Simplifying Radicals

Is the radical $\sqrt{9}$ simplified?

The answer is No: the radicand contains a perfect square, 9, so by rule 1, the radical is not simplified.

Is the radical $\sqrt{\frac{1}{3}}$ simplified?

The answer is again No: the radicand contains a fraction, so by rule 2, the radical is not simplified.

Is the radical $\frac{1}{\sqrt{3}}$ simplified?

The answer is yet again No: there is a radical in the denominator of the fraction, so by rule 3, the radical is not simplified.

Simplify the following radical: $\sqrt{\frac{20}{9}}$.

Solution:

The radicand contains a fraction, so by rule 2. We have to get rid of it. To do so, use the quotient property to rewrite this radical as $\frac{\sqrt{20}}{\sqrt{9}}$.

Now we have to simplify separately the numerator and the denominator.

To simplify the numerator, we have to use the product property to isolate perfect squares, so that rule 1 is fulfilled. Note that $20=4·5$, so the numerator can be rewritten as

$\sqrt{20}=\sqrt{4·5}=\sqrt{4}·\sqrt{5}=2\sqrt{5}$

Now to the denominator: the square root of 9 is 3, so we can get rid of the radical in the denominator of the fraction, fulfilling rule 3, and the final result is

$\boxed{\sqrt{\frac{\mathbf{20}}{\mathbf{9}}}\mathbf{=}\frac{\mathbf{2}\sqrt{\mathbf{5}}}{\mathbf{3}}}$

Simplify the following radical: $\frac{2}{\sqrt{3}}$

Solution:

Rule 3 is violated since there is a radical in the denominator. This cannot be simplified directly, since 3 is not a perfect square. This let's multiply both the numerator and the denominator by $\sqrt{3}$:

$\frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}}·\frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}·\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3·3}}=\frac{2\sqrt{3}}{\sqrt{9}}=\frac{2\sqrt{3}}{3}$

Magic! The radical in the denominator vanished, mission completed.

Simplify the following radical: $\sqrt[3]{\frac{24}{125}}$

Solution:

Using the quotient property, we can split this radical into a fraction with two cube roots

$\sqrt[3]{\frac{24}{125}}=\frac{\sqrt[3]{24}}{\sqrt[3]{125}}$

Now, the denominator of this fraction is equal to 5, since$5^3=125$.

What about the numerator? To simplify it, use the product property to split it and isolate a perfect cube:

$\frac{\sqrt[3]{24}}{\sqrt[3]{125}}=\frac{\sqrt[3]{8}·\sqrt[3]{3}}{5}$

We know that $\sqrt[3]{8}=2$, so the final result is

$\sqrt[3]{\frac{24}{125}}=\frac{2\sqrt[3]{3}}{5}$

Are the following radicals fully simplified? If not, simplify them further.

a) $\sqrt[3]{30}$

b) $\sqrt{8}$

c)$\frac{1}{\sqrt{3}}$

d)$\sqrt{\frac{9}{16}}$

Solutions:

a) The radical obeys all three rules and cannot be simplified further. It is fully simplified.

b) A perfect square is present in the radicand as$8=2·4$, and 4 is a perfect square, so this must be simplified further: $\sqrt{8}=\sqrt{4}·\sqrt{2}=2\sqrt{2}$.

c) There is a radicand in the denominator of a fraction, so this expression is not fully simplified. To get rid of this radicand, we must multiply the top and bottom of the fraction by the same radicand, $\sqrt{3}$. This gives $\frac{1·\sqrt{3}}{\sqrt{3}·\sqrt{3}}=\frac{\sqrt{3}}{3}$. This expression is now fully simplified.

d) There is a fraction inside the radicand, so you must use the quotient property to separate this:

$\sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}}=\frac{3}{4}$

a) Simplify $\sqrt[3]{x^3{y}^{15}{z}^{20}}$

Solution:

Step 1: In this example, we have a cubic root, as the index is 3. We have 3 variables x , y and z, and their corresponding exponents are 3, 15 and 20.

Steps 2 and 3:

• The exponent of the variable x is 3, which is equal to the index 3. Therefore, we can divide the exponent by the index.

$3÷3=1,witharemainderof0$

We can write x outside of the radical with an exponent of 1, which is the same as writing x, and nothing inside the radical $\Rightarrow x\sqrt[3]{}$

• The exponent of the variable y is 15, which is greater than the index 3. Therefore, we can divide the exponent by the index.

$15÷3=5,witharemainderof0$

We can write y outside of the radical with an exponent of 5, and nothing inside the radical $\Rightarrow x{y}^5\sqrt[3]{}$

• The exponent of the variable z is 20, which is greater than the index 3. Therefore, we can divide the exponent by the index.

