Math, asked by ps6817593, 3 months ago

If x=2-√3, find the value of x^2+1/x^2.​

Answers

Answered by kumarijyotisp18
3

Answer:

x=2-√3

1/x = 2+√3

x^2+1/x^2

(2-√3)^2 +( 2+√3)^2

4-4√3+3 +4+4√3+3

4+3+4+3

14

Answered by Salmonpanna2022
4

Answer:

Hence, the value of x^2 + 1/x^2 is 14.

Step-by-step explanation:

Given:-

x = 2 - √3

To find out:-

Value of x^2 + 1 / x^2

Solution:-

We have,

x = 2 - √3

∴ 1/x = 1/2-√3

The denomination = 2-√3

We know that

Rationalising factor of a-√b = a+√b

So, the rationalising factor of 2-√3 = 2+√3

On rationalising the denominator then

1/x = [1/(2-√3)]×[(2+√3)/2+√3)]

1/x = [1(2+√3)]/[(2+√3)(2+√3)]

1/x = (2+√3)/[(2-√3)(2+√3)]

Now, we will apply algebraic Identity in denominator because the denominator is in the form of

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

Where we have to put a = 2 and b = 3, we get

1/x = (2+√3)/[(2)^2 - (√3)^2]

1/x = (2+√3)/(4 - 3)

1/x = (2+√3)/1

1/x = 2+√3

Now, we have to add both values x and 1/x, we get

∴ x + 1/x = 2-√3+2+√3

3 will be cancel out

x + 1/x = 2 + 2

x + 1/x = 4

Now, squaring on both, we get

(x +1/x)^2 = (4)^2

Using algebraic Identity,

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

Where we have to put a = x and b = 1/x , we get

➟ x^2 + 2(x)(1/x) + (1/x)^2 = 16

➟ x^2 + 2 + (1/x)^2 = 16

➟ x^2 + 2 + 1/x^2 = 16

➟ x^2 + 1/x^2 = 16 - 2

➟ x^2 + 1/x^2 = 14

Answer:-

Hence, the value of x^2 + 1/x^2 is 14.

Used formulae:-

Rationalising factor of a-√b = a+√b

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

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

  • I hope it's help you. ☺️
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