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Addition & Subtraction Of Square Roots

What are square roots?

A square root of a number is a part of the number that can be multiplied by itself to get back to its original value. For example, if we solve 2 x 2, we will get 4. Thus, we can say that 4 is the square of 2 and that 2 is the square root of 4. 

Some square roots can be easily identified, like 2 and 4, while others can be more complex. For instance, if we try to find the square root of 2, we will get an ”irrational” number. This means that there is no simple way to write it except for its radical form, which is $\sqrt2$.

How to simplify square roots?

Just like simplifying other mathematical expressions, we can simplify square roots in radical form by combining “like terms” or “like radicals” through addition or subtraction. 

One essential key when combining radicals by addition or subtraction is to look at the radicand. If the given square roots have the same radicand, they are like radicals, and combining them is feasible. If not, then you cannot combine the two square roots. This is like combining like terms in an algebraic expression. You can treat the square roots in radical form as variables and combine like terms by adding or subtracting their numerical coefficients and attaching their common variable. For example, to combine $\sqrt{x}$ and 2$\sqrt{x}$, we can have $\sqrt{x}$+ 2$\sqrt{xy}$ . The radical $\sqrt{x}$ has a numerical coefficient of 1 so we can visualize it as 1$\sqrt{xy}$ while the radical 2$\sqrt{x}$ has a numerical coefficient of 2. To combine, simply add their numerical coefficients. So, $\sqrt{x}+ 2\sqrt{x}=(1+2)\sqrt{x}=3\sqrt{x}$.

Simplifying square roots is the process of writing them in the most efficient and compact structure possible while maintaining the value of the original expression. It is a useful mathematical skill because it transforms complicated or difficult-to-read radical expressions into simpler ones. Below are some rules and steps in simplifying square roots expressions.

Simplifying square roots without variables

  1. Factor the radicand whenever possible. Find any factors of the radicand that is a perfect square. For example, in the expression $\sqrt{50}$ we can factor 50 such that one factor is a perfect square, that is, $\sqrt{50} = \sqrt{25 x 2}$. The perfect square here is 25 because 25 = 5 5. 
  2. Bring out the square root of the perfect square factor. Get the square root of the perfect square and bring it in front of the radical sign. Leave the remaining factor inside the radical. So, in the example above, $\sqrt{50} = \sqrt{25 x 2}=5\sqrt{2}$.

Simplifying square roots with variables

Let us assume that all variables represent nonnegative real numbers to make the simplification rules simpler. 

Simplifying square roots with variables is like simplifying square roots without variables. We can treat the variable as a factor. If the variable appears twice, such as x2, we can bring out the variable x to the front of the radical sign. If the variable appears three times, such as x3 , we can factor it as x2 × x , and then bring out the variable x from x2 to the front of the radical sign leaving the single x inside the radical sign.

In general,

  1. If the exponent of a variable is even, divide the exponent by two and write the result in front of the radical sign leaving no variable in the radicand. For example, in the radical expression $\sqrt{b^8}$, the exponent of the variable is even, which is 8. Hence, to find the square root of b8, simply bring out b in front of the radical sign, and its exponent will be half of the exponent of b in the radicand. Thus, $\sqrt{b^8}$= b4.
  2. If the exponent of the variable is odd, subtract 1 from the exponent, divide it by two, and write the result in front of the radical sign, leaving the variable inside the radical sign without an exponent. For example, in the radical expression $\sqrt{c^7}$, the exponent of the variable is odd, which is 7. Hence, to find the square root of c7, we must factor it as c6 ×c and bring out the square root of  c6 in front of the radical sign leaving just c inside the radical sign Thus, $\sqrt{c^7}= c^3\sqrt{c}$.

Example 

Simplify $\sqrt{45 x^5 y^6}$.

Solution

$\sqrt{45 x^5 y^6}=\sqrt{(9 × 5)(x^4 × x){y^6}}\sqrt{45 x^5 y^6}=(3)(x^2)(y^3)\sqrt{(5)(x)}\sqrt{45 x^5 y^6}=3x^{2}y^{3}\sqrt{(5)(x)}$

How to add square roots?

In adding square roots, we combine the like radicals, and the unlike radicals are written as they are. So, if we add 4$\sqrt{2}$ and 2$\sqrt{2}$, it is possible because they are like radicals. We can think of this as follows: “If we have four square roots of 2, and we add that to two square roots of 2, then how many square roots of 2 in total do we have?” Well, four of them plus two of them are a total of six of them. So, we get 6$\sqrt{2}$. In other words, we simply added their coefficient and attached it to their common radical.

