{\frac{1}{2}} \cdot 3 2 6 2 5 2 8 ^{1}}{2 6 2 5 2 8}

$(a^{m})^{n} = (a^{n})^{m}$

v = \frac{K \cdot ( ( 2 6 2 5 2 8 ^{2} ) ^{\frac{1}{2}} ) ^{\frac{1}{3}} \cdot 3 2 6 2 5 2 8 ^{1}}{2 6 2 5 2 8}

$(a^{m})^{n} = a^{m \cdot n}$

v = \frac{K \cdot ( 2 6 2 5 2 8 ^{2 \cdot \frac{1}{2}} ) ^{\frac{1}{3}} \cdot 3 2 6 2 5 2 8 ^{1}}{2 6 2 5 2 8}

$a \cdot \frac{1}{a} = 1$

v = \frac{K \cdot ( 2 6 2 5 2 8 ^{1} ) ^{\frac{1}{3}} \cdot 3 2 6 2 5 2 8 ^{1}}{2 6 2 5 2 8}

$a^{1} = a$

v = \frac{K \cdot ( 2 6 2 5 2 8 ) ^{\frac{1}{3}} \cdot 3 2 6 2 5 2 8}{2 6 2 5 2 8}

PowToRootD

$a^{\frac{1}{n}} = n a$

v = \frac{K \cdot 3 2 6 2 5 2 8 \cdot 3 2 6 2 5 2 8}{2 6 2 5 2 8}

$3 a \cdot 3 b = 3 a \cdot b$

v = \frac{K \cdot 3 2 6 2 5 2 8 \cdot 2 6 2 5 2 8}{2 6 2 5 2 8}

ProdToPowTwoFac

$a \cdot a = a^{2}$

v = \frac{K \cdot 3 2 6 2 5 2 8 ^{2}}{2 6 2 5 2 8}

Now that the denominator contains only an integer number, it has been successfully rationalized. The calculated velocity corresponds to option B.

b It is given that the distance

r_{2}

of the second satellite to the center of the Earth is

(3 - 2)^{2}

times greater than the distance

r

of the first satellite considered in Part A.

r_{2} = (3 - 2)^{2} r

To write

v_{2}

in terms of

v,

start with the formula for

v_{2}

by using the given formula.

v_{2} = \frac{K}{r _{2}}

Next, the Division Property of Equality will be applied to divide both sides of the equation by

v .

This will allow the expression to be written in terms of

v .

Then,

v = \frac{K}{r}

can be substituted on the right-hand side.

\frac{v _{2}}{v} = \frac{\frac{K}{r _{2}}}{v} \Rightarrow \frac{v _{2}}{v} = \frac{\frac{K}{r _{2}}}{\frac{K}{r}}

Now, substitute

(3 - 2)^{2} r

for

r_{2}

in the obtained quotient and evaluate the right-hand side.

\frac{v _{2}}{v} = \frac{\frac{K}{r _{2}}}{\frac{K}{r}}

$r_{2} = (3 - 2)^{2} r$

\frac{v _{2}}{v} = \frac{\frac{K}{( 3 - 2 ) ^{2} r}}{\frac{K}{r}}

▼

Evaluate right-hand side

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

\frac{v _{2}}{v} = \frac{\frac{K}{( 3 - 2 ) ^{2} r}}{\frac{K}{r}}

DivFracByFracD

$\frac{a / b}{c / d} = \frac{a}{b} \cdot \frac{d}{c}$

\frac{v _{2}}{v} = \frac{K}{( 3 - 2 ) ^{2} r} \cdot \frac{r}{K}

MultFrac

Multiply fractions

\frac{v _{2}}{v} = \frac{K \cdot r}{( 3 - 2 ) ^{2} r \cdot K}

$a \cdot b = a \cdot b$

\frac{v _{2}}{v} = \frac{K \cdot r}{( 3 - 2 ) ^{2} \cdot r \cdot K}

CancelCommonFac

Cancel out common factors

\frac{v _{2}}{v} = \frac{K \cdot r}{( 3 - 2 ) ^{2} \cdot r \cdot K}

SimpQuot

Simplify quotient

\frac{v _{2}}{v} = \frac{1}{( 3 - 2 ) ^{2}}

$a^{2} = ∣ a ∣$

\frac{v _{2}}{v} = \frac{1}{∣ ∣ ∣ 3 - 2 ∣ ∣ ∣}

$∣ ∣ ∣ 3 - 2 ∣ ∣ ∣ = 3 - 2$

\frac{v _{2}}{v} = \frac{1}{3 - 2}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot ( 3 + 2 )}{b \cdot ( 3 + 2 )}$

