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| Student Learning Objectives: |
|---|
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| | 15 Theory slides |
| | 11 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Here are a few practice exercises before getting started with this lesson.
sqrt(8x^3)*(x^()12)^5/sqrt(2) * x^3
What is the simplest form of the expression?Ignacio is planning to build an astronomical observatory in his garden. The building will be enclosed by a fence with a triangular shape. The dimensions of Ignacio's garden are presented in the following diagram.
In this diagram, all dimensions are measured in meters. Although some side lengths are still not decided, help Ignacio calculate the length of the fence L(x) with respect to x. What is the value of L(5)?
The n^(th) root of a real number a expresses another real number that, when multiplied by itself n times, will result in a. In addition to the radical symbol, the notation is made up of the radicand a and the index n.
fourth rootof 16. Notice that sqrt(16) simplifies to 2 because 2 multiplied by itself 4 times equals 16. sqrt(16) = sqrt(2^4) = 2 The general expression sqrt(a) represents a number which equals a when multiplied by itself n times.
sqrt(a) * sqrt(a) * ... * sqrt(a)_(ntimes)=a or ( sqrt(a) )^n=a
Some numbers have more than one real n^\text{th} root. For example, 16 has two fourth roots, 2 and -2, because both 2^4 and (-2)^4 are equal to 16. The number of real n^\text{th} roots depends on the sign of a radicand a and an integer n.
| n is Even | n is Odd | |
|---|---|---|
| a>0 | Two unique real n^\text{th} roots, sqrt(a) and -sqrt(a) | One real n^\text{th} root, sqrt(a) |
| a=0 | One real n^\text{th} root, sqrt(0) = 0 | One real n^\text{th} root, sqrt(0) = 0 |
| a<0 | No real n^\text{th} roots | One real n^\text{th} root, sqrt(a) |
By the definition of an n^\text{th} root, calculating the n^\text{th} power of the n^\text{th} root of a number a results in the same number a. The following formula shows what happens if these two operations are swapped.
To simplify an n^(th) root, the radicand must first be expressed as a power. If the index of the radical and the power of the radicand are equal such that sqrt(a^n), the radical expression can be simplified as follows.
sqrt(a^n)= a, & if n is odd |a|, & if n is even
The absolute value of a number is always non-negative, so when n is even, the result will always be non-negative. Consider a few example n^\text{th} roots that can be simplified by using the formula.
| a | n | sqrt(a^n) | Is n even? | Simplify |
|---|---|---|---|---|
| 4 | 3 | sqrt(4^3) | * | 4 |
| -6 | 5 | sqrt((-6)^5) | * | -6 |
| 2 | 6 | sqrt(2^6) | ✓ | |2| = 2 |
| -3 | 4 | sqrt((-3)^4) | ✓ | |-3| = 3 |
Both cases will be considered one at a time.
Start by noting that since a is non-negative, a^n is also non-negative. This means that the n^\text{th} root can be rewritten using a rational exponent. sqrt(a^n) = ( a^n )^(1n) Since both n and 1n are rational numbers, the Power of a Power Property for Rational Exponents can be applied to simplify the obtained expression.
Therefore, sqrt(a^n) is equal to a if a is non-negative.
In case of a negative value of a, there are also two cases two consider.
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 sqrt(a^(n_e)).
(- a)^(n_e) = a^(n_e)
sqrt(a)=a^(1n)
(a^m)^n=a^(m* n)
a * 1/a=1
a^1=a
A root with an odd index n_o is defined for all real numbers. By the definition of the n^\text{th} root, the expression sqrt(a^(n_o)) is the number y that, when multiplied by itself n_o times, will result in a^(n_o). y * y * ... * y_(n_otimes)=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. sqrt(a^n) = a
If a is non-negative, sqrt(a^n) is always equal to a. However, in case of negative a, the value of sqrt(a^n) depends on the parity of n.
| a≥ 0 | a<0 | |
|---|---|---|
| Even n | sqrt(a^n) = a | sqrt(a^n) = - a |
| Odd n | sqrt(a^n) = a | sqrt(a^n) = a |
To conclude, for odd values of n, the expression sqrt(a^n) is equal to a. On the other hand, if n is even, sqrt(a^n) can be written as |a|.
sqrt(a^n)= a, & if n is odd |a|, & if n is even
If n is even, sqrt(a^m) is defined only for non-negative a^m.