If the radicands of the square roots are not the same, then they are not like radicals, and we cannot combine them. So, if we have $\sqrt{2}$ and $\sqrt{3}$​​, we cannot add or subtract one of them from the other. The radicands are different (one is 2 and the other is 3), so they are not like radicals and cannot be combined.

It is not always true that square roots with different radicands cannot be added or subtracted. Sometimes, we need to simplify radicals to end up with like radicals. For example, it appears that $\sqrt{3}+\sqrt{27}$ cannot be simplified since the radicands of the terms are different. But $\sqrt{27}$ can be simplified as 3$\sqrt{3}$. Now, $\sqrt{3}$ and 3$\sqrt{3}$ are like radicals and can be therefore combined as 4$\sqrt{3}$.

Even the radicands are different; sometimes, one or more radicals in an expression can be rewritten such that the radicands will be the same. So, it is important to look out for opportunities to rewrite the radicals before concluding that they are not like radicals and cannot be combined.

Example #1

What is the sum of 2$\sqrt{5}$ and 5$\sqrt{5}$?

Solution

Square Roots Addition ProcessStep-by-step Explanation
2$\sqrt{5}+ 5\sqrt{5}$Set up the addition.
7$\sqrt{5}$Combine the like radicals into a single term by adding their coefficients.
Therefore, the sum of 2$\sqrt{5}$ and 5$\sqrt{5}$ is 7$\sqrt{5}$.

Example #2

Find the sum of 2$\sqrt{x}$ and 2$\sqrt{y}$?

Solution

Square Roots Addition ProcessStep-by-step Explanation
2$\sqrt{x}+ 2\sqrt{y}$Set up the addition.
2$\sqrt{x}+ 2\sqrt{y}$Notice that the two terms are unlike radicals, so we cannot combine them into a single term. 
Therefore, the sum of $2\sqrt{x}\:and\:2\sqrt{y}$ is 2$\sqrt{x}+ 2\sqrt{y}$.

Example #3

Add $x\sqrt{3}\:and\:2\sqrt{12}$?

Solution

Square Roots Addition ProcessStep-by-step Explanation
$x\sqrt{3} + 2\sqrt{12}$Set up the addition.
$x\sqrt{3} + 2\sqrt{4×3}$Factor the radicands whenever possible such that at least one factor is a perfect square. In this case, the radicand 12 can be factored as 4 x 3, where 4 is a perfect square.
$x\sqrt{3} + (2×2)\sqrt{3}$Get the square root of 4 and bring it in front of the radical sign.
$x\sqrt{3} + 4\sqrt{3}$Simplify any multiplication problem in the expression.
$(x+4)\sqrt{3}$Combine the like radicals into a single term by adding their coefficients.
Therefore, the sum of $x\sqrt{3}\:and\:2\sqrt{12}\:is\:(x + 4)\sqrt{3}$.

Example #4

What is the result of adding $\sqrt{9a^{2}bc^{5}}- 2\sqrt{ab^2}+ 7\sqrt{8a^{2}b}\:and\:3b\sqrt{a^3}- \sqrt{b^{3}c}$?

Solution

Square Roots Addition ProcessStep-by-step Explanation
$\sqrt{9a^{2}bc^{5}}- 2\sqrt{ab^2}+ 7\sqrt{8a^{2}b} + 3b\sqrt{a^3}- \sqrt{b^{3}c}$Set up the addition.
$\sqrt{9a^{2}bc^{4}c}- 2\sqrt{ab^2}+ 7\sqrt{4(2)a^{2}b} + 3b\sqrt{a^{2}a}- \sqrt{b^{2}bc}$Factor the radicands whenever possible such that there are perfect squares as factors 
$(3ac^{2}\sqrt{bc}- 2b\sqrt{a}+ 7(2)a\sqrt{2b})+(3ab\sqrt{a}-b\sqrt{bc})$Get the square roots of the perfect squares and bring them to the front of the radical sign
$3ac^2\sqrt{bc}- 2b\sqrt{a}+ 14a\sqrt{2b}+ 3ab\sqrt{a}- b\sqrt{bc}$Remove the grouping symbols 
$3ac^2\sqrt{bc}- b\sqrt{bc}- 2b\sqrt{a}+ 3ab\sqrt{a}+ 14a\sqrt{2b}$Arrange the terms such that like radicals are placed beside each other
$(3ac^2-b)\sqrt{bc} + (-2b + 3ab)\sqrt{a} + 14a\sqrt{2b}$Combine like radicals by adding their coefficients.
Therefore, the result of adding $\sqrt{9 a^2 b c^5} – 2\sqrt{ab^2} + 7\sqrt{8a^{2}b}\:and\:3b\sqrt{a^3} – \sqrt{b^{3}c}\:is\:(3ac^2 – b) \sqrt{bc} + (-2b + 3ab) \sqrt{a} + 14a\sqrt{2b}$.

