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Math  /  Algebra

Question10) If x2+1x2=18x^{2}+\frac{1}{x^{2}}=18, find the value of x−1xx-\frac{1}{x}.

Studdy Solution

STEP 1

1. We are given the equation x2+1x2=18 x^2 + \frac{1}{x^2} = 18 .
2. We need to find the value of x−1x x - \frac{1}{x} .
3. We can use algebraic identities to transform the given expression into the desired form.

STEP 2

1. Use algebraic identities to express x2+1x2 x^2 + \frac{1}{x^2} in terms of x−1x x - \frac{1}{x} .
2. Solve for x−1x x - \frac{1}{x} .

STEP 3

Recall the identity for the square of a difference:
(x−1x)2=x2−2+1x2 \left( x - \frac{1}{x} \right)^2 = x^2 - 2 + \frac{1}{x^2}

STEP 4

Rearrange the identity to express x2+1x2 x^2 + \frac{1}{x^2} in terms of x−1x x - \frac{1}{x} :
x2+1x2=(x−1x)2+2 x^2 + \frac{1}{x^2} = \left( x - \frac{1}{x} \right)^2 + 2

STEP 5

Substitute the given value x2+1x2=18 x^2 + \frac{1}{x^2} = 18 into the rearranged identity:
18=(x−1x)2+2 18 = \left( x - \frac{1}{x} \right)^2 + 2

STEP 6

Solve for (x−1x)2 \left( x - \frac{1}{x} \right)^2 by subtracting 2 from both sides:
(x−1x)2=18−2 \left( x - \frac{1}{x} \right)^2 = 18 - 2 (x−1x)2=16 \left( x - \frac{1}{x} \right)^2 = 16

STEP 7

Take the square root of both sides to solve for x−1x x - \frac{1}{x} :
x−1x=±16 x - \frac{1}{x} = \pm \sqrt{16} x−1x=±4 x - \frac{1}{x} = \pm 4
The possible values of x−1x x - \frac{1}{x} are:
4and−4 \boxed{4} \quad \text{and} \quad \boxed{-4}

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