A PLUS B WHOLE SQUARE FORMULA WITH RADICALS

The formula for 

(a + b)² = a² + 2ab + b ²

Expand and simplify the following perfect squares :

Problem 1 :

(1 + √2)²

Solution :

(1 + √2)²

Using algebraic identities.

= (1)² + (√2)² + 2(1)(√2)

= 1 + 2 + 2√2

= 3 + 2√2

Problem 2 :

(√3 + 2)²

Solution :

(√3 + 2)²

Using algebraic identities.

= (√3)² + (2)² + 2(√3)(2)

= 3 + 4 + 4√3

= 7 + 4√3

Problem 3 :

(1 + √5)²

Solution :

(1 + √5)²

Using algebraic identities.

= (1)² + (√5)² + 2(1)(√5)

= 1 + 5 + 2√5

= 6 + 2√5

Problem 4 :

(1 + 3√2)²

Solution :

(1 + 3√2)²

Using algebraic identities.

= (1)² + (3√2)² + 2(1)(3√2)

= 1 + 9(2) + 6√2

= 1 + 18 + 6√2

= 19 + 6√2

Problem 5 :

(4√3 + 5)²

Solution :

(4√3 + 5)²

Using algebraic identities.

= (4√3)² + (5)² + 2(4√3)(5)

= 16(3) + 25 + 40(√3)

= 48 + 25 + 40(√3)

= 73 + 40(√3)

Problem 6 :

(2√11 + 3)²

Solution :

(2√11 + 3)²

Using algebraic identities.

= (2√11)² + (3)² + 2(2√11)(3)

= 4(11) + 9 + 12(√11)

= 44 + 9 + 12(√11)

= 53 + 12(√11)

Problem 7 :

1/(√7 + 1)

Solution :

1/(√7 + 1)

Conjugate of √7 + 1 is √7 - 1

= [1/(√7 + 1)] ⋅ [(√7 - 1)/(√7 - 1)]

= [(√7 - 1)/(√7 - 1)(√7 + 1)]

= [(√7 - 1)/(√7² - 1²)]

= [(√7 - 1)/(7 - 1)]

= (√7 - 1)/6

Problem 8 :

1/(√6 - 2)

Solution :

1/(√6 - 2)

Conjugate of √6 - 2 is √6 + 2

= [1/(√6 - 2)] ⋅ [(√6 + 2)/(√6 + 2)]

= [(√6 + 2)/(√6 - 2)(√6 + 2)]

= [(√6 + 2)/(√6² - 2²)

= (√6 + 2)/(6 - 4)

= (√6 + 2)/2

Problem 9 :

What is the square root of (8 + 2√15) ?

Solution :

= (8 + 2√15)

= √(5 + 3 + 2√15)

= √(√5² + √3² + 2√(5 x 3)

= √[√5² + √3² + 2√5 x √3]

= √(√5 + √3)²

= √5 + √3

Problem 10 :

√(5x + 3) = √(4x + 5)

Solution :

√(5x + 3) = √(4x + 5)

Squaring on both sides, we get

5x + 3 = 4x + 5

5x - 4x = 5 - 3

x = 2

So, the value of x is 2.

Problem 11 :

√(x - 1) + 5 = x - 2

Solution :

√(x - 1) + 5 = x - 2

√(x - 1) = x - 2 - 5

√(x - 1) = x - 7

Squaring on both sides, we get

(x - 1) = (x - 7)²

x - 1 = x² - 14x + 49

x - 1 = x² - 14x + 49

x² - 14x - x + 49 + 1 = 0

x² - 15x + 50 = 0

(x - 10)(x - 5) = 0

Equating each factor to 0, we get

x = 10 and x = 5.

Problem 12 :

Simplify :

√(49a² + 56a + 16)

Solution :

= √(49a² + 56a + 16)

= √((7a)² + 56a + 4²)

= √(7a)² + 2(7a)(4) + 4²)

= √(7a + 4)²

= 7a + 4

Problem 13 :

Simplify :

(5√2 + √3)(2√2 - 3√3)

Solution :

= (5√2 + √3)(2√2 - 3√3)

Using distributive property, we get

= (5√2) (2√2) - 5√2 (3√3) + √3(2√2) - 3√3√3

= 10(2) - 15√(2 x 3) + 2√(2 x 3) - 3(3)

= 20 - 15√6 + 2√6 - 9

= 20 - 9 - 15√6 + 2√6 

= 11 - 13√6

Problem 13 :

Solve

√(d² - 19) - 2d + 11 = 0

Solution :

√(d² - 19) - 2d + 11 = 0

√(d² - 19) = 2d - 11

Squaring on both sides, we get

(d² - 19) = (2d - 11)²

(d² - 19) = (2d)² - 2(2d)(11) + (11)²

d² - 19 = 4d² - 44d + 121

4d² - d² - 44d + 121 + 19 = 0

3d² - 44d + 140 = 0

3d² - 30d - 14d + 140 = 0

3d(d - 10) - 14(d - 10) = 0

(3d - 14)(d - 10) = 0

3d = 14 and d - 10 = 0

d = 14/3 and d = 10

Problem 14 :

Solve

√(3x - 5) = -5

Solution :

√(3x - 5) = -5

Squaring on both sides, we get

3x - 5 = (-5)²

3x - 5 = 25

3x = 25 + 5

3x = 30

x = 30/5

x = 10

So, the solution is 10.