Find the Intersection of the Functions f(x)=2x-40/x , f(a)=-2

$f (x) = 2 x - \frac{40}{x}$ , $f (a) = - 2$

Step 1

Substitute $- 2$ for $f (x)$ .

$- 2 = 2 x - \frac{40}{x}$

Step 2

Solve for $x$ .

Rewrite the equation as $2 x - \frac{40}{x} = - 2$ .

$2 x - \frac{40}{x} = - 2$

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.

$1, x, 1$

The LCM of one and any expression is the expression.

$x$

Multiply each term in $2 x - \frac{40}{x} = - 2$ by $x$ to eliminate the fractions.

Multiply each term in $2 x - \frac{40}{x} = - 2$ by $x$ .

$2 x \cdot x - \frac{40}{x} x = - 2 x$

Simplify the left side.

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$2 (x \cdot x) - \frac{40}{x} x = - 2 x$

Multiply $x$ by $x$ .

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

Cancel the common factor of $x$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Solve the equation.

Add $2 x$ to both sides of the equation.

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

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 40 + 2 x$ .

Factor $2$ out of $2 x^{2}$ .

$2 (x^{2}) - 40 + 2 x = 0$

Factor $2$ out of $- 40$ .

$2 (x^{2}) + 2 \cdot - 20 + 2 x = 0$

Factor $2$ out of $2 (x^{2}) + 2 \cdot - 20$ .

$2 (x^{2} - 20) + 2 x = 0$

Factor $2$ out of $2 (x^{2} - 20) + 2 x$ .

$2 (x^{2} - 20 + x) = 0$

Factor.

Factor $x^{2} - 20 + x$ 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 $- 20$ and whose sum is $1$ .

$- 4, 5$

Write the factored form using these integers.

$2 ((x - 4) (x + 5)) = 0$

Remove unnecessary parentheses.

$2 (x - 4) (x + 5) = 0$

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

$x - 4 = 0$

$x + 5 = 0$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $2 (x - 4) (x + 5) = 0$ true.

$x = 4, - 5$

Find the Intersection of the Functions f(x)=2x-40/x , f(a)=-2

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