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Math  /  Algebra

Questionx−1x=2 бил x2+1x2=?(2 01100 )x-\frac{1}{x}=2 \text { бил } x^{2}+\frac{1}{x^{2}}=?(2 \text { 01100 }) A. 1 B. -2 C. 6 D. 8 E. 2

Studdy Solution

STEP 1

1. We are given the equation x−1x=2 x - \frac{1}{x} = 2 .
2. We need to find the value of x2+1x2 x^2 + \frac{1}{x^2} .
3. We will use algebraic identities to solve the problem.

STEP 2

1. Square the given equation to find an expression for x2+1x2 x^2 + \frac{1}{x^2} .
2. Simplify the expression to find the desired value.

STEP 3

Start by squaring both sides of the equation x−1x=2 x - \frac{1}{x} = 2 :
(x−1x)2=22 \left(x - \frac{1}{x}\right)^2 = 2^2

STEP 4

Expand the left side using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2:
x2−2⋅x⋅1x+1x2=4 x^2 - 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 4
Simplify:
x2−2+1x2=4 x^2 - 2 + \frac{1}{x^2} = 4

STEP 5

Rearrange the equation to solve for x2+1x2 x^2 + \frac{1}{x^2} :
x2+1x2=4+2 x^2 + \frac{1}{x^2} = 4 + 2
x2+1x2=6 x^2 + \frac{1}{x^2} = 6
The value of x2+1x2 x^2 + \frac{1}{x^2} is:
6 \boxed{6}

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