Question

If x =√2−1/√2+1, y =√2+1/√2−1, then find the value of x2+5xy+y2

Answer 1

Answer:

x^2 + 5xy + y^2 = 5

Step-by-step explanation:

GIVEN:

x = √2 - 1 / √2 + 1

y = √2 + 1 / √2 - 1

to find: x^2 + 5xy + y^2

→ (√2 - 1 / √2 + 1)^2 + 5[ (√2 - 1 / √2 + 1)(√2 + 1 / √2 - 1) ] + (√2 + 1 / √2 - 1)^2

→(√2 - 1)^2 / (√2 + 1)^2 + 5[ (√2 - 1)(√2 + 1) / (√2+ 1)(√2 - 1) ] + (√2+1)^2/(√-1)^2

[ since ,

(a - b)^2 =a^2 + b^2 - 2ab;

(a + b)^2 = a^2 + b^2 +2ab;

a^2 - b^2 = (a + b)(a -b) ]

→ (√2)^2 + (1)^2 -2(√2)(1) / (√2)^2 + (1)^2 +2(√2)(1)  +   5 [ (√2)^2 - (1)^2 / (√2)^2     + (1)^2 + 2(√2)(1) +  (√2)^2 + (1)^2 +2(√2)(1) /  (√2)^2 + (1)^2 -2(√2)(1)  

→ 2 + 1 - 2√2 / 2+1+2√2 + 5(2-1/2-1) + 2+1+2√2/2+1-2√2

→ 3 - 2√2 / 3+ 2√2 + 5 + 3 +2√2 / 3-2√2

[ Taking LCM ]

→ ( 3- 2√2)(3-2√2) + 5(3 +2√2)(3-2√2) + (3+2√2)(3+2√2) / (3+2√2)(3-2√2)

→ (3-2√2) + 5 (3)^2 - (2√2)^2 + ( 3+ 2√2)

→ (3 - 2√2) + (3+2√2) + 5( 9- 8)

→ 3 - 2√2 + 3 + 2√2 + 5

→ 5