Question

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If $\rm \: x = \frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } + \frac{ \sqrt{3} - 1 }{ \sqrt{3} + 1 } + \frac{ \sqrt{3} - 2 }{ \sqrt{3} + 2}$ then, find the value of $\sf {x}^{2} + \bigg( \frac{39}{x} \bigg)^{2}$

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Answer 1

Step-by-step explanation:

Given :-

x =(√3+1)/(√3-1) +(√3-1)/(√3+1) +(√3-2)/(√3+2)

To find :-

Find the value of x^2+(39/x)^2 ?

Solution :-

Given that :

x = (√3+1)/(√3-1) + (√3-1)/(√3+1) + (√3-2)/(√3+2)

We know that

The Rationalising factor of a+√b is a-√b

The Rationalising factor of a-√b is a+√b

It has three parts

(I)

On taking (√3+1)/(√3-1)

The Rationalising factor of √3-1 = √3+1

On Rationalising the denominator then

=> [(√3+1)/(√3-1)]×[(√3+1)/(√3+1)]

=> [(√3+1)(√3+1)]/[(√3-1)(√3+1)]

=> [(√3+1)^2]/[(√3)^2-1^2]

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

=> (√3+1)^2/(3-1)

=> (√3+1)^2/2

=> [(√3)^2+2(√3)(1)+1^2]/2

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

=> (3+1+2√3)/2

=> (4+2√3)/2

=>2(2+√3)/2

=> 2+√3-----------(1)

(ii)

On taking (√3-1)/(√3+1)

The Rationalising factor of √3+1 is √3-1

On Rationalising the denominator then

=> [(√3-1)/(√3+1)]×[(√3-1)/(√3-1)]

=> [(√3-1)(√3+1)]/[(√3+1)(√3+1)]

=> [(√3-1)^2]/[(√3)^2-1^2]

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

=> (√3-1)^2/(3-1)

=> (√3-1)^2/2

=> [(√3)^2-2(√3)(1)+1^2]/2

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

=> (3+1-2√3)/2

=> (4-2√3)/2

=>2(2-√3)/2

=> 2-√3-----------(2)

(iii)

On taking (√3-2)/(√3+2)

The Rationalising factor of √3+2 is √3-2

On Rationalising the denominator then

=> [(√3-2)/(√3+2)]×[(√3-2)/(√3-2)]

=> [(√3-2)(√3-2)]/[(√3+2)(√3-2)]

=> [(√3-2)^2]/[(√3)^2-2^2]

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

=> (√3-2)^2/(3-4)

=> (√3-2)^2/(-1)

=> [(√3)^2-2(√3)(2)+2^2]/(-1)

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

=> (3+4-4√3)/(-1)

=> (7-4√3)/(-1)

=> 4√3-7 -----------(3)

On adding (1),(2)&(3)

x = (2+√3)+(2-√3)+(4√3-7)

=>x = 4+4√3-7

=> x = 4√3-3 ---------(4)

On squaring both sides then

=> x^2 = (4√3-3)^2

=> x^2=(4√3)^2-2(4√3)(3)+3^2

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

=> x^2=48-24√3+9

=>x^2 = 57-24√3--------(5)

We have

=>1/x = 1/(4√3-3)

The Rationalising factor of 4√3-3 is 4√3+3

On Rationalising the denominator then

=> 1/x =[1/(4√3-3)]×[(4√3+3)/(4√3+3)]

=> 1/x = (4√3+3)/[(4√3-3)(4√3+3)]

=> 1/x = (4√3+3)/[(4√3)^2-3^2]

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

=> 1/x = (4√3+3)/(48-9)

=> 1/x = (4√3+3)/39 -----------(5)

On squaring both sides

=> (1/x)^2 = [(4√3+3)/39]^2

=> (1/x^2) = (4√3)^2+2(4√3)(3)+3^2)/(39)^2

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

=> (1/x)^2 = (48+9+24√3)/39^2

=> (1/x)^2 = (57+24√3)/39^2 -----(6)

now,

(39/x)^2 = (39^2)(57+24√3)/39^2

=> (39/x)^2 = 57+24√3--------(7)

Now on adding (5)&(7)

=> x^2+(39/x)^2

=> 57-24√3+57+24√3

=>57+57

=> 114

Answer:-

The value of x^2+(39/x)^2 for the given problem is 114

Used formulae:-

• The Rationalising factor of a+√b is a-√b
• The Rationalising factor of a-√b is a+√b

• (a+b)^2 = a^2+2ab+b^2

• (a+b)(a-b) = a^2-b^2

• (a-b)^2 = a^2-2ab+b^2

Answer 2

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