Solve for x (x-40/x)^2-3(x-40/x)-18=0

${(x - \frac{40}{x})}^{2} - 3 (x - \frac{40}{x}) - 18 = 0$

Step 1

Factor the left side of the equation.

Let $u = x - \frac{40}{x}$ . Substitute $u$ for all occurrences of $x - \frac{40}{x}$ .

$u^{2} - 3 u - 18 = 0$

Factor $u^{2} - 3 u - 18$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 18$ and whose sum is $- 3$ .

$- 6, 3$

Write the factored form using these integers.

$(u - 6) (u + 3) = 0$

Replace all occurrences of $u$ with $x - \frac{40}{x}$ .

$(x - \frac{40}{x} - 6) (x - \frac{40}{x} + 3) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - \frac{40}{x} - 6 = 0$

$x - \frac{40}{x} + 3 = 0$

Step 3

Set $x - \frac{40}{x} - 6$ equal to $0$ and solve for $x$ .

Set $x - \frac{40}{x} - 6$ equal to $0$ .

$x - \frac{40}{x} - 6 = 0$

Solve $x - \frac{40}{x} - 6 = 0$ for $x$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x, 1, 1$

The LCM of one and any expression is the expression.

$x$

Multiply each term in $x - \frac{40}{x} - 6 = 0$ by $x$ to eliminate the fractions.

Multiply each term in $x - \frac{40}{x} - 6 = 0$ by $x$ .

$x \cdot x - \frac{40}{x} x - 6 x = 0 x$

Simplify the left side.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} - \frac{40}{x} x - 6 x = 0 x$

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{40}{x}$ into the numerator.

$x^{2} + \frac{- 40}{x} x - 6 x = 0 x$

Cancel the common factor.

$x^{2} + \frac{- 40}{x} x - 6 x = 0 x$

Rewrite the expression.

$x^{2} - 40 - 6 x = 0 x$

Simplify the right side.

Multiply $0$ by $x$ .

$x^{2} - 40 - 6 x = 0$

Solve the equation.

Factor $x^{2} - 40 - 6 x$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 40$ and whose sum is $- 6$ .

$- 10, 4$

Write the factored form using these integers.

$(x - 10) (x + 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 10 = 0$

$x + 4 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 10) (x + 4) = 0$ true.

$x = 10, - 4$

Step 4

Set $x - \frac{40}{x} + 3$ equal to $0$ and solve for $x$ .

Set $x - \frac{40}{x} + 3$ equal to $0$ .

$x - \frac{40}{x} + 3 = 0$

Solve $x - \frac{40}{x} + 3 = 0$ for $x$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x, 1, 1$

The LCM of one and any expression is the expression.

$x$

Multiply each term in $x - \frac{40}{x} + 3 = 0$ by $x$ to eliminate the fractions.

Multiply each term in $x - \frac{40}{x} + 3 = 0$ by $x$ .

$x \cdot x - \frac{40}{x} x + 3 x = 0 x$

Simplify the left side.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} - \frac{40}{x} x + 3 x = 0 x$

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{40}{x}$ into the numerator.

$x^{2} + \frac{- 40}{x} x + 3 x = 0 x$

Cancel the common factor.

$x^{2} + \frac{- 40}{x} x + 3 x = 0 x$

Rewrite the expression.

$x^{2} - 40 + 3 x = 0 x$

Simplify the right side.

Multiply $0$ by $x$ .

$x^{2} - 40 + 3 x = 0$

Solve the equation.

Factor $x^{2} - 40 + 3 x$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 40$ and whose sum is $3$ .

$- 5, 8$

Write the factored form using these integers.

$(x - 5) (x + 8) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x + 8 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

The final solution is all the values that make $(x - 5) (x + 8) = 0$ true.

$x = 5, - 8$

Step 5

The final solution is all the values that make $(x - \frac{40}{x} - 6) (x - \frac{40}{x} + 3) = 0$ true.

$x = 10, - 4, 5, - 8$

Solve for x (x-40/x)^2-3(x-40/x)-18=0

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