Solve for k k+12/k=8

$k + \frac{12}{k} = 8$

Step 1

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.

$1, k, 1$

The LCM of one and any expression is the expression.

$k$

Step 2

Multiply each term in $k + \frac{12}{k} = 8$ by $k$ to eliminate the fractions.

Multiply each term in $k + \frac{12}{k} = 8$ by $k$ .

$k \cdot k + \frac{12}{k} k = 8 k$

Simplify the left side.

Simplify each term.

Multiply $k$ by $k$ .

$k^{2} + \frac{12}{k} k = 8 k$

Cancel the common factor of $k$ .

Cancel the common factor.

$k^{2} + \frac{12}{k} k = 8 k$

Rewrite the expression.

$k^{2} + 12 = 8 k$

Step 3

Solve the equation.

Subtract $8 k$ from both sides of the equation.

$k^{2} + 12 - 8 k = 0$

Factor $k^{2} + 12 - 8 k$ 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 $12$ and whose sum is $- 8$ .

$- 6, - 2$

Write the factored form using these integers.

$(k - 6) (k - 2) = 0$

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

$k - 6 = 0$

$k - 2 = 0$

Set $k - 6$ equal to $0$ and solve for $k$ .

Set $k - 6$ equal to $0$ .

$k - 6 = 0$

Add $6$ to both sides of the equation.

$k = 6$

Set $k - 2$ equal to $0$ and solve for $k$ .

Set $k - 2$ equal to $0$ .

$k - 2 = 0$

Add $2$ to both sides of the equation.

$k = 2$

The final solution is all the values that make $(k - 6) (k - 2) = 0$ true.

$k = 6, 2$

Solve for k k+12/k=8

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