Solve 4/k^2-8k+12=k/k-2+1/k-6

$$\frac { 4 } { k ^ { 2 } - 8 k + 12 } = \frac { k } { k - 2 } + \frac { 1 } { k - 6 }$$

Solve for k

$$k=-1$$

Steps Using Factoring

Variable k cannot be equal to any of the values 2,6 since division by zero is not defined. Multiply both sides of the equation by \left(k-6\right)\left(k-2\right), the least common multiple of k^{2}-8k+12,k-2,k-6.

$$4=\left(k-6\right)k+k-2$$

Use the distributive property to multiply k-6 by k.

$$4=k^{2}-6k+k-2$$

Combine -6k and k to get -5k.

$$4=k^{2}-5k-2$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}-5k-2=4$$

Subtract 4 from both sides.

$$k^{2}-5k-2-4=0$$

Subtract 4 from -2 to get -6.

$$k^{2}-5k-6=0$$

To solve the equation, factor k^{2}-5k-6 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.

$$a+b=-5 ab=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

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

$$\left(k-6\right)\left(k+1\right)$$

To find equation solutions, solve k-6=0 and k+1=0.

$$k=6 k=-1$$

Variable k cannot be equal to 6.

$$k=-1$$

Steps Using Factoring By Grouping

Variable k cannot be equal to any of the values 2,6 since division by zero is not defined. Multiply both sides of the equation by \left(k-6\right)\left(k-2\right), the least common multiple of k^{2}-8k+12,k-2,k-6.

$$4=\left(k-6\right)k+k-2$$

Use the distributive property to multiply k-6 by k.

$$4=k^{2}-6k+k-2$$

Combine -6k and k to get -5k.

$$4=k^{2}-5k-2$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}-5k-2=4$$

Subtract 4 from both sides.

$$k^{2}-5k-2-4=0$$

Subtract 4 from -2 to get -6.

$$k^{2}-5k-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite k^{2}-5k-6 as \left(k^{2}-6k\right)+\left(k-6\right).

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

Factor out k in k^{2}-6k.

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

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

$$\left(k-6\right)\left(k+1\right)$$

To find equation solutions, solve k-6=0 and k+1=0.

$$k=6 k=-1$$

Variable k cannot be equal to 6.

$$k=-1$$

Steps Using the Quadratic Formula

Variable k cannot be equal to any of the values 2,6 since division by zero is not defined. Multiply both sides of the equation by \left(k-6\right)\left(k-2\right), the least common multiple of k^{2}-8k+12,k-2,k-6.

$$4=\left(k-6\right)k+k-2$$

Use the distributive property to multiply k-6 by k.

$$4=k^{2}-6k+k-2$$

Combine -6k and k to get -5k.

$$4=k^{2}-5k-2$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}-5k-2=4$$

Subtract 4 from both sides.

$$k^{2}-5k-2-4=0$$

Subtract 4 from -2 to get -6.

$$k^{2}-5k-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-6\right)}}{2}$$

Square -5.

$$k=\frac{-\left(-5\right)±\sqrt{25-4\left(-6\right)}}{2}$$

Multiply -4 times -6.

$$k=\frac{-\left(-5\right)±\sqrt{25+24}}{2}$$

Add 25 to 24.

$$k=\frac{-\left(-5\right)±\sqrt{49}}{2}$$

Take the square root of 49.

$$k=\frac{-\left(-5\right)±7}{2}$$

The opposite of -5 is 5.

$$k=\frac{5±7}{2}$$

Now solve the equation k=\frac{5±7}{2} when ± is plus. Add 5 to 7.

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

Divide 12 by 2.

$$k=6$$

Now solve the equation k=\frac{5±7}{2} when ± is minus. Subtract 7 from 5.

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

Divide -2 by 2.

$$k=-1$$

The equation is now solved.

$$k=6 k=-1$$

Variable k cannot be equal to 6.

$$k=-1$$

Steps for Completing the Square

Variable k cannot be equal to any of the values 2,6 since division by zero is not defined. Multiply both sides of the equation by \left(k-6\right)\left(k-2\right), the least common multiple of k^{2}-8k+12,k-2,k-6.

$$4=\left(k-6\right)k+k-2$$

Use the distributive property to multiply k-6 by k.

$$4=k^{2}-6k+k-2$$

Combine -6k and k to get -5k.

$$4=k^{2}-5k-2$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}-5k-2=4$$

Add 2 to both sides.

$$k^{2}-5k=4+2$$

Add 4 and 2 to get 6.

$$k^{2}-5k=6$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-5k+\left(-\frac{5}{2}\right)^{2}=6+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-5k+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$k^{2}-5k+\frac{25}{4}=\frac{49}{4}$$

Factor k^{2}-5k+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$k-\frac{5}{2}=\frac{7}{2} k-\frac{5}{2}=-\frac{7}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$k=6 k=-1$$

Variable k cannot be equal to 6.

$$k=-1$$