How to subtract square roots?

The process for subtracting square roots is the same as the process for adding them. The only difference is that when subtracting one square root term from another, you must change the signs of each term in the expression being subtracted and then combine like radicals.

But why do we have to change the signs of a mathematical expression when being subject to subtraction? Let us take an example of two numbers, 2 and −1. Suppose if we must subtract −1 from 2, we write it as 2−(−1).  We know that the product of two positive signs or two negative signs is positive, and the product of two unlike signs is negative. In the example: 2−(−1) = 2 + 1 = 3. If we do not change the sign, we will have 2 – 1 = 1, which is a completely different result. So, changing the signs of the subtrahend when subtracting expressions is necessary. In other words, the subtraction symbol (−) must be distributed to each term of the subtrahends before combining mathematical terms.

Example #1

Subtract -$\sqrt{3}$ from $3\sqrt{3}$.

Solution

Square Roots Subtraction ProcessStep-by-step Explanation
$3\sqrt{3}-(-\sqrt{3})$Set-up subtraction.
$3\sqrt{3} + \sqrt{3}$Distribute the minus sign to the term in the subtrahend. Thus, -$\sqrt{3}$ will become $\sqrt{3}$. Then, proceed like in the process of addition.
$4\sqrt{3}$Combine the like radicals into a single term by adding their coefficients. Note that if a radical has no written coefficient, it is understood that it has 1 as its coefficient.
Therefore, subtracting -$\sqrt{3}$ from $3\sqrt{3}$ will result to $4\sqrt{3}$.

Example #2

What is the result of subtracting $\sqrt{2y}$ from $\sqrt{2x}$?

Solution

Square Roots Subtraction ProcessStep-by-step Explanation
$\sqrt{2x}-(\sqrt{2y})$Set-up subtraction.
$\sqrt{2x}-\sqrt{2y}$Distribute the minus sign to the term in the subtrahend. Thus, $\sqrt{2y}$ will become -$\sqrt{2y}$. Then, proceed like in the process of addition.
$\sqrt{2x}-\sqrt{2y}$Notice that the two terms are unlike radicals, so we cannot combine them into a single term.
Therefore, the result of subtracting $\sqrt{2x}\:from\:\sqrt{2y}\:is\:\sqrt{2x}-\sqrt{2y}$.

Example #3

Find the difference between $2\sqrt{8x^2}\:and\:3x\sqrt{32}$?

Solution

Square Roots Subtraction ProcessStep-by-step Explanation
$2\sqrt{8x^2} – (3x\sqrt{32})$Set up the subtraction.
$2\sqrt{4(2)x^2} – ( 3x \sqrt{16(2)})$Factor the radicands whenever possible such that there are perfect squares as factors.
$2(2)(x)\sqrt{2} -( 3x (4)\sqrt{2})$Get the square roots of the perfect squares and bring them to the front of the radical sign
$4x\sqrt{2} – (12x\sqrt{2})$Simplify any multiplication problem in the expression.
$4x\sqrt{2} – 12x\sqrt{2}$Distribute the minus sign to the term in the subtrahend. Thus, $12x\sqrt{2}$ will become -$12x\sqrt{2}$. Then, proceed like in the process of addition.
$-8x\sqrt{2}$Combine the like radicals into a single term by adding their coefficients. 
Therefore, the sum of $x\sqrt{3}\:and\:2\sqrt{12}\:is\:(x + 4)\sqrt{3}$.

Example #4

Solve  $(\sqrt{9abc^{7}}- 2\sqrt{ab^{3}c}+ 7a\sqrt{a^{2}b}) – (3\sqrt{a^{4}b} – \sqrt{b^{2}c^{8}})$.

