Radicals - Algebra 2
Card 1 of 1760
Evaluate: 
Evaluate:
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The exponent can be distributed through both terms in the parentheses.
![(\frac{1}{8})^{\frac{1}{3}} = \frac{1^{\frac{1}{3}}}{8^{\frac{1}{3}} } = \frac{\sqrt[3]{1}}{\sqrt[3]{8}}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/833035/gif.latex)
Simplify each radical. The cube root is identifying a number that multiplies itself three times to produce the value inside the radical.
The answer is: 
The exponent can be distributed through both terms in the parentheses.
Simplify each radical. The cube root is identifying a number that multiplies itself three times to produce the value inside the radical.
The answer is:
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Solve for x: 
Solve for x:
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To solve, perform inverse opperations, keeping in mind order of opperations:
first, square both sides
subtract 1
divide by 2

To solve, perform inverse opperations, keeping in mind order of opperations:
first, square both sides
subtract 1
divide by 2
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Solve for x: 
Solve for x:
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To solve, perform inverse opperations, keeping in mind order of opperations:
take the square root of both sides
subtract 19 from both sides
square both sides

To solve, perform inverse opperations, keeping in mind order of opperations:
take the square root of both sides
subtract 19 from both sides
square both sides
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Evaluate: 
Evaluate:
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In order to add the two terms, we must first find the values of each term in the expression.
Rewrite the fractional exponents as an expression of a square root.
![64^{\frac{1}{6}}=\sqrt[6]{64}=2](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/761186/gif.latex)
![8^{\frac{2}{3}} =( \sqrt[3]{8})^2 = (2)^2 =4](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/761187/gif.latex)
Add the two values.

The answer is: 
In order to add the two terms, we must first find the values of each term in the expression.
Rewrite the fractional exponents as an expression of a square root.
Add the two values.
The answer is:
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Fractional exponents have the power as the numerator and the root as the denominator.

![\sqrt{x}=\sqrt[2]{x^1}=x^\frac{1}{2}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/763022/gif.latex)
Fractional exponents have the power as the numerator and the root as the denominator.
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Fractional exponents have the power as the numerator and the root as the denominator.
In this case, the power is 5, and the root is 3.

![x^\frac{5}{3}=\sqrt[3]{x^5}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/763011/gif.latex)
Fractional exponents have the power as the numerator and the root as the denominator.
In this case, the power is 5, and the root is 3.
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Solve: 
Solve:
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The numbers with the fractional exponents can be rewritten as radicals.
![144^\frac{3}{2}-64^\frac{1}{6}=( \sqrt{144})^3 - \sqrt[6]{64}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/784902/gif.latex)
Simplify both radicals. The sixth root of 64 is two.

The answer is: 
The numbers with the fractional exponents can be rewritten as radicals.
Simplify both radicals. The sixth root of 64 is two.
The answer is:
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Solve: 
Solve:
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This can be rewritten as the fourth root.
![-64^{\frac{1}{4}} = -\sqrt[4]{64}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/803837/gif.latex)
Rewrite the inner quantity with factors of numbers to the fourth root.

![-\sqrt[4]{64} = -\sqrt[4]{16\times 4} = -\sqrt[4]{16} \cdot \sqrt[4]{4}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/803839/gif.latex)
Simplify both terms.
![-\sqrt[4]{16} =-2](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/803840/gif.latex)
![\sqrt[4]{4} = (4^\frac{1}{2})^{\frac{1}{2}} = 2^{\frac{1}{2}} = \sqrt 2](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/803841/gif.latex)
The answer is: 
This can be rewritten as the fourth root.
Rewrite the inner quantity with factors of numbers to the fourth root.
Simplify both terms.
The answer is:
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Solve for x: 
Solve for x:
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To solve, use inverse opperations keeping in mind order of opperations:
divide both sides by 5
square both sides
add 12 to both sides

To solve, use inverse opperations keeping in mind order of opperations:
divide both sides by 5
square both sides
add 12 to both sides
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Solve for
:

Solve for :
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When working with radicals, a helpful step is to square both sides of an equation so that you can remove the radical sign and deal with a more classic linear or quadratic equation. But of course if you were to simply square both sides first here, you would still end up with radical signs, as were you to FOIL the left side you wouldn't eliminate the radicals. So a good first step is to add
and subtract
from both sides so that you get:

Now when you square both sides, you'll eliminate the radical on the right-hand side and yield:

Then when you subtract
from both sides to set up a quadratic equalling zero, you have a factorable quadratic:

This factors to:

