Solve k+12/k=8

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

Solve for k

$$k=2$$

$$k=6$$

Steps Using Factoring

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$kk+12=8k$$

Multiply k and k to get k^{2}.

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

Subtract 8k from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor k^{2}-8k+12 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=-8 ab=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.

$$\left(k-6\right)\left(k-2\right)$$

To find equation solutions, solve k-6=0 and k-2=0.

$$k=6 k=2$$

Steps Using Factoring By Grouping

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$kk+12=8k$$

Multiply k and k to get k^{2}.

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

Subtract 8k from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+12. To find a and b, set up a system to be solved.

$$a+b=-8 ab=1\times 12=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-6 b=-2$$

Rewrite k^{2}-8k+12 as \left(k^{2}-6k\right)+\left(-2k+12\right).

$$\left(k^{2}-6k\right)+\left(-2k+12\right)$$

Factor out k in the first and -2 in the second group.

$$k\left(k-6\right)-2\left(k-6\right)$$

Factor out common term k-6 by using distributive property.

$$\left(k-6\right)\left(k-2\right)$$

To find equation solutions, solve k-6=0 and k-2=0.

$$k=6 k=2$$

Steps Using the Quadratic Formula

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$kk+12=8k$$

Multiply k and k to get k^{2}.

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

Subtract 8k from both sides.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 12}}{2}$$

Square -8.

$$k=\frac{-\left(-8\right)±\sqrt{64-4\times 12}}{2}$$

Multiply -4 times 12.

$$k=\frac{-\left(-8\right)±\sqrt{64-48}}{2}$$

Add 64 to -48.

$$k=\frac{-\left(-8\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$k=\frac{-\left(-8\right)±4}{2}$$

The opposite of -8 is 8.

$$k=\frac{8±4}{2}$$

Now solve the equation k=\frac{8±4}{2} when ± is plus. Add 8 to 4.

$$k=\frac{12}{2}$$

Divide 12 by 2.

$$k=6$$

Now solve the equation k=\frac{8±4}{2} when ± is minus. Subtract 4 from 8.

$$k=\frac{4}{2}$$

Divide 4 by 2.

$$k=2$$

The equation is now solved.

$$k=6 k=2$$

Steps for Completing the Square

Variable k cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by k.

$$kk+12=8k$$

Multiply k and k to get k^{2}.

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

Subtract 8k from both sides.

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

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

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

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-8k+\left(-4\right)^{2}=-12+\left(-4\right)^{2}$$

Square -4.

$$k^{2}-8k+16=-12+16$$

Add -12 to 16.

$$k^{2}-8k+16=4$$

Factor k^{2}-8k+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-4\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-4\right)^{2}}=\sqrt{4}$$

Simplify.

$$k-4=2 k-4=-2$$

Add 4 to both sides of the equation.

$$k=6 k=2$$