\frac{v _{2}}{v} = \frac{1 ( 3 + 2 )}{( 3 - 2 ) ( 3 + 2 )}

IdPropMult

Identity Property of Multiplication

\frac{v _{2}}{v} = \frac{3 + 2}{( 3 - 2 ) ( 3 + 2 )}

$(a - b) (a + b) = a^{2} - b^{2}$

\frac{v _{2}}{v} = \frac{3 + 2}{3 ^{2} - ( 2 ) ^{2}}

CalcPow

Calculate power

\frac{v _{2}}{v} = \frac{3 + 2}{9 - ( 2 ) ^{2}}

$(a)^{2} = a$

\frac{v _{2}}{v} = \frac{3 + 2}{9 - 2}

SubTerm

Subtract term

\frac{v _{2}}{v} = \frac{3 + 2}{7}

MultEqn

$LHS \cdot v = RHS \cdot v$

v_{2} = \frac{3 + 2}{7} v

Finally, it can concluded that

v_{2}

is equal to

\frac{3 + 2}{7} v .

Discussion

Adding and Subtracting Like Radical Expressions and Roots

Concept

Like Radical Expressions

Radical expressions are called like radical expressions or like radicals if both the index and the radicand of the corresponding roots are identical.

Use the Distributive Property to add or subtract like radical expressions.

$a n x + b n x = (a + b) n x$
$a n x - b n x = (a - b) n x$

Rule

Adding and Subtracting Radicals

A numeric or algebraic expression that contains two or more radical terms with the same radicand and the same index — called like radical expressions — can be simplified by adding or subtracting the corresponding coefficients.

$a n x \pm b n x = (a \pm b) n x$

Here, $a, b,$ and $x$ are real numbers and $n$ is a natural number. If $n$ is even, then $x$ must be greater than or equal to zero.

Proof

n

is an even number, then

x

is greater than or equal to zero. If

n

is odd, then

x

can be any real number. In both cases,

n x

is a real number. This means that the Distributive Property can be used to factor out

n x .

a n x \pm b n x

FactorOut

Factor out $n x$

(a \pm b) n x

Method

Adding and Subtracting Radicals

Two radicals can be added or subtracted if it is possible to rewrite them as like radical expressions. Use the Distributive Property to add or subtract like radicals.

$a n x + b n x = (a + b) n x$
$a n x - b n x = (a - b) n x$

For example, consider the following radical expressions.

4 x^{5} y^{3} and 4 16 x y^{7}

In order to add these expressions, there are three steps to follow. Note that subtraction of the radicals can be performed by applying the same three steps, only instead of adding the like radicals, they will be subtracted.

Simplify the Radicals to Have the Same Index and Radicand

Start by rewriting the expressions as like radicals. To do so, split the radicands into factors and use the properties of radicals.

4 x^{5} y^{3} + 4 16 x y^{7}

▼

Rewrite

Split into factors

4 x^{5} \cdot y^{3} + 4 16 \cdot x \cdot y^{7}

WriteSum

Write as a sum

4 x^{4 + 1} \cdot y^{3} + 4 16 \cdot x \cdot y^{4 + 3}

SumInExponent

$a^{m + n} = a^{m} \cdot a^{n}$

4 x^{4} \cdot x^{1} \cdot y^{3} + 4 16 \cdot x \cdot y^{4} \cdot y^{3}

CommutativePropMult

Commutative Property of Multiplication

4 x^{4} \cdot x^{1} \cdot y^{3} + 4 16 \cdot y^{4} \cdot x \cdot y^{3}

RootProd

$4 a \cdot b = 4 a \cdot 4 b$

4 x^{4} \cdot 4 x^{1} \cdot y^{3} + 4 16 \cdot y^{4} \cdot 4 x \cdot y^{3}

$a^{1} = a$

4 x^{4} \cdot 4 x \cdot y^{3} + 4 16 \cdot y^{4} \cdot 4 x \cdot y^{3}

$4 a^{4} = ∣ a ∣$

∣ x ∣ \cdot 4 x \cdot y^{3} + 4 16 \cdot y^{4} \cdot 4 x \cdot y^{3}

Write as a power

∣ x ∣ \cdot 4 x \cdot y^{3} + 4 2^{4} \cdot y^{4} \cdot 4 x \cdot y^{3}

ProdPow

$a^{m} \cdot b^{m} = (a \cdot b)^{m}$

∣ x ∣ \cdot 4 x \cdot y^{3} + 4 (2 y)^{4} \cdot 4 x \cdot y^{3}

$4 a^{4} = ∣ a ∣$

∣ x ∣ \cdot 4 x \cdot y^{3} + ∣ 2 y ∣ \cdot 4 x \cdot y^{3}

∣ x ∣ 4 x y^{3} + ∣ 2 y ∣ 4 x y^{3}

Since both the index and the radicand of the roots are identical, they are like radical expressions.

Add or Subtract Like Radicals

Now, like radical expressions can by added by using the Distributive Property.

