# EQUALITY OF SURDS

Suppose √k is irrational and a, b, c and d are rational

if

a + b√k = c + d√k

then

a = c and b = d

Solve for x and y given that they are irrational.

Problem 1 :

x + y √2 = 5 - 6√2

Solution :

x + y √2 = 5 - 6√2

Comparing the corresponding terms, we get x = 5 and y = -6

Problem 2 :

(x + y√2)(3 - √2) = - 2√2

Solution :

(x + y√2)(3 - √2) = - 2√2

Using distributive property,

3x - x √2 + 3y √2 - y √2 √2 = -2√2

3x - x √2 + 3y √2 - 2y = -2√2

(3x - 2y) + √2(-x + 3y) = -2√2

By comparing the corresponding terms,

3x - 2y = 0 -----(1)

-x + 3y = -2 -----(2)

While solving the system of equation, we can get the values of x and y

(1) + 3(2)

3x - 3x - 2y + 9y = 0 - 6

7y = -6

y = -6/7

Applying the value of y in (2), we get the value of x.

-x + 3(-6/7) = -2

-x - (18/7) = -2

-x = -2 + (18/7)

-x = (-14 + 18)/7

-x = 4/7

x = -4/7

Problem 3 :

Find rationals a and b such that

(a + 2√2)(3 - √2) = 5 + b√2

Solution :

(a + 2√2)(3 - √2) = 5 + b√2

Using distributive property,

3a - a√2 + 3(2√2) - 2√2√2 = 5 + b√2

3a - a√2 + 6√2 - 4 = 5 + b√2

(3a - 4) + √2(-a + 6) = 5 + b√2

Comparing the corresponding terms,

3a - 4 = 5 ----(1)

-a + 6 = b -----(2)

 From (1)3a = 5 + 43a = 9a = 9/3a = 3 Applying the value of a,-3 + 6 = bb = 3

So, the value of and b are 3 and 3 respectively.

Problem 4 :

Write √(1/7) in the form k√7. Then find the value of k.

Solution :

Comparing the coefficient of √7, we get k = 1/7

Problem 5 :

(a + b√2)2 = 33 + 20√2, find the values of a and b.

Solution :

(a + b√2)2 = 33 + 20√2

Using the algebraic identity

(a + b)2 = a2 + 2ab + b2

a2 + 2ab√2 + b2 √22 = 33 + 20√2

a2 + 2ab√2 + 2 b2 = 33 + 20√2

a2 + 2b2 = 33 ----(1)

2ab = 20

ab = 10

b = 10/a

Applying the value of b in (1), we get

a2 + 2(10/a)2 = 33

a4 + 200 = 33a2

a4 - 33a2+ 200 = 0

Let a2 = t

t2 - 33t + 200 = 0

(t - 25) (t - 8) = 0

t = 25 and t = 8

a2 = 25 and a2 = 8

a = 5 and a = 2√2

When a = 5, b = 10/5 ==> 2

When a = 2√2, b = 2√2/5

Problem 6 :

(a + b√2)2 = 41 - 24√2, find the values of a and b.

Solution :

(a + b√2)2 = 41 - 24√2

Using the algebraic identity

(a + b)2 = a2 + 2ab + b2

a2 + 2ab√2 + b2 √22 = 41 - 24√2

a2 + 2ab√2 + 2 b2 = 41 - 24√2

a2 + 2b2 = 41 ----(1)

2ab = -24

ab = -12

b = -12/a

Applying the value of b in (1), we get

a2 + 2(-12/a)2 = 41

a4 - 288 = 41a2

a4 - 41a2- 288 = 0

Let a2 = t

t2 - 33t - 200 = 0

(t - 32) (t + 9) = 0

t = 32 and t = -9

a2 = 32 and a2 = -9 (not possible)

a = 4√2

When a = 4√2, b = -12/4√2 ==> -3/√2

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