Equiradical Surds

If two or more surds are of the same order they are said to be equiradical.

Surds are not equiradical when their surd indices are different.

Thus, √5, √7, 2√5, √x and 10^1/2 are equiradical surds.

But √2, ∛7, ∜6 and 9^2/5 are not equiradical.

Note: Non-equiradical surds can be reduced to equiradical surds.

Thus, non-equiradical surds √3, ∛3, ∜3 become $\sqrt[12]{729}$, $\sqrt[12]{81}$, $\sqrt[12]{27}$ respectively when they are reduced to equiradical surds.

If x is a positive integer with nth root, then $\sqrt[n]{x}$ is a surd of nth order when the value of $\sqrt[n]{x}$  is irrational. In $\sqrt[n]{x}$  expression n is the order of surd and x is called as radicand. For example ∛7 is surd of order 3.

When two or more surds have the same order, they called as Equiradical Surds. For example √2, √3, √5, √7, √x are the surds of order 2. So these surds are equiradical.

When two or more surds don’t have the same order they are called non-equiradical surds. For example√5, √7, ∛10, ∛17, ∜9, ∜20 these surds are non-equiradical surds as they have different orders as 2,3 and 4. 

Non-equiradical surds can be expressed in the form of equiradical surds. For example √2, ∛3 and ∜5 are non-equiradical surds with order 2,3 and 4. If we can change the indices of surds such that all the surds  can be converted in to a same order, then non-equiradical surds can be expressed in the form of equiradical surds. For this case with orders of 2, 3, 4 we can change surds in the a same order if we change it to the LCM (Lowest Common Multiple) of order and that is 12.

Changing the order of first surd from 2 to 12, √2 = 2$^{1/2}$ = 2$^{6/123}$ = 64$^{1/12}$ = $\sqrt[12]{64}$

Changing the order of second surd from 3 to 12, ∛3 = 3$^{1/3}$ = 3 $^{4/12}$ = 81$^{1/12}$ = $\sqrt[12]{81}$

Changing the order of third surd from 4 to 12, ∜5 = 5$^{1/4}$ = 5$^{3/12}$ = 125$^{1/12}$ = $\sqrt[12]{125}$

So √2, ∛3 and ∜5 are the non-equiradical surds which can be expressed in the form of equiradical surds as $\sqrt[12]{64}$, $\sqrt[12]{81}$, $\sqrt[12]{125}$

In two equiradical surds $\sqrt[n]{x}$ and $\sqrt[n]{y}$, $\sqrt[n]{x}$ > $\sqrt[n]{y}$ when x > y.  For example ∛7 and ∛5 are the two equiradical surds, as 7 > 5, so ∛7 > ∛5. The same comparison can done for more than two equiradical surds also. 

For non-equiradical surds if we change it to the form of equiradical surds, then similarly we can compare the values of surds like it is compared for the case of two equiradical numbers. For example ∛7 and ∜5 are two non-equiradical surds. If we need find out ∛7 > ∜5 or ∜5 > ∛7, then we first need to express the surds in to equiradical surds. As the orders of the surds are 3 and 4, LCM of 3 and 4 is 12, so if we make the surds in order 12 we can find out which one is greater value.

∛7 = 7$^{1/3}$ = 7$^{4/12}$ = 2401$^{1/12}$ = $\sqrt[12]{2401}$

∜5 = 5$^{1/4}$ = 5$^{3/12}$ = 125$^{1/12}$ = $\sqrt[12]{125}$

As 2401 > 125, so ∛7 > ∜5.

Solved Example:

Arrange the surds in descending order.

√10, ∛25, ∜40

Solution:

√10, ∛25, ∜40

Surds are in the order of 2, 3, and 4. So the surds are non-equiradical surds. To arrange the surds in descending order, the surds need to be expressed in the form of equiradical surds. As the LCM of 2, 3 and 4 is 12, so the order of the equiradical surds will be 12.

√10 = 10$^{1/2}$ = 10$^{6/12}$ = 1000000$^{1/12}$

 = $\sqrt[12]{1000000}$

∛25 = 25$^{1/3}$ = 25$^{4/12}$ = 390625$^{1/12}$ = $\sqrt[12]{390625}$

∜40 = 40$^{1/4}$ = 40$^{3/12}$ = 64000$^{1/12}$

= $\sqrt[12]{64000}$

As 1000000 > 390625 > 64000, the ascending order will be √10, ∛25, ∜40.

11 and 12 Grade Math

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