# How to add Square Roots with Variables?

The **Square root **of a number is the factor of a number when multiplied by itself gives the original number. Simply it was an inverse operation of squaring a number. It is represented by the symbol **√**. This symbol is called **Radical**. The term under the square root is called the **radicand**. Instead of representing the square root with a symbol, It can also be represented in numeric form by representing 1/2 as an exponent for a number.

**Examples for symbolic notation: **√2, √(5x), √(8x^{3}).

In the above examples 2, 5x, 8x^{3} are radicands.

**Examples for numeric representation of square root:** 4^{(1/2) }and 36^{(1/2)}.

The square root of a number is the inverse operation of squaring a number. The square of a number is obtained by multiplying the number by itself but the square root of a number is the factor of a number when multiplied by itself gives the original number.

### What is Square root with variables?

Square root with variables is nothing but the term that holds variables also. i.e., Radicand contains the variables. To perform any arithmetic operation between two expressions having square roots with variables, the radicands should be the same for both expressions. Let’s look into an example of a square root with variables-

Example:√(8x).

In the above expression, radicand holds variable x along with constant 8. So it can be called a square root with variable.

### How to add Square Roots with variables?

Addition between Square roots with variables can only be done if and only if there radicands are same. Below are the steps that need to be followed while adding square roots with variables.

Steps to add Square Roots with variables:

- Simplify each radical.
- Identify similar radicals.
- Perform addition between like radicals by adding their coefficients.

Let’s look into a few examples of how to add the square roots with variables and how to make radicands the same.

### Sample Problems

**Problem 1: Solve 4√(8x ^{3}) + 2x√(2x).**

**Solution:**

4√(8x

^{3}) + 2x√(2x) = 4 √(2×2×2×x×x×x) + 2x √2xThere are three 2’s & x’s inside square root. Each of those two 2’s and x’s can be taken out from square root as one number i.e., 2, x

=4×2×x √(2x) + 2×x √(2x)

=8x √(2x) + 2x √(2x)

=10x √(2x)

Hence 4√(8x^{3}) + 2x√(2x) = 10x √(2x)

**Problem 2: Perform addition between √(x ^{5}) and √(x^{9}).**

**Solution:**

√x

^{5}+ √x^{9}= √(x^{2}. x^{2}. x) + √(x^{2}.x^{2}.x^{2}.x^{2}x)=(x×x)√x+(x×x×x×x)√x

=x

^{2}√x+x^{4}√x=(x

^{2}+x^{4})√x

√(x^{5}) and √(x^{9}) = (x^{2}+x^{4})√x

**Problem 3: Solve √(100x)+√(64x).**

**Solution:**

√100x + √64x = √102x + √4

^{2}.2^{2}.x=10√x+4×2√x

=10√x+8√x

=18√x

Hence,

√(100x)+√(64x) = 18√x

**Problem 4: Perform addition between 2xy√(16x ^{5}y^{7}) and √(x^{7}y^{9}).**

**Solution:**

2xy 16x 5y

^{7} + x^{7}y^{9} = 2xy4^{2}.x^{2}.x^{2}.y^{2}.y^{2}.y^{2}.x.y + x^{2}.x^{2}.x^{2}.y^{2}.y^{2}.y^{2}.y^{2}.x.y=2xy×4×x

^{2}×y^{3}√(xy) +x^{3}y^{4}√(xy)=8x

^{3}y^{4}√(xy) +x^{3}y^{4}√(xy)=9x

^{3}y^{4}√(xy)

So, 2xy√(16x^{5}y^{7}) and √(x^{7}y^{9}) =9x^{3}y^{4}√(xy)

**Problem 5: Solve 2 √(5x ^{3}) + √(x) + √(75x^{3}).**

**Solution:**

2 √5x

^{3}+ √x + √75x^{3}= 2 √(5.x^{2}.x) + √x + √(5^{2}.5.x^{2}.x)= 2x √(5x) + √x + 5x √(5x)

= (2x √(5x) + 5x √(5x) ) + √x

= 7x √(5x) +√(x)

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