Adding and Subtracting Radicals
In mathematics, a radical is defined as an expression with a root. The term “radical” originated from the Latin word “radix”, which means “root”. A radical is an expression that involves any kind of roots, such as a square root, a cube root, a fourth root, or an nth root. A root is considered to be a square root if the index or degree of the radical is not mentioned. A radical has three components, namely, an index or degree, a radical symbol, and a radicand. The index number, or a degree of the radical, is the number written before it. Radicand is the number under the vinculum, where the vinculum is the horizontal line covering the number. “(√)” is the radical symbol. For example, “(n√x)” is read as “the nth root of x.”
In the figure given above, “3” is the co-efficient of the radical, “5” is the index, “8” is the radicand, and “(√)” is the radical symbol.
Some rules for solving Radicals
Product Rule: If “(n√a)” and “(n√b)”are real numbers and n is a natural number, then
n√(a × b) = n√a × n√b, where a, b ∈ R, n ∈ N.
Quotient rule: If “(n√a)” and “(n√b)”are real numbers and n is a natural number, then
n√(a /b) = n√a /n√b, where a, b ∈ R, n ∈ N, b≠0
Fractional exponent: If “(n√a)” is a real number, n is a natural number and m is an integer, then,
n√(am) = a(m/n)
Addition and Subtraction of Radicals
The addition and subtraction of radical expressions are similar to the addition and subtraction of like terms. When two radicals have the same radicand and index, they are said to be similar radicals or like radicals. We cannot add or subtract “unlike” radical terms, i.e., radicals having different radicands.
Let us consider another example to understand the concept better.
Example 1: Solve: [3√2 + 7√2].
Solution:
Here,
Since the radicands are the same, so the radicals are like radicals.
That means we can add both terms.
So,
[3√2 + 7√2] = (3 + 7)√2
= 10√2.
Example 2: Solve 2√27 + 6√16 – 5√3.
Solution:
The given expression is 2√27 + 6√16 – 5√3.
Here, the radicands of the three terms are different. But the radicands can be simplified further.
2√(33) + 6√(42) + 5√3 = 2 × 3√3 + 6 × 4 – 5√3
= 6√3 + 24 – 5√3
Now, we have like-radicals.
= 24 + (6 – 5)√3 = 24 + √3.
Hence,
2√27 + 6√16 – 5√3 = 24 + √3.
Steps while adding or subtracting radicals:
In order to add or subtract radicals, they must be “like radicals”. So, the coefficients of “like radicals” are added or subtracted to add or subtract radicals.
- First, break down the given radicals and then simplify each term.
- Now, find the like-radicals.
- Finally, add or subtract the like radicals by adding or subtracting their coefficients.
Example: Solve 21√2 + 10√5 −9√2 + 7√5.
Solution:
Given expression: 21√3 + 10√7 −9√3 + 7√7
The radicals are in their simplified form. So, identify the like-radicals.
= (21√3 – 9√3) + (10√7 + 7√7)
Now, add or subtract the coefficients of the like terms.
= 12√3 + 17√7
We cannot simplify any further as √3 and √7 are not like radicals since their radicands are different.
Hence, 21√3 + 10√7 −9√3 + 7√7 = 12√3 + 17√7.
Sample Problems
Problem 1: Solve 6√27 − 2√12 + √48.
Solution:
Given expression: 6√27 − 2√12 + √48
= 6√(3 × 3 × 3) − 2√(2 × 2 × 3) + √(4 × 4 × 3)
= 6 [√(3×3) × √3] − 2[√(2 × 2) × √3] + √(4 × 4) × √3 {Since, n√(a × b)= n√a × n√b}
= 6 × 3√3 − 2 × 2√3 + 4√3
= 18√3 − 4√3 + 4√3
= 18√3
Thus, 6√27 – 2√12 + √48 = 18√3.
Problem 2: Solve: 2 3√162 − 5 3√40 + 3 3√48 +7 3√135
Solution:
Given expression: 2 3√162 − 5 3√40 + 3 3√48 + 7 3√135
= 2 3√(27 × 6) − 5 3√(8 × 5) + 3 3√(8 × 6) + 7 3√(27 × 5)
= 2 [3√27 × 3√6] – 5[3√8 × 3√5] + 3[3√8 × 3√6] + 7[3√27 × 3√5] {Since, n√(a × b)= n√a × n√b}
= (2 × 3 × 3√6) − (5 × 2 × 3√5) + (3 × 2 × 3√6) + (7 × 3 × 3√5)
= 6 3√6 − 10 3√5 + 6 3√6 + 21 3√5
= 12 3√6 + 11 3√5
Hence, 2 3√162 − 5 3√40 + 3 3√48 +7 3√135 = 12 3√6 + 11 3√5
Problem 3: Solve: 4√(x6y) + 7√(xy2) − √(y2x).
Solution:
Given expression: 4√(x6y) + 7√(xy2) − √(y2x)
= 4 [√x6 × √y] + 7[√x × √y2] − [√y2 × √x] {Since, n√(a × b) = n√a × n√b}
= 4 [√(x3)2 × √y] + 7[√x × y] −[√y2 × √x] {Since, amn = (am)n}
= 4x3√y + 7y√x − y√x
= 4x3√y + 6y√x
Thus, 4√(x6y) + 7√(xy2) − √(y8x) = 4x3√y + 6y√x.
Problem 4: Solve: 13√11 −12√17 − 22√11 + 20√17 + 18√11.
Solution:
Given expression: 13√11 −12√17 − 22√11 + 20√17 + 18√11
= (13√11 – 22√11 + 18√11) + (-12√17 + 20√17)
= (31√11 – 22√11) + (8√17)
= 9√11 + 8√17.
Thus, 13√11 −12√17 − 22√11 + 20√17 + 18√11 = 9√11 + 8√17.
Problem 5: Solve: 2[√a + 5√(3b)]− 8[√a – 11√(3b)] + 14[3√a + 2√(3b)]
Solution:
Given expression: 2[√a + 5√(3b)]− 8[√a – 11√(3b)] + 14[3√a – 2√(3b)]
= 2√a + 10√(3b) -8√a + 88√(3b) + 42√a – 28√(3b)
= [2√a – 8√a + 42√a] + [10√(3b) + 88√(3b) – 28√(3b)]
= 36√a + 70√(3b)
Thus, 2[√a + 5√(3b)]− 8[√a – 11√(3b)] + 14[3√a + 2√(3b)] = 36√a + 70√(3b).
Problem 6: Solve: −11 5√(2xy) + 23 5√(2xy) − 5√(64xy) + 15 5√(2xy).
Solution:
Given expression: −11 5√(2xy) + 23 5√(2xy) − 5√(64xy) + 15 5√(2xy)
= −11 5√(2xy) + 23 5√(2xy) − 5√(32 × 2xy) +15 5√(2xy)
= −11 5√(2xy) − 5√32 × 5√(2xy) + 23 5√(2xy) + 15 5√(2xy)
= −11 5√(2xy) − 2 5√(2xy) + 38 5√(2xy)
= −13 5√(2xy) + 38 5√(2xy)
= 25 5√(2xy)
Hence, −11 5√(2xy) + 23 5√(2xy) − 5√(64xy) + 15 5√(2xy) = 25 5√(2xy)