Question

if x=root(2-root3) and x+1/x=k. find k square​

Answer 1

Step-by-step explanation:

Given :-

x = √(2-√3)

x + (1/x) = k

To find :-

Find the value of k ?

Solution:-

Given that :

x = √(2-√3)

On multiplying the numerator and the denominator with 2 then

=> x = [√2(2-√3)]/√2

=>x =[√(4-2√3)]/√2

=>x = [√{(√3)^1+(√1)^2-2(√3)(√1)}]/√2

It is in the form of a^2-2ab+b^2

Where a = √3 and b = √1 = 1

We know that

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

=> x =[√{√3-1}^2]/√2

=> x = (√3-1)/√2

On multiplying the numerator and the denominator with √2

=> x = (√3-1)√2/(√2×√2)

=>x = (√3-1)√2/2

=>x =(√(3×2)-√2)/2

=> x = (√6-√2)/2

=> 1/x = 1/[(√6-√2)/2]

=>1/x = 2/(√6-√2)

On multiplying the numerator and the denominator with √6+√2

=> 1/x = [2/(√6-√2)]×[(√6+√2)/(√6+√2)]

=> 1/x = 2(√6+√2)/[(√6)^2-(√2)^2]

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

=> 1/x = 2(√6+√2)/(6-2)

=> 1/x = 2(√6+√2)/4

On cancelling 2 then

=> 1/x =(√6+√2)/2

Now given that

x+ 1/x = k

=> (√6-√2)/2 + (√6+√2)/2 = k

=> (√6-√2+√6+√2)/2 = k

=> (√6+√6)/2 = k

=> 2√6/2 = k

=> k = √6

Answer:-

The value of k for the given problem is √6

Used formulae:-

• (a-b)^2 = a^2-2ab+b^2

• (a+b)(a-b)=a^2-b^2