∣ x ∣ 4 x y^{3} + ∣ 2 y ∣ 4 x y^{3} ⇕ (∣ x ∣ + ∣ 2 y ∣) 4 x y^{3}

Simplify the Result as Necessary

The signs of the

x -

and

y -

variables are not given. This means that the expression in parentheses cannot be further simplified and the result of the addition is as follows.

(∣ x ∣ + ∣ 2 y ∣) 4 x y^{3}

Example

The Area of Triangles on a Door

The last step in designing the observatory is to come up with a new logo. Ignacio has sketched the following prototype of his logo.

The logo is made of equilateral triangles. The side of a large triangle is

6

inches and the side of a small triangle is

332

inches. Find the exact total area of the design.

Hint

How many large and small triangles form the logo? Use the formula for the area of a triangle using sine. Also, keep in mind that $sin 6 0^{\circ} = \frac{3}{2} .$

Solution

It is given that the logo is made of two types of equilateral triangles. To find the area of the design, start by determining the number of triangles of each type.

There are

6

large and

4

small triangles. Now, the area of a single triangle of each type will be found. Recall the formula for the area of a triangle involving the sine ratio. Start with the area

A_{1}

of a large triangle.

Area = \frac{1}{2} a b sin C

The length of all sides is

6

inches and the measure of all angles in the triangle is

6 0^{\circ} .

This means that

a = 6,

b = 6,

and

C = 6 0^{\circ}

can be substituted into the formula to find the area.

A_{1} = \frac{1}{2} a b sin C

SubstituteValues

Substitute values

A_{1} = \frac{1}{2} (6) (6) sin 6 0^{\circ}

▼

Evaluate right-hand side

A_{1} = \frac{1}{2} (36) sin 6 0^{\circ}

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

A_{1} = \frac{3 6}{2} sin 6 0^{\circ}

A_{1} = \frac{3 6}{2} (\frac{3}{2})

MultFrac

Multiply fractions

A_{1} = \frac{3 6 3}{2 ( 2 )}

A_{1} = \frac{3 6 3}{4}

CalcQuot

Calculate quotient

A_{1} = 93

Similarly, the area

A_{2}

of a small triangle can be calculated. In this case, the side length is

332

inches and the measure of the included angle is

6 0^{\circ}

as previously.

A_{2} = \frac{1}{2} a b sin C

SubstituteValues

Substitute values

A_{2} = \frac{1}{2} (332) (332) sin 6 0^{\circ}

▼

Evaluate right-hand side

$3 a \cdot 3 b = 3 a \cdot b$

A_{2} = \frac{1}{2} (31024) sin 6 0^{\circ}

MoveNumRight

$\frac{a}{b} = \frac{1}{b} \cdot a$

A_{2} = \frac{3 1 0 2 4}{2} sin 6 0^{\circ}

A_{2} = \frac{3 1 0 2 4}{2} (\frac{3}{2})

MultFrac

Multiply fractions

A_{2} = \frac{3 1 0 2 4 \cdot 3}{4}

In this expression, it is important not to make a mistake by using the Product Property of Radicals. The indices of roots are not equal, so they cannot be combined. Next, the total area

A

can be found by adding

6

times the area of a large triangle to

4

times the area of a small triangle.

A = 6 A_{1} + 4 A_{2}

Finally, substitute the simplified areas of a single triangle of each type and evaluate the area by adding like radical expressions.

A = 6 A_{1} + 4 A_{2}

$A_{1} = 93$ , $A_{2} = \frac{3 1 0 2 4 \cdot 3}{4}$

A = 6 (93) + 4 (\frac{3 1 0 2 4 \cdot 3}{4})

▼

Evaluate right-hand side

A = 543 + 4 (\frac{3 1 0 2 4 \cdot 3}{4})

DenomMultFracToNumber

$4 \cdot \frac{a}{4} = a$

A = 543 + 31024 \cdot 3

FactorOut

Factor out $3$

A = (54 + 31024) 3

Split into factors

A = (54 + 3 512 \cdot 2) 3

RootProd

$3 a \cdot b = 3 a \cdot 3 b$

A = (54 + 3512 \cdot 32) 3

Calculate root

A = (54 + 832) 3

Therefore, the total area of the logo is

(54 + 832) 3

square inches.

Closure

Finding the Length of a Fencing

In the challenge presented at the beginning of this lesson, the dimensions of Ignacio's garden were given. He wants to fence in a triangular area of the garden in which to build his observatory. Notice that some side lengths are missing in the diagram. They can be calculated by using the given lengths. Also, unknown side lengths of an interior triangles will be marked.

The length of fencing required to build the observatory is the perimeter of an inscribed triangle. Each side of this triangle is the hypotenuse of a right triangle. Since the length of all these triangle's legs are known, their hypotenuse can be found by applying the Pythagorean Theorem one at a time.