Solution

Square Roots Subtraction ProcessStep-by-step Explanation
$(\sqrt{9abc^{7}}- 2\sqrt{ab^{3}c}+ 7a\sqrt{a^{2}b}) – (3\sqrt{a^{4}b} – \sqrt{b^{2}c^{8}})$ Set up the subtraction.
$(\sqrt{9abc^{6}c}- 2\sqrt{ab^{2}bc}+ 7a\sqrt{a^{2}b}) – (3\sqrt{a^{4}b} – \sqrt{b^{2}c^{8}})$ Factor the radicands whenever possible such that there are perfect squares as factors.
$(3c^3\sqrt{abc}- 2b\sqrt{abc}+ 7a(a)\sqrt{b})- (3(a^2)\sqrt{b} – b^{2}c^{4})$Get the square roots of the perfect squares and bring them to the front of the radical sign
$(3c^3\sqrt{abc}- 2b\sqrt{abc}+ 7a^{2}\sqrt{b})- (3a^{2}\sqrt{b} – b^{2}c^{4})$ Simplify any multiplication problem in the expression.
$(3c^3\sqrt{abc}- 2b\sqrt{abc}+ 7a^{2}\sqrt{b})- 3a^{2}\sqrt{b} – b^{2}c^{4}$Distribute the minus sign to the term in the subtrahend. Thus, $3a^{2}\sqrt{b}$ will become -$3a^{2}\sqrt{b}\:and\: -b^2c^4$ will become $b^{2}c^{4}$.  Then, proceed like in the process of addition.
$(3c^{3} – 2b)\sqrt{abc} + 4a^{2}\sqrt{b} + b^{2}c^{4}$Combine the like radicals into a single term by adding their coefficients. 
Therefore, solving $(\sqrt{9abc^{7}}- 2\sqrt{ab^{3}c}+ 7a\sqrt{a^{2}b}) – (3\sqrt{a^{4}b} – \sqrt{b^{2}c^{8}})$ will give us
$(3c^{3} – 2b)\sqrt{abc} + 4a^{2}\sqrt{b} + b^{2}c^{4}$

How to solve problems involving addition and subtraction of square roots?

Problem #1

Mark and John both have a square garden. The area of Marks’s garden is 18x + 9 square feet, while John’s garden has an area of 8x + 4 square feet. What is the difference between the length of the sides of their garden?

Solution

ProcessStep-by-step explanation
$T= \sqrt{M} – \sqrt{J}$Set up the working formula. To get the length of the sides of a square, we must get the square root of its area. Let T be the difference of the length of the sides of the terrariums, let M be the area of Mark’s garden, and let J be the area of John’s garden.
$T= \sqrt{18x + 9} – \sqrt{8x  + 4}$Plugin the values into the working formula.
$T= \sqrt{9(2x + 1)} – \sqrt{4(2x + 1)}$Factor the radicands if possible. In this case, we can factor 18x + 9 as 9 (2x + 1) and 8x + 4 as 4 (2x + 1)
$T= 3\sqrt{(2x + 1)} – \sqrt{2(2x + 1)}$Get the square root of 9 and 4, so it can get out of the radical sign. So,  $\sqrt{9 (2x + 1)} – \sqrt{4 (2x + 1)} = 3\sqrt{(2x + 1)} – 2\sqrt{(2x + 1)}$
$T= \sqrt{(2x + 1)}$Combine the like radicals into a single term by adding their numerical coefficients.
Therefore, the difference between the length of the sides of Marks’s and John’s garden is $\sqrt{(2x + 1)}$ inches.

Problem #2

Find the perimeter of an isosceles triangle whose legs measures $2\sqrt{2ab}$ inches and a base that is $\sqrt{8a^{3}b}$ inches.

Solution

ProcessStep-by-step explanation
P =  2L+B.Set- up the formula for the perimeter of an isosceles triangle. The perimeter of an isosceles triangle can be solved by adding the length of its two congruent sides and the length of its base. Let P be the perimeter, let L be the measure of the legs, and let B be the measure of the base.
$P= 2(2\sqrt{2ab} )+ \sqrt{8a^{3}b}$Plugin the values to the working formula
$P= 4\sqrt{2ab} + \sqrt{8a^{3}b}$Simplify the multiplication problem in the first term of the expression on the right-hand side of the equation.
$P= 4\sqrt{2ab} + \sqrt{4(2)a^{2}ab}$ Factor the radicands whenever possible such that there are perfect squares as factors.
$P= 4\sqrt{2ab} + 2a\sqrt{2ab}$Get the square roots of the perfect squares and bring them out to the front of the radical sign.
$P=(4 + 2a)\sqrt{2ab}$Combine the like radicals into a single term by adding their numerical coefficients.
Therefore, the perimeter of the isosceles triangle is $(4 + 2a)\sqrt{2ab}$ inches.

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