SOLVE FOR UNKNOWN INVOLVING RADICALS

Solve x and y given that they are rational.

Example 1  :

(x+y√2) (2-√2)  =  1+√2

Solution  :

(x+y√2) (2-√2)  =  1+√2

By distribution, we get

2x-√2x+2√2y–2y  =  1+√2

By grouping, we get

-√2x + 2√2y + 2x – 2y  =  1+√2

(2x–2y) + (-x+2y)√2  =  1+√2

By equating corresponding terms, we get

2x-2y  =  1  ----(1)

-x+2y  =  1 ----(2)

Add (1) + (2), we get

2x-2y-x+2y  =  1+1

x  =  2

By applying the value of x in (2), we get

-2+2y  =  1

2y  =  3

y  =  3/2

So, x  =  2 and y  =  3/2

Example 2  :

(2 - 3√2) (x + y√2)  =  √2

Solution  :

(2-3√2) (x+y√2)  =  √2

By distribution, we get

2x+2√2y-3√2x–6y  =  √2

By grouping like terms

2x–6y-3√2x+2√2y  =  √2

(2x-6y) + √2(-3x+2y)  =  0+√2

Equate the corresponding terms.

2x–6y  =  0  ----(1)

-3x+2y  =  1  ----(2)

(1)+3(2)

2x-6y-9x+6y  =  0+3

-7x  =  0+3

x  =  -3/7

By applying the value of x in (1), we get

2(-3/7) - 6y  =  0

-6/7  =  6y

y  =  -1/7

So, the value of x is -3/7 and y  =  -1/7.

Example 3  :

(x+y√2) (3+√2)  =  1

Solution  :

(x+y√2) (3+√2)  =  1

By distribution, we get

3x+√2x+3y√2+2y  =  1

By grouping the like terms.

(3x+2y) + (x+3y)√2  =  1 + 0

3x + 2y  =  1  -----(1)

x + 3y  =  0 -----(2)

(1)-3(2)  ==>

3x+2y - 3x-9y  =  1-0

-7y  =  1

y  =  -1/7

We get,

Now, y  =  - 1/7

By applying the value of x in (1)

We get,

x + 3y  =  0

x + 3(-1/7)  =  0

x – 3/7  =  0

x  =  3/7

So, x  =  3/7 and y  =  - 1/7

Find rational a and b such that :

Example 4  :

(a+√2) (2-√2)  =  4-b√2

Solution  :

(a+√2) (2-√2)  =  4-b√2

By distribution, we get

2a-√2a+2√2–2  =  4-b√2

(2a-2)+(2-a)√  2  =  4-b√2

2a–2  =  4  -----(1)

2-a  =  -b -----(2)

From (1),

2a  =  6

a  =  3

By applying, the value of a in (1)

2-3  =  -b

-1  =  -b

So, a  =  3 and b  =  - 1

Example 5  :

(a + b√2)²  =  33 + 20√2

Solution  :

Given, (a + b√2)²  =  33 + 20√2

By using algebraic identity, we get

(a + b)²  =  a² + 2ab + b²

a² + 2ab√2 + 2b²  =  33 + 20√2

(a²+2b²)+ 2ab√2  =  33+20√2

a²+2b²  =  33  ----(1)

2ab  =  20

ab  =  10  ----(2)

By applying the value of b in (1),

We get,

a² + 2 (10/a)²  =  33

a² + 2 (100/a²)  =  33

a² + 200/a²  =  33

a⁴ + 200  =  33a²

a⁴ -33a²+ 200  =  0

Let t  =  a²

t² – 33t + 200  =  0

By factorization, we get

(t-25)(t-8)  =  0

t  =  25 and t  =  8

a²  =  25 and a²  =  8

a  =  5, a  =  2

Example 6  :

(x+y√2) (3-√2)  =  -4√2

Solution  :

By distribution, we get

3x-√2x+3y√2–2y  =  -4√2

(3x-2y)+(-x+3y)√2  =  0-4√2

Equating corresponding terms, we get

3x–2y  =  0  -----(1)

-x+3y  =  -4  -----(2)

(1)+3(2)

3x-2y-3x+9y  =  0-12

7y  =  -12

y  =  -12/7

By applying the value of y in (1), we get

3x-2(-12/7)  =  0

3x+24/7  =  0

3x  =  -24/7

x  =  -8/7

So, the value of x is -8/7 and y is -12/7

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