Question

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

Answer 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