Which would seem to yield solutions of
and
. However, when you're solving for quadratics it's always important to plug your solutions back into the original equation to check for extraneous solutions. Here if you plug in
the math holds:
because
, and
.
But if you plug in
, you'll see that the original equation is not satisfied:
because
does not equal
. Therefore the only proper answer is
.
When working with radicals, a helpful step is to square both sides of an equation so that you can remove the radical sign and deal with a more classic linear or quadratic equation. But of course if you were to simply square both sides first here, you would still end up with radical signs, as were you to FOIL the left side you wouldn't eliminate the radicals. So a good first step is to add and subtract
from both sides so that you get:
Now when you square both sides, you'll eliminate the radical on the right-hand side and yield:
Then when you subtract from both sides to set up a quadratic equalling zero, you have a factorable quadratic:
This factors to:
Which would seem to yield solutions of and
. However, when you're solving for quadratics it's always important to plug your solutions back into the original equation to check for extraneous solutions. Here if you plug in
the math holds:
because
, and
.
But if you plug in , you'll see that the original equation is not satisfied:
because
does not equal
. Therefore the only proper answer is
.
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Solve for
:

Solve for :
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To solve, first square both sides:
squaring the left side just givs x - 3. To square the left side, use the distributite property and multiply
:
This is a quadratic, we just need to combine like terms and get it equal to 0
now we can solve using the quadratic formula:

This gives us 2 potential answers:
and

To solve, first square both sides:
squaring the left side just givs x - 3. To square the left side, use the distributite property and multiply
:
This is a quadratic, we just need to combine like terms and get it equal to 0
now we can solve using the quadratic formula:
This gives us 2 potential answers:
and
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Solve for
.

Solve for .
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To get rid of the radical, we square both sides.

To get rid of the radical, we square both sides.
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Solve for
.

Solve for .
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To get rid of the radical, we need to square both sides. The issue is radicals don't generate negative numbers unless we talk about imaginary numbers. In this case, our answer choice should be no answer.
To get rid of the radical, we need to square both sides. The issue is radicals don't generate negative numbers unless we talk about imaginary numbers. In this case, our answer choice should be no answer.
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Solve for
.

Solve for .
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Square both sides to get rid of the radical.
Subtract
on both sides.

Square both sides to get rid of the radical.
Subtract
on both sides.
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Solve the equation: 
Solve the equation:
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Subtract two from both sides.

Simplify the equation.

Square both sides.


Add
on both sides.


Subtract
on both sides.

The answer is: 
Subtract two from both sides.
Simplify the equation.
Square both sides.
Add on both sides.
Subtract on both sides.
The answer is:
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Solve for
.

Solve for .
Tap to reveal answer
Square both sides to get rid of the radical.
Add
on both sides.

Square both sides to get rid of the radical.
Add
on both sides.
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Solve for
.

Solve for .
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Square both sides to get rid of the radical.
Divide
on both sides.

Square both sides to get rid of the radical.
Divide
on both sides.
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Solve for
.

Solve for .
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Square both sides to get rid of the radical.
Multiply
on both sides.

Square both sides to get rid of the radical.
Multiply
on both sides.
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Rewrite the radical as an exponent:
![\sqrt[5]{x^{4}}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/750522/gif.latex)
Rewrite the radical as an exponent:
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In order to rewrite a radical as an exponent, the number in the radical that indicates the root, gets written as a fractional exponent as shown below:
![\sqrt[5]{x^{4}}=x^{\frac{4}{5}}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/750523/gif.latex)
In this case, we are done because there are no further simplification steps.
In order to rewrite a radical as an exponent, the number in the radical that indicates the root, gets written as a fractional exponent as shown below:
In this case, we are done because there are no further simplification steps.
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What is the value of:
?
What is the value of: ?
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In order to solve this expression, we will need to rewrite each square root using one half as the exponent. This will allow us to solve for the value of the inner terms.
![2\sqrt{\sqrt{32}}=2((\sqrt{32})^\frac{1}{2})^\frac{1}{2} = 2\sqrt[4]{32}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/830062/gif.latex)
We can then rewrite the radical using factors of perfect numbers raised to the fourth power, such as: 
![2\sqrt[4]{32} = 2\cdot \sqrt[4]{16}\cdot \sqrt[4]{2} = 2\cdot 2\cdot \sqrt[4]{2}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/830064/gif.latex)
The answer is: ![4 \sqrt[4]{2}](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/830065/gif.latex)
In order to solve this expression, we will need to rewrite each square root using one half as the exponent. This will allow us to solve for the value of the inner terms.
We can then rewrite the radical using factors of perfect numbers raised to the fourth power, such as:
The answer is:
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