Question

Which of the following is a like radical to RootIndex 3 StartRoot 7x EndRoot?

Answer 1

Answer:

A. 4 (RootIndex 3 StartRoot 7 x EndRoot)  or $4(\sqrt[3]{7x})$

Step-by-step explanation:

Given:

A radical whose value is, $r_1=\sqrt[3]{7x}$

Now, we need to find the like radical for $r_1$.

Let the like radical be $r_2$.

As per the definition of like radicals, like radicals are those that can be expressed as multiples of each other.

So, if two radicals $r_1\ and\ r_2$ are like radicals, then

$r_1 = n \times r_2$

Where, 'n' is a real number.

Here, $r_1=\sqrt[3]{7x}$

Now, let us check all the options .

Option A:

4 (RootIndex 3 StartRoot 7 x EndRoot) or $r_2=4\sqrt[3]{7x}$

Now, we observe that $r_2$ is a multiple of $r_1$ because

$r_2=4\times \sqrt[3]{7x}\\\\ r_2=4\times r_1..............(r_1=\sqrt[3]{7x})$

Therefore, option A is correct.

Option B:

StartRoot 7 x EndRoot or $r_2=\sqrt{7x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Option C:

x (RootIndex 3 StartRoot 7 EndRoot) or $r_2=x\sqrt[3]{7}$

As the term inside the cubic root is not same as that of $r_1$, this option is also incorrect.

Option D:

7 StartRoot x EndRoot or $r_2=7\sqrt{x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Therefore, the like radical is option (A) only.

Answer 2

Answer:

option a

Step-by-step explanation:

on edge