if value of x= 3root5+2root2 and value of y=3root5–2root2 then find (x²–y²)²​

1 thought on “if value of x= 3root5+2root2 and value of y=3root5–2root2 then find (x²–y²)²​”

1. Step-by-step explanation:

   Given :–

   x = 3√5+2√2

   y = 3√5-2√2

   To find :–

   Find the value of (x^2-y^2)^2 ?

   Solution:–

   Given that :-

   x = 3√5+2√2 ——-(1)

   y = 3√5-2√2 ——–(2)

   Method-1:–

   x = 3√5+2√2 ——-(1)

   On squaring both sides then

   => x^2 = (3√5+2√2)^2

   => x^2 = (3√5)^2+2(3√5)(2√2)+(2√2)^2

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

   =>x^2 = 45+12√10+8

   =>x^2 = 53+12√10———–(3)

   y = 3√5-2√2 ——–(2)

   On squaring both sides then

   => y^2 = (3√5-2√2)^2

   => y^2 = (3√5)^2-2(3√5)(2√2)+(2√2)^2

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

   =>y^2 = 45-12√10+8

   =>y^2 = 53-12√10———–(4)

   Now,

   (3)-(4)=>

   x^2-y^2

   =>( 53+12√10)-(53-12√10)

   => 53+12√10-53+12√10

   => (53-53)+(12√10+12√10)

   => 0+24√10

   =>x^2-y^2 = 24√10——–(5)

   Now,

   (x^2-y^2)^2

   From (5)

   => (24√10)^2

   => 24^2×(√10)^2

   => 576×10

   => 5760

   Method –2:–

   Given that :-

   x = 3√5+2√2 ——-(1)

   y = 3√5-2√2 ——–(2)

   On adding (1)&(2)

   x+y = 3√5+2√2+3√5-2√2

   => x+y = 3√5+3√5

   => x+y = (3+3)√5

   x+y = 6√5———–(3)

   on Subtracting (2) from (1)

   x-y = (3√5+2√2)-(3√5-2√2)

   =>x-y = 3√5+2√2-3√5+2√2

   => x-y = 2√2+2√2

   =>x-y = (2+2)√2

   =>x-y = 4√2——–(4)

   On multiplying (3)&(4)

   =>(x+y)(x-y)

   => (6√5)(4√2)

   => (x+y)(x-y) = (6×4×√5×√2)

   => x^2-y^2 = 24√10——(5)

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

   Now ,

   (x^2-y^2)^2

   => [24√10]^2

   => 24^2×(√10)^2

   => 576×10

   => 5760

   Answer:–

   The value of (x^2-y^2)^2 for the given problem is 5760

   Used formulae:–

   • (a+b)^2 = a^2+2ab+b^2

   • (a-b)^2 = a^2-2ab+b^2

   • (a+b)(a-b)=a^2-b^2

   • (ab)^m = a^m × a^n

   Reply