The following applet presents a decision tree to simplify sqrt(a^m). In this applet, k is the greatest common factor of m and n.
Consider example n^\text{th} roots sqrt(a^m) that can be simplified by using the decision tree.
| a | m | n | sqrt(a^m) | Simplify |
|---|---|---|---|---|
| 2 | 8 | 4 | sqrt(2^8) | 2^()84 = 2^2 = 4 |
| -3 | 6 | 2 | sqrt((-3)^6) | | (-3)^()62 | = | (-3)^3 | = |-27| = 27 |
| x | 6 | 3 | sqrt(x^6) | x^()63 = x^2 |
| -3 | 2 | 8 | sqrt((-3)^2) | sqrt(|-3|) = sqrt(3) |
| 2 | 3 | 6 | sqrt(2^3) | sqrt(2) |
| -2 | 3 | 9 | sqrt((-2)^3) | sqrt(-2) |
| 2 | 6 | 8 | sqrt(2^6) | sqrt(2^3) = sqrt(8) |
| -2 | 6 | 8 | sqrt((-2)^6) | sqrt(|-2|^3) = sqrt(2^3) = sqrt(8) |
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.
| Radical Expression | Simplified Form |
|---|---|
| sqrt(x^2-4x+4) | ? |
| sqrt((x-2)^5) | x-2 |
| sqrt(16x^8) | ? |
| sqrt(4x^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- 4x+ 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.
Therefore, the first expression simplifies to |x-2|.
| Radical Expression | Simplified Form |
|---|---|
| sqrt(x^2-4x+4) | |x-2| |
| sqrt((x-2)^5) | x-2 |
| sqrt(16x^8) | ? |
| sqrt(4x^2) | ? |
Next, properties of exponents will be used to write the radicand 16x^8 as a power with the exponent of 4.
Write as a power
Split into factors
a^(m* n)=(a^m)^n
a^m b^m = (a b)^m
sqrt(a^4)=|a|
Since x^2 is non-negative, 2x^2 is also non-negative. This means that | 2x^2 | is equal to 2x^2.
| Radical Expression | Simplified Form |
|---|---|
| sqrt(x^2-4x+4) | |x-2| |
| sqrt((x-2)^5) | x-2 |
| sqrt(16x^8) | 2x^2 |
| sqrt(4x^2) | ? |
Now the final expression sqrt(4x^2) will be examined. Similar to the previous expression, properties of exponents will be used to rewrite the radicand as a square.
This time, it is not known whether 2x 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 |
|---|---|
| sqrt(x^2-4x+4) | |x-2| |
| sqrt((x-2)^5) | x-2 |
| sqrt(16x^8) | 2x^2 |
| sqrt(4x^2) | |2x| |
Usually, the n^\text{th} Roots of n^\text{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.
Given two non-negative numbers a and b, the n^\text{th} root of their product equals the product of the n^\text{th} root of each number.
sqrt(ab) = sqrt(a)* sqrt(b), for a≥ 0 and b≥ 0
If n is an odd number, the n^\text{th} root of a negative number is defined. In this case, the Product Property of Radicals for negative a and b is also true.
y^n= b
ab= z^n
a^m b^m = (a b)^m
Rearrange equation
Since xy and z are of the same sign, the final equation implies that z=xy. z^n=(xy)^n ⇒ z=xy The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=xy ⇔ sqrt(ab)=sqrt(a)* sqrt(b)
The following property indicates how to work with roots of a quotient.
Let a be a non-negative number and b be a positive number. The n^\text{th} root of the quotient ab equals the quotient of the n^\text{th} roots of a and b.
sqrt(a/b) = sqrt(a)/sqrt(b), for a≥ 0, b > 0
If n is an odd number, the n^\text{th} root of a negative number is defined. In this case, the Quotient Property of Radicals for negative a and b is also true.
y^n= b
a/b= z^n
a^m/b^m=(a/b)^m
Rearrange equation
Since xy and z are of the same sign, the final equation implies that z= xy. z^n=(x/y)^n ⇒ z=x/y The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=x/y ⇔ sqrt(a/b)=sqrt(a)/sqrt(b)
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.
The voltage V required for a circuit is given by V = sqrt(P)*sqrt(R). In this formula, P is the power in watts and R is the resistance in ohms.
