Question

solve the equation by making an appropriate substitution.
\\(\left(x - \dfrac{18}{x}\
ight)^2 - 4\left(x - \dfrac{18}{x}\
ight) - 21 = 0\\)

Explanation:

Step1: Let \( u = x - \frac{18}{x} \)

Substitute \( u \) into the equation, we get \( u^2 - 4u - 21 = 0 \)

Step2: Solve the quadratic equation \( u^2 - 4u - 21 = 0 \)

Factor the quadratic: \( (u - 7)(u + 3) = 0 \)
Set each factor to zero: \( u - 7 = 0 \) or \( u + 3 = 0 \)
So \( u = 7 \) or \( u = -3 \)

Step3: Substitute back \( u = x - \frac{18}{x} \)

Case 1: \( x - \frac{18}{x} = 7 \)

Multiply both sides by \( x \) ( \( x
eq 0 \) ): \( x^2 - 18 = 7x \)
Rearrange: \( x^2 - 7x - 18 = 0 \)
Factor: \( (x - 9)(x + 2) = 0 \)
So \( x = 9 \) or \( x = -2 \)

Case 2: \( x - \frac{18}{x} = -3 \)

Multiply both sides by \( x \) ( \( x
eq 0 \) ): \( x^2 - 18 = -3x \)
Rearrange: \( x^2 + 3x - 18 = 0 \)
Factor: \( (x + 6)(x - 3) = 0 \)
So \( x = -6 \) or \( x = 3 \)

Step4: Check for extraneous solutions

Substitute \( x = 9, -2, -6, 3 \) back into the original equation to verify. All satisfy the equation.

Answer:

\( x = -6, -2, 3, 9 \)