ADD AND SUBTRACT RADICALS

Add or subtract. Assume all variables are positive. Answers must be simplified.

Problem 1 :

5√6 + 3√6

Solution :

The terms which are inside the radical sign is the same, so both are like radicals.

= 8√6

Problem 2 :

4√20 - 2√5

Solution :

The terms which are inside the radical sign are not same, immediately cannot decide they are not like radicals, because we can decompose 20.

√20 = √(2 ⋅ 2 ⋅ 5)

= 2√5

4√20 = 4(2√5) ==> 8√5

4√20 - 2√5 = 8√5 - 2√5

= 6√5

Problem 3 :

3√(32x²) + 5x√8

Solution :

= 3√(32x²) + 5x√8

Decomposing 32x² and 8 as much as possible, we get

= 3√(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2x²) + 5x√(2 ⋅ 2 ⋅ 2)

= 3(2 ⋅ 2 ⋅ x) √2 + (5x ⋅ 2) √2

= 3(2 ⋅ 2 ⋅ x) √2 + (5x ⋅ 2) √2

= 12x√2 + 10x√2

=  22x√2

Problem 4 :

7√4x² + 2√25x - √16x

Solution : 

= 7√4x² + 2√25x - √16x

Decomposing the radicands as much as possible.

= 7√(2 ⋅ 2 ⋅ x ⋅ x) + 2√(5 ⋅ 5 ⋅ x) - √(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ x)

= 7⋅2⋅x + 2⋅5√x - (2⋅ 2)√x

= 14x + 10√x - 4√x

= 14x + 6√x

Problem 5 :

5∛x²y + ∛27x⁵y⁴ 

Solution : 

= 5∛x²y + ∛(3 ⋅ 3 ⋅ 3 x⁵y⁴)

= 5∛x²y + 3xy∛x²y

Factoring ∛x²y, we get

= (5 + 3xy) ∛x²y

Problem 6 :

3√9y³ - 3y√16y + √25y³

Solution : 

= 3√9y³ - 3y√16y + √25y³

= (3 ⋅ 3y)√y - (3y ⋅ 4)√y + 5y√y

= 9y√y - 12y√y+ 5y√y

= 14y√y - 12y√y

= 2y√y

Problem 7 :

√(248 + (√(51 + √169))

Solution : 

= √(248 + (√(51 + √169))

= √(248 + (√(51 + √13 ⋅ 13))

= √(248 + (√(51 + 13))

= √(248 + (√64))

= √(248 + (√8 ⋅ 8))

= √(248 + 8)

= √256

= √16 ⋅ 16

= 16

So, the value of the expression √(248 + (√(51 + √169)) is 16.

Problem 8 :

If a * b * c = √(a + 2) (b + 3) / (c + 1) then find the value of 6 * 15 * 3

Solution : 

Comparing the given numbers, we know that a = 6, b = 15, c = 3

6 * 15 * 3 = √(6 + 2) (15 + 3) / (3 + 1)

= √8 (18) / 4

= √144 / 4

= √(12 ⋅ 12) / 4

= 12/4

= 3

Problem 9 :

What will come in the place of question mark in each of the following :

i)  √(32.4/?) = 2

ii)  √86.49 + √(5 + ?²) = 12

Solution : 

i)  √(32.4/?) = 2

To remove the square root, we need to square on both sides. Let x be the unknown.

32.4/x = 2²

32.4/x = 4

x = 32.4/4

x = 8.1

So, the value of x is 8.1.

ii)  √86.49 + √(5 + ?²) = 12

Let x be the unknown.

√86.49 + √(5 + x²) = 12

√(9.3 ⋅  9.3) + √(5 + x²) = 12

9.3 + √(5 + x²) = 12

√(5 + x²) = 12.3 - 9.3

√(5 + x²) = 12.3 - 9.3

√(5 + x²) = 3

Squaring both sides

(5 + x²) = 3²

(5 + x²) = 9

x² = 9 - 5

x² = 4

x = √4

x = 2

Problem 10 :

If √1 + (x/144) = 13/12, find the value of x

Solution : 

√1 + (x/144) = 13/12

Squaring on both sides, we get

1 + (x/144) = (13/12)²

1 + (x/144) = 169/144

(x/144) = (169/144) - 1

(x/144) = (169 - 144)/144

(x/144) = 25/144

x = 25

So, the value of x is 25.

Problem 11 :

If x = 1 + √2 and y = 1 - √2, find the value of x² + y²

Solution : 

x = 1 + √2 and y = 1 - √2

x² + y² = (1 + √2)²  + (1 - √2)²

= 1 + √2²  + 2√2 + 1 + √2² - 2√2

= 1 + 2 + 1 + 2

= 6

Problem 12 :

√(10 + (√25 + (√108 + (√154 + (√225))))) is 

Solution : 

= √(10 + (√25 + (√108 + (√154 + (√225)))))

= √(10 + (√25 + (√108 + (√154 + (√15 ⋅ 15)))))

= √(10 + (√25 + (√108 + (√154 + 15))))

= √(10 + (√25 + (√108 + √169))

= √(10 + (√25 + (√108 + √13 ⋅ 13))

= √(10 + (√25 + (√108 + 13))

= √(10 + (√25 + (√121))

= √(10 + (√25 + (√11 ⋅ 11))

= √(10 + (√25 + 11))

= √(10 + √36)

= √(10 + √6 ⋅ 6)

= √(10 + 6)

= √16

= √4 ⋅ 4

= 4