Question

If x = 3^1/4 + 3^-1/4 and y = 3^1/4 – 3^-1/4, then the value of 3(x² + y^2)^2
will be​

Answer 1

Answer:

value of 3(x²+y²)²=16

Explanation:

Given:-

x = 3^1/4 + 3^-1/4 and y = 3^1/4 – 3^-1/4,

To find:-

the value of 3(x² + y²)²

Solution:-

x=3^1/4 + 3^-1/4------(1)

y=3^1/4 - 3^-1/4------(2)

x+y=3^1/4 + 3^-1/4+3^1/4 - 3^-1/4

=>x+y=3^1/4+3^1/4

=>x+y=2(3^1/4) ------(3)

xy=(3^1/4+3^-1/4)(3^1/4-3^-1/4)

xy=[3^1/4]²-[3^-1/4]²

xy=3^2/4-3^-2/4

xy=3^1/2-3^-1/2----(4)

we know that (a+b)²=a²+2ab+b²

=>a²+b²=(a+b)²-2ab

=>x²+y²=(x+y)²-2xy

now from (3)&(4)

=>x²+y²=[2(3^1/4)]²-2[3^1/2-3^-1/2]

=>x²+y²=4(3^2/4)-2(3^1/2)-2(3^-1/2)

=>x²+y²=4(3^1/2)-2(3^1/2)-2(3^-1/2)

=>x²+y²=2(3^1/2-2(3^-1/2)

=>x²+y²=2[3^1/2-3^-1/2]

(x²+y²)²=[2(3^1/2-3^-1/2)]²

=>(x²+y²)²=4(3^2/2-2(3^1/2)(3^-1/2)+3^-2/2)

=>(x²+y²)²=4(3-2(1)+3^-1)

=>(x²+y²)²=4(3-2+1/3)

=>(x²+y²)²=4(1+1/3)

=>(x²+y²)²=4(4/3)

=>(x²+y²)²=16/3

Now value of 3(x²+y²)²

=>3(16/3)

=>16

Answer:-

value of 3(x²+y²)²=16

Used formulae:-

• (a+b)²=a²+2ab+b²
• (a-b)²=a²-2ab+b²
• a^-n=1/a^n

Answer 2

Answer:

64

Explanation:

x²=[3^(1/4)+3^(-1/4)]²

=3^(1/2)+3^(-1/2)+2

y²=[3^(1/4)-3^(-1/4)]²

=3^(1/2)+3^(-1/2)-2

3[x²+y²]²= 3[2(3^(1/2)+3^(-1/2)+2-2]²

3×2²[(3^(1/2)+3^(-1/2)]²

12×[3+ (1/3) + 2]

12×16 ÷ 3

64