Math, asked by ashok96248, 9 months ago

If x-1/x=√10 then. Find the value x^(2)+1/x^(2)

Answers

Answered by rsultana331
23

Step-by-step explanation:

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

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

= (10^2 - 2) (x - 1/x) (10)

= 980 (x -1/x) → Equation 1

Given x + 1/x = 10,

=> (x - 1/x)^2 = ((x+ 1/x)^2 - 2) - 2

= 10^2 - 2 - 2

= 96

=> (x- 1/x) = 96^ 1/2 → Equation 2

Therefore, from Equation 1 and 2,

(x^4 - 1/x^4) = 980 (x - 1/x)

= 980 (96)^1/2

= 980 (16x6)^1/2

= 980 (4)(6)^1/2

= 3920(6)^1/2.

Answered by Anonymous
2

Step-by-step explanation:

(1) The first is a much longer and strenuous approach. Most people would solve this problem using this method.

Let’s start with what you were given, x+1/x=10x+1/x=10 .

We can multiply both sides of the equation by xx to get the following: x2+1=10xx2+1=10x .

We want to make this a quadratic, so let’s move the 10x10x to the left side by subtracting 10x10x from both sides. We get x2−10x+1=0x2−10x+1=0 .

Using the quadratic formula, we get that x=5±26–√x=5±26 .

We can just plug the value of xx into x4−1/x4x4−1/x4 .

We have

(5+26–√)4–1(5+26√)4(5+26)4–1(5+26)4

=4801+19606–√−14801+19606√=4801+19606−14801+19606

=4801+19606–√−(14801+19606√⋅4801–19606√4801–19606√)=4801+19606−(14801+19606⋅4801–196064801–19606)

=4801+19606–√−4801–19606√23049601–23049600=4801+19606−4801–1960623049601–23049600

=4801+19606–√−(4801–19606–√)=4801+19606−(4801–19606)

=19606–√+19606–√=39206–√=19606+19606=39206 .

Your answer is 39206–√39206 .

(2) A much quicker method for more-experienced students:

x4−1x4x4−1x4

=(x2−1x2)(x2+1x2)=(x2−1x2)(x2+1x2)

=(x−1x)(x+1x)[(x+1x)2−2]=(x−1x)(x+1x)[(x+1x)2−2]

=980(x−1x)=980(x−1x) .

To get x−1xx−1x , note that

(x−1x)2(x−1x)2

=(x+1x)2−4=(x+1x)2−4

=96=96 .

We have, x−1x=±46–√x−1x=±46 .

Therefore, x4−1x4x4−1x4

=39206–√=39206 .

I hope that this helps! :)

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