V=sqrt(P)*sqrt(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.
Therefore, a 24-volt battery is needed to power the light bulb.
This means that a battery with a higher voltage is needed to light the 20-watt light bulb.
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 169 times its height.
w = 16/9h ⇕ h = 9/16w 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 + (9/16w )^2 Now, the equation will be solved for w. Keep in mind that the absolute value of a positive number is equal to itself.
(a b)^m=a^m b^m
(a/b)^m=a^m/b^m
a/c* b = a* b/c
a = 256* a/256
Add fractions
Add terms
a* b/c=a/c* b
LHS * 256/337=RHS* 256/337
Rearrange equation
sqrt(LHS)=sqrt(RHS)
sqrt(a^2)=|a|
sqrt(a)*sqrt(b)=sqrt(a* b)
sqrt(a/b)=sqrt(a)/sqrt(b)
Calculate root
sqrt(a^2)=|a|
|w|=w &|d|=d
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= 916w.
Finally, the area of the TV screen can be found.
Therefore, the area of the TV screen is 1296 square inches.
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 4π3 cubic inches.
The volume of a sphere is given by the formula V= 43π 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.
Volume:& V= 43π r^3 Surface Area:& S=4π 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.
a/c* b = a* b/c
LHS * 3/4π=RHS* 3/4π
a/c* b = a* b/c
Rearrange equation
sqrt(LHS)=sqrt(RHS)
sqrt(a^3)=a
Now, the rewritten expression in terms of r will be substituted into the formula for the surface area S=4π r^2. Use the properties of exponents and radicals to simplify the formula.
r= sqrt(3V/4π)
sqrt(a)=a^(1n)
(a^m)^n=a^(m* n)
1/b* a = a/b
(a/b)^m=a^m/b^m
a * b/c= a/c* b
a=a^1
a^m/a^n= a^(m-n)
Rewrite 1 as 3/3
Subtract fractions
a/b=a* 1/b
a^(m* n)=(a^m)^n
a^(1n)=sqrt(a)
This corresponds to option C.
V= 4π/3
3 * a/3= a
sqrt(a)*sqrt(b)=sqrt(a* b)
Multiply
sqrt(a^3)=a
Therefore, the surface area of the model Earth is 4π square inches.
Some irrational numbers are written as expressions involving n^(th) roots of powers. Previously, the Product Property of Radicals and the n^\text{th} Roots of n^\text{th} Powers Property were used to eliminate the root. Consider the following example. 3sqrt(16) = 3sqrt(4^2) Given an irrational number in this form, it is possible to find another irrational number by changing the power in the radicand.
Let a, b, and x be rational numbers and n a natural number so that the n^\text{th} root sqrt(b^x) is irrational. The irrational conjugate of an n^\text{th} root asqrt(b^x) is obtained by raising b to the power of n-x in the radicand. ccc Number & & Conjugate [0.15cm] asqrt(b^x) & Change Power & asqrt(b^(n-x)) For example, the conjugate of 4sqrt(7^2) is 4sqrt(7^3). 4sqrt(7^2) Change Power 4sqrt(7^(5-2)) = 4sqrt(7^3)
The irrational conjugate can be used to rationalize a denominator containing an n^\text{th} root.Find the irrational conjugate of a+bsqrt(c) or the irrational conjugate of an n^\text{th} root.
When a numeric expression has a denominator that involves an irrational n^\text{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. 4/sqrt(27) Since the denominator involves an n^\text{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.
sqrt(a)*sqrt(b)=sqrt(a* b)
a^m*a^n=a^(m+n)
Add terms
sqrt(a^5)=a
Calculate power and product
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^\text{th} roots with different indexes, rationalizing is a more complicated task.