(ℓ_{1})^{2} = (10 - x)^{2} + (40 x)^{2}

▼

Solve for

ℓ_{1}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

(ℓ_{1})^{2} = 1 0^{2} - 2 (10) x + x^{2} + (40 x)^{2}

Calculate power and product

(ℓ_{1})^{2} = 100 - 20 x + x^{2} + (40 x)^{2}

$(a)^{2} = a$

(ℓ_{1})^{2} = 100 - 20 x + x^{2} + 40 x

Add terms

(ℓ_{1})^{2} = 100 + 20 x + x^{2}

Write as a power

(ℓ_{1})^{2} = 1 0^{2} + 20 x + x^{2}

Split into factors

(ℓ_{1})^{2} = 1 0^{2} + 2 (10) x + x^{2}

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

(ℓ_{1})^{2} = (10 + x)^{2}

$LHS = RHS$

(ℓ_{1})^{2} = (10 + x)^{2}

$a^{2} = ∣ a ∣$

∣ ℓ_{1} ∣ = ∣ 10 + x ∣

$∣ ℓ_{1} ∣ = ℓ_{1}$

ℓ_{1} = ∣ 10 + x ∣

Since

x

is one of side lengths, it must be positive. Consequently,

10 + x

is also positive and

ℓ_{1}

is equal to

10 + x .

Next,

ℓ_{2}

will be calculated.

(ℓ_{2})^{2} = 1 0^{2} + (360 x - 40 x)^{2}

▼

Solve for

ℓ_{2}

CalcPow

Calculate power

(ℓ_{2})^{2} = 100 + (360 x - 40 x)^{2}

Split into factors

(ℓ_{2})^{2} = 100 + (36 \cdot 10 x - 4 \cdot 10 x)^{2}

$a \cdot b = a \cdot b$

(ℓ_{2})^{2} = 100 + (36 \cdot 10 x - 4 \cdot 10 x)^{2}

Calculate root

(ℓ_{2})^{2} = 100 + (6 10 x - 2 10 x)^{2}

SubTerm

Subtract term

(ℓ_{2})^{2} = 100 + (4 10 x)^{2}

PowProdII

$(a b)^{m} = a^{m} b^{m}$

(ℓ_{2})^{2} = 100 + 4^{2} (10 x)^{2}

CalcPow

Calculate power

(ℓ_{2})^{2} = 100 + 16 (10 x)^{2}

$(a)^{2} = a$

(ℓ_{2})^{2} = 100 + 16 (10 x)

(ℓ_{2})^{2} = 100 + 160 x

$LHS = RHS$

(ℓ_{2})^{2} = 100 + 160 x

$a^{2} = ∣ a ∣$

∣ ℓ_{2} ∣ = 100 + 160 x

$∣ ℓ_{2} ∣ = ℓ_{2}$

ℓ_{2} = 100 + 160 x

FactorOut

Factor out $4$

ℓ_{2} = 4 (25 + 40 x)

$a \cdot b = a \cdot b$

ℓ_{2} = 4 \cdot 25 + 40 x

Calculate power and product

ℓ_{2} = 2 25 + 40 x

Now, the last side length will be calculated.

(ℓ_{3})^{2} = x^{2} + (360 x)^{2}

▼

Solve for

ℓ_{3}

$(a)^{2} = a$

(ℓ_{3})^{2} = x^{2} + 360 x

$LHS = RHS$

(ℓ_{3})^{2} = x^{2} + 360 x

$a^{2} = ∣ a ∣$

∣ ℓ_{3} ∣ = x^{2} + 360 x

$∣ ℓ_{3} ∣ = ℓ_{3}$

ℓ_{3} = x^{2} + 360 x

The length of the fencing required is given by the perimeter of the inscribed triangle. The perimeter

L (x)

will be calculated by summing up the side lengths.

L (x) = ℓ_{1} + ℓ_{2} + ℓ_{3} ⇓ L (x) = 10 + x + 2 25 + 40 x + x^{2} + 360 x

Unfortunately, since there are no like radicals, the obtained expression cannot be further simplified. Finally, substitute

x = 5

into the expression to find

L (5) .

L (x) = 10 + x + 2 25 + 40 x + x^{2} + 360 x

$x = 5$

L (5) = 10 + 5 + 2 25 + 40 (5) + 5^{2} + 360 (5)

▼

Evaluate right-hand side

Calculate power and product

L (5) = 10 + 5 + 2 25 + 200 + 25 + 1800

Add terms

L (5) = 15 + 2225 + 1825

Calculate root

L (5) = 15 + 2 (15) + 1825

L (5) = 15 + 30 + 1825

Add terms

L (5) = 45 + 1825

Split into factors

L (5) = 45 + 25 \cdot 73