USE THE FOIL TO MULTIPLY BINOMIALS WITH RADICALS

Expand and simplify :

Problem 1 :

(1 + √2) (2 + √2)

Solution :

(1 + √2) (2 + √2)

= 2 + √2 + 2√2 + (√2)²

= 2 + 3√2 + 2

= 4 + 3√2

So, the answer is 4 + 3√2.

Problem 2 :

(2 + √3) (2 + √3)

Solution :

(2 + √3) (2 + √3)

= 4 + 2√3 + 2√3 + (√3)²

= 4 + 4√3 + 3

= 7 + 4√3

So, the answer is 7 + 4√3.

Problem 3 :

(√3 + 2) (√3 - 1)

Solution :

(√3 + 2) (√3 - 1)

= (√3)² - √3 + 2√3 - 2

= 3 + √3 - 2

= 1 + √3

So, the answer is 1 + √3.

Problem 4  :

(4 - √2) (3 + √2)

Solution :

(4 - √2) (3 + √2)

= 12 + 4√2 - 3√2 - (√2)²

= 12 + √2 - 2

= 10 + √2

So, the answer is 10 + √2.

Problem 5 :

(1 + √3) (1 - √3)

Solution :

(1 + √3) (1 - √3)

= 1 - √3 + √3 - (√3)²

= 1 - 3

= -2

So, the answer is -2.

Problem 6 :

(5 + √7) (2 - √7)

Solution :

(5 + √7) (2 - √7)

= 10 - 5√7 + 2√7 - (√7)²

= 10 - 3√7 - 7

 = 3 - 3√7

So, the answer is 3 - 3√7.

Problem 7 :

(√5 + 2) (√5 - 3)

Solution :

(√5 + 2) (√5 - 3)

= (√5)² - 3√5 + 2√5 - 6

= 5 - √5 - 6

= -1 - √5

So, the answer is -1 - √5.

Problem 8 :

(√7 - √3) (√7 + √3)

Solution :

(√7 -√3) (√7 + √3)

= (√7)² + (√7) (√3) - (√3) (√7) - (√3)²

= 7 - 3

= 4

So, the answer is 4.

Problem 9 :

(2√2 + √3) (2√2 - √3)

Solution :

(2√2 + √3) (2√2 - √3)

= 4(√2)² - 2√6 + 2√6 - (√3)²

= 8 - 3

= 5

So, the answer is 5.

Problem 10 :

(4 - √2) (3 - √2)

Solution :

(4 - √2) (3 - √2)

= 12 - 4√2 - 3√2 + (√2)²

= 12 - 7√2 + 2

= 14 - 7√2

So, the answer is 14 - 7√2.

Problem 11 :

Find the integer part of the sum (√2 + √3)⁴

Solution :

= (√2 + √3)⁴

= [(√2 + √3)²]² -----(1)

a = √2 and b = √3

a² + 2ab + b² = √2² + 2(√2)(√3) + √3²

= 2 + 2√6 + 3

= 5 + 2√6

Applying this expansion in (1), we get

= [5 + 2√6]²

= 5² + 2(5)(2√6) + (2√6)²

= 25 + 20√6 + 4(6)

= 25 + 20√6 + 24

= 49 + 20√6

Integers are 49 and 20, sum of them = 49 + 20

= 69

So, the answer is 69.

Problem 12 :

Solve for x in the following equation 2√(x² + 1) = 9

Solution :

2√(x² + 1) = 9

√(x² + 1) = 9/2

Squaring on both sides, we get

x² + 1 = (9/2)²

x² + 1 = 81/4

x² = (81/4) - 1

x² = (81 - 4)/4

x² = 77/4

x = √(77/4)

x = √77/2

Problem 13 :

Simplify the following 

(³√x⁷) (⁵√x⁵)

Solution :

= (³√x⁷) (⁵√x⁵)

= (x⁷)^1/3 . (x⁵)^1/5

= x^7/3 . x^5/5

= x^7/3 . x¹

= x^{(7/3) + 1}

= x^{(7 + 3)/3}

= x^10/3

Problem 14 :

Solve the equation 

2^x+2 = 4√2

Solution :

2^x+2 = 4√2

2^x+2 = 2²√2

2^x+2 = 2²2^1/2

2^x+2 = 2^{2 + 1/2}

2^x+2 = 2^5/2

Since the bases are equal, we have to equate the powers.

x + 2 = 5/2

x = (5/2) - 2

x = (5 - 4)/2

x = 1/2

Problem 15 :

If √4x + √9x² = 3, find the value of x.

Solution :

√4x + √9x² = 3

√9x² = 3 - √4x

Squaring on both sides, we get

(√9x²)² = (3 - √4x)²

9x² = (3 - √4x)²

√9x² = 3 - √4x

3x = 3 - √4x

3x - 3 = -√4x

√4x = 3 - 3x

4x = (3 - 3x)²

4x = 9 - 2(3)(3x) + (3x)²

4x = 9 - 18x + 9x²

9x²- 18x - 4x + 9 = 0

9x²- 22x + 9 = 0

To solve this quadratic equation, we have to use the quadratic formula.