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 = sqrt(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.
v = sqrt(K/r) = sqrt(K)/sqrt(r) It is given that the distance from the Earth to the satellite is sqrt(262 528^2) miles. Therefore, r = sqrt(262 528^2) will be substituted into the formula and the denominator rationalized.
r= sqrt(262 528^2)
a/b=a * sqrt(sqrt(262 528^2))/b * sqrt(sqrt(262 528^2))
sqrt(a)* sqrt(a)= a
a/b=a * sqrt(262 528^1)/b * sqrt(262 528^1)
sqrt(a)*sqrt(b)=sqrt(a* b)
a^m*a^n=a^(m+n)
sqrt(a^3)=a
Now that the denominator has been rationalized, the numerator of the obtained expression will be further simplified.
sqrt(a)=a^(12)
sqrt(a)=a^(1n)
(a^m)^n= (a^n)^m
(a^m)^n=a^(m* n)
a * 1/a=1
a^1=a
a^(1n)=sqrt(a)
sqrt(a)*sqrt(b)=sqrt(a* b)
a* a=a^2
Now that the denominator contains only an integer number, it has been successfully rationalized. The calculated velocity corresponds to option B.
r_2 = (3-sqrt(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 = sqrt(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=sqrt(Kr) can be substituted on the right-hand side. v_2/v = sqrt(Kr_2)/v ⇒ v_2/v = sqrt(Kr_2)/sqrt(Kr) Now, substitute (3-sqrt(2))^2 r for r_2 in the obtained quotient and evaluate the right-hand side.
r_2= (3-sqrt(2))^2 r
sqrt(a/b)=sqrt(a)/sqrt(b)
.a /b./.c /d.=a/b*d/c
Multiply fractions
sqrt(a* b)=sqrt(a)*sqrt(b)
Cancel out common factors
Simplify quotient
sqrt(a^2)=|a|
|3-sqrt(2)|=3-sqrt(2)
a/b=a * (3+sqrt(2))/b * (3+sqrt(2))
Identity Property of Multiplication
(a-b)(a+b)=a^2-b^2
Calculate power
( sqrt(a) )^2 = a
Subtract term
LHS * v=RHS* v
Finally, it can concluded that v_2 is equal to 3+sqrt(2)7 v.
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.
asqrt(x) + bsqrt(x) = (a + b) sqrt(x)
asqrt(x) - bsqrt(x) = (a - b) sqrt(x)
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.
asqrt(x)± bsqrt(x)=(a± b)sqrt(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.
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.
asqrt(x) + bsqrt(x) = ( a + b) sqrt(x)
asqrt(x) - bsqrt(x) = ( a - b) sqrt(x)
For example, consider the following radical expressions. sqrt(x^5 y^3) and sqrt(16xy^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.
Split into factors
Write as a sum
a^(m+n)=a^m*a^n
Commutative Property of Multiplication
sqrt(a* b)=sqrt(a)*sqrt(b)
a^1=a
sqrt(a^4)=|a|
Write as a power
a^m* b^m=(a * b)^m
sqrt(a^4)=|a|
Multiply
Since both the index and the radicand of the roots are identical, they are like radical expressions.
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 sqrt(32) inches. Find the exact total area of the design.
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 = 1/2absin C The length of all sides is 6 inches and the measure of all angles in the triangle is 60^(∘). This means that a= 6, b= 6, and C = 60^(∘) can be substituted into the formula to find the area.
Substitute values
Multiply
a/b=1/b* a
\ifnumequal{60}{0}{\sin\left(0^\circ\right)=0}{}\ifnumequal{60}{30}{\sin\left(30^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{45}{\sin\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{60}{\sin\left(60^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{90}{\sin\left(90^\circ\right)=1}{}\ifnumequal{60}{120}{\sin\left(120^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{135}{\sin\left(135^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{150}{\sin\left(150^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{180}{\sin\left(180^\circ\right)=0}{}\ifnumequal{60}{210}{\sin\left(210^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{225}{\sin\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{240}{\sin\left(240^\circ\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{60}{270}{\sin\left(270^\circ\right)=\text{-}1}{}\ifnumequal{60}{300}{\sin\left(300^\circ\right)=\text{-}\dfrac {\sqrt 3}2}{}\ifnumequal{60}{315}{\sin\left(315^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{330}{\sin\left(330^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{360}{\sin\left(360^\circ\right)=0}{}
Multiply fractions
Multiply
Calculate quotient
Similarly, the area A_2 of a small triangle can be calculated. In this case, the side length is sqrt(32) inches and the measure of the included angle is 60^(∘) as previously.