$20÷3=6,witharemainderof2$

We can write z outside of the radical with an exponent of 6, and z with an exponent of 2 inside the radical $\Rightarrow x{y}^5{z}^6\sqrt[3]{z^2}$

So, the radical fully simplified is:

$\boxed{\sqrt[\mathbf{3}]{{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{15}}{\mathbf{z}}^{\mathbf{20}}}\mathbf{=}\mathbf{x}{\mathbf{y}}^{\mathbf{5}}{\mathbf{z}}^{\mathbf{6}}\mathbf{}\sqrt[\mathbf{3}]{{\mathbf{z}}^{\mathbf{2}}}}$

b) Simplify $\sqrt{18x^7{y}^{8}z}$

Solution:

Step 1: In this example, we have a square root, as the index is 2. We have 3 variables x , y and z, and their corresponding exponents are 7, 8 and 14.

We also have the number 18 inside the radical. Isolating perfect squares, we can say that $18=9·2$, so we can write $\sqrt{18}$ as $\sqrt{9·2}=\sqrt{9}·\sqrt{2}=3\sqrt{2}$. Therefore, we can write 3 outside the radical and 2 stays inside. Now let's focus on the variables.

Steps 2 and 3:

• The exponent of the variable x is 7, which is greater than the index 2. Therefore, we can divide the exponent by the index.

$7÷2=3,withareminderof1$

We can write x outside of the radical with an exponent of 3, and x with an exponent of 1 inside the radical$\Rightarrow 3x^3\sqrt{2x}$

• The exponent of the variable y is 8, which is greater than the index 2. Therefore, we can divide the exponent by the index.

$8÷2=4,withareminderof0$

We can write y outside of the radical with an exponent of 4, and nothing inside the radical $\Rightarrow 3x^3{y}^4\sqrt{2x}$

• The exponent of the variable z is 1, which is less than the index 2. Therefore, we leave it inside the radical $\Rightarrow 3x^3{y}^4\sqrt{2xz}$

Our final answer in this case is:

$\boxed{\sqrt{\mathbf{18}{\mathbf{x}}^{\mathbf{7}}{\mathbf{y}}^{\mathbf{8}}\mathbf{z}}\mathbf{=}\mathbf{3}{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{4}}\mathbf{}\sqrt{\mathbf{2}\mathbf{x}\mathbf{z}}}$

Simplify $\sqrt[3]{\frac{8x^{10}{y}^5}{x^3{y}^2}}$

Solution:

Step 1: In this example, we have a cubic root, as the index is 3. We have the number 8 inside the cube root, but we know that $\sqrt[3]{8}=2$, so we can write 2 outside of the radical $\Rightarrow 2\sqrt[3]{}$

We also have 2 variables x and y. However, both variables appear in the numerator and denominator of the fraction.

Using the same-base division property, we get the following:

$2\sqrt[3]{x^{10-3}{y}^{5-2}}=2\sqrt[3]{x^7{y}^3}$

Now we can focus on simplifying the variables.

Steps 2 and 3:

• The exponent of the variable x is 7, which is greater than the index 3. Therefore, we can divide the exponent by the index.

$7÷3=2,witharemainderof1$

We can write x outside of the radical with an exponent of 3, and x with an exponent of 1 inside the radical

$\Rightarrow 2x^2\sqrt[3]{x}$

• The exponent of the variable y is 3, which is equal to the index 3. Therefore, we can divide the exponent by the index.

$3÷3=1,witharemainderof0$

We can write y outside of the radical with an exponent of 1, or just y, and nothing inside the radical $\Rightarrow 2x^{2}y\sqrt[3]{x}$

The radical fully simplified is:

$\boxed{\sqrt[\mathbf{3}]{\frac{\mathbf{8}{\mathbf{x}}^{\mathbf{10}}{\mathbf{y}}^{\mathbf{5}}}{{\mathbf{x}}^{\mathbf{3}}{\mathbf{y}}^{\mathbf{2}}}}\mathbf{=}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{}\sqrt[\mathbf{3}]{\mathbf{x}}}$

$\sqrt{-25}\ne -5$ No real solution!

a) Simplify $\sqrt[3]{-8}$

Solution:

$\boxed{\sqrt[\mathbf{3}]{\mathbf{-}\mathbf{8}}\mathbf{=}\mathbf{-}\mathbf{2}}$

Since ${(-2)}^3=\left(-2\right)·\left(-2\right)·\left(-2\right)=-8$

When you multiply a negative number by itself three times, the result is also negative.

Therefore, you can in fact simplify the cube root of negative numbers using real numbers!

b) Simplify $\sqrt[3]{-27}$

Solution:

$\boxed{\sqrt[\mathbf{3}]{\mathbf{-}\mathbf{27}}\mathbf{=}\mathbf{-}\mathbf{3}}$

Since ${(-3)}^3=\left(-3\right)·\left(-3\right)·\left(-3\right)=-27$