Substitute values
sqrt(a)*sqrt(b)=sqrt(a* b)
a/b=1/b* a
\ifnumequal{60}{0}{\sin\left(0^\circ\right)=0}{}\ifnumequal{60}{30}{\sin\left(30^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{45}{\sin\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{60}{\sin\left(60^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{90}{\sin\left(90^\circ\right)=1}{}\ifnumequal{60}{120}{\sin\left(120^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{135}{\sin\left(135^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{150}{\sin\left(150^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{180}{\sin\left(180^\circ\right)=0}{}\ifnumequal{60}{210}{\sin\left(210^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{225}{\sin\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{240}{\sin\left(240^\circ\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{60}{270}{\sin\left(270^\circ\right)=\text{-}1}{}\ifnumequal{60}{300}{\sin\left(300^\circ\right)=\text{-}\dfrac {\sqrt 3}2}{}\ifnumequal{60}{315}{\sin\left(315^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{330}{\sin\left(330^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{360}{\sin\left(360^\circ\right)=0}{}
Multiply fractions
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 = 6A_1 + 4A_2 Finally, substitute the simplified areas of a single triangle of each type and evaluate the area by adding like radical expressions.
A_1= 9sqrt(3), A_2= sqrt(1024)* sqrt(3)/4
Multiply
4 * a/4= a
Factor out sqrt(3)
Split into factors
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root
Therefore, the total area of the logo is ( 54 + 8sqrt(2)) sqrt(3) square inches.
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.
(a-b)^2=a^2-2ab+b^2
Calculate power and product
( sqrt(a) )^2 = a
Add terms
Write as a power
Split into factors
a^2-2ab+b^2=(a-b)^2
sqrt(LHS)=sqrt(RHS)
sqrt(a^2)=|a|
|l_1|=l_1
Calculate power
Split into factors
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root
Subtract term
(a b)^m=a^m b^m
Calculate power
( sqrt(a) )^2 = a
Multiply
sqrt(LHS)=sqrt(RHS)
sqrt(a^2)=|a|
|l_2|=l_2
Factor out 4
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate power and product
x= 5
Calculate power and product
Add terms
Calculate root
Multiply
Add terms
Split into factors
sqrt(a* b)=sqrt(a)*sqrt(b)
Calculate root and product
According to n^\text{th} Roots of n^\text{th} Powers Property, for any real number a, the radical expression sqrt(a^n) can be simplified as follows. sqrt(a^n)= a, & if n is odd |a|, & if n is even Because the given root has an even index and the variables have even exponents, we will use the absolute value symbol to simplify the expression.
Any number raised to an even exponent will always be non-negative. The expression 9x^2y^4 will always be non-negative because there are only even exponents. The product of non-negative numbers is also always non-negative. Therefore, the absolute value symbol is not needed in this case. |9x^2y^4| ⇔ 9x^2y^4
Just as in Part A, n^\text{th} Roots of n^\text{th} Powers Property will be applied.
sqrt(a^n)=
a, & if n is odd
|a|, & if n is even
Because the given root has an even index and the variable has an even exponent, we will use the absolute value symbol to simplify the expression.
Since the exponent of a is odd, it is necessary to use the absolute value symbol because the expression 2a^3 could be negative for negative values of a.
To simplify the given expression, we will use the Properties of Radicals. Let's start by applying the Product Property of Radicals.
Next, use the Product of Powers Property to multiply the π-factors.
To simplify the given expression, we will use the Properties of Radicals. We will start by factoring out perfect cubes and using the Product Property of Radicals. Moreover, consider the following two cases for any real number a. sqrt(a^n)= a, &if n is odd |a|, &if n is even Because our radical has an odd index, we will not need to use absolute value symbols to simplify our expression.
Simplify the given rational expression. Make assumptions on the variables x, y, and z so that the square roots are real. sqrt(75x^3yz^3)/sqrt(48x)* sqrt(z^5)
Let's start by rewriting the given expression by applying the Product Property of Radicals. sqrt(75x^3yz^3)/sqrt(48x)* sqrt(z^5) = sqrt(75x^3yz^3)/sqrt(48xz^5) Next, before we can divide this expression, we need to answer three questions.
Let's start with the first question. For the square roots to be real, their radicands should be non-negative because otherwise, the square roots would be imaginary. The radicand in the denominator is non-negative only if both x and z are non-negative or both x and z are non-positive. Denominator [0.1cm] 48xz^5≥ 0 ⇔ ( x, z≥ 0 or x, z ≤ 0 ) Recall that the square roots in the denominator had been separated before applying the Product Property of Radicals. This means that both x and z must be non-negative. For the square root in the numerator to be also real, the radicand 75x^3yz^3 should be non-negative. Since x and z are non-negative, y must be non-negative as well. Numerator [0.1cm] 75x^3yz^3≥ 0 x, z≥ 0 ⇒ y≥ 0 Therefore, all three variables must be non-negative. Now let's answer the second question. The expressions can be divided only if the denominator is not equal to 0. This puts an additional restriction on the x- and z-variables, as neither can be equal to 0. Since we already established that these variables are non-negative, this implies that x and z must be positive. x, z≥ 0 x, z≠ 0 ⇒ x, z>0 Finally, to answer the third question, consider the rule regarding absolute value symbols for any real number a. sqrt(a^n)= a, & if n is odd |a|, &if n is even In our case, the index is 2, an even number. According to the rule, we need to use absolute value bars in our answer. However, we also found that all of the variables can only take non-negative values. Since the absolute value of a non-negative value is the value itself, we do not need absolute value symbols after all. Let's sum up all the information we have found.
| Question | Answer |
|---|---|
| What are the assumptions on the variables? | x, y, z≥ 0 |
| Can the expressions be divided? | Yes, if x, z≠ 0. |
| Do absolute value symbols need to be added to the answer? | No, as x, z>0 and y≥ 0. |
Now we can use the Quotient Property of Radicals and reduce the fraction.
Next, let's simplify the radical expression by finding all of the perfect squares inside the radicals. Then we will apply the n^\text{th} Roots of n^\text{th} Powers Property.
We need to rationalize a fraction with an irrational binomial denominator. To do so, we multiply the numerator and denominator by the irrational conjugate of the denominator. We find the conjugate by changing the sign of the second term of the radical expression.
| Binomial | Conjugate |
|---|---|
| asqrt(c) + bsqrt(d) | asqrt(c) - bsqrt(d) |
| asqrt(c) - bsqrt(d) | asqrt(c) + bsqrt(d) |
In this case, the conjugate of the denominator is sqrt(2)-2sqrt(6).
Now that the denominator has been rationalized, the expression can be further simplified.
To simplify the given expression, we can first rewrite the radical as a quotient of two radicals.
sqrt(9x^3y^2/4a^3b) = sqrt(9x^3y^2)/sqrt(4a^3b)
Now, we can rationalize the denominator of the quotient. To do so, we will first multiply the numerator and denominator by the irrational conjugate of an n^\text{th} root and use the fact that we can multiply the radicands of n^\text{th} roots if they have the same index.
If sqrt(a) and sqrt(b) are real numbers, then sqrt(a)* sqrt(b) is equal to sqrt(ab).
Let's start by finding the exponents necessary to create perfect 4^(th) powers in the denominator. Our goal is to have each factor raised to the 4^\text{th} power.
Now that we have found the factors that make the radicand of the denominator perfect 4^(th) powers, we can begin to simplify the quotient. Consider the index of the radicals to see how we should format our solution. The n^\text{th} Roots of n^\text{th} Powers Property can help with this. sqrt(a^n)= a, &if n is odd |a|, &if n is even Since both radicals are real numbers and the roots are even, the expressions underneath the radicals must be positive & mdash; if they were not, the radicals would be imaginary. With this in mind, let's consider the possible values of the variables a, b, x, and y.
This means that to calculate the fourth root of a or b, we will need absolute value symbols.
Simplify the following radical expression. 3sqrt(32) + sqrt(500) - 3sqrt(108) + 2sqrt(128)
To simplify the given radical expression, we first need to rewrite the radicands so that they have exponents that match the index of the radicals. Then we will consider the properties for combining radical expressions when they are part of a sum or difference. asqrt(x)+bsqrt(x)=(a+b)sqrt(x) asqrt(x)-bsqrt(x)=(a-b)sqrt(x) Notice that radicals can only be added or subtracted when the index and the radicand are exactly the same. Let's simplify our radicals to see if we can create like terms.
Now three terms have the same radical expression. Let's add and subtract them!
Now, let's consider the remaining radicals. 2sqrt(4) + 8sqrt(2) Because the radicands are not the same, it is not possible to add them. Therefore, the simplest form of the radical expression is 2sqrt(4) + 8sqrt(2).