Solve 1/4k^2-1=(k/2-1)(k-2)

$$\frac { 1 } { 4 } k ^ { 2 } - 1 = ( \frac { k } { 2 } - 1 ) ( k - 2 )$$

Solve for k

$$k=2$$

$$k=6$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$1k^{2}-4=4\left(\frac{k}{2}-1\right)\left(k-2\right)$$

Use the distributive property to multiply 4 by \frac{k}{2}-1.

$$1k^{2}-4=\left(4\times \frac{k}{2}-4\right)\left(k-2\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$1k^{2}-4=\left(2k-4\right)\left(k-2\right)$$

Use the distributive property to multiply 2k-4 by k-2 and combine like terms.

$$1k^{2}-4=2k^{2}-8k+8$$

Subtract 2k^{2} from both sides.

$$1k^{2}-4-2k^{2}=-8k+8$$

Combine 1k^{2} and -2k^{2} to get -k^{2}.

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

Add 8k to both sides.

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

Subtract 8 from both sides.

$$-k^{2}-4+8k-8=0$$

Subtract 8 from -4 to get -12.

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

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.

$$a+b=8 ab=-\left(-12\right)=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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

Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$1k^{2}-4=4\left(\frac{k}{2}-1\right)\left(k-2\right)$$

Use the distributive property to multiply 4 by \frac{k}{2}-1.

$$1k^{2}-4=\left(4\times \frac{k}{2}-4\right)\left(k-2\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$1k^{2}-4=\left(2k-4\right)\left(k-2\right)$$

Use the distributive property to multiply 2k-4 by k-2 and combine like terms.

$$1k^{2}-4=2k^{2}-8k+8$$

Subtract 2k^{2} from both sides.

$$1k^{2}-4-2k^{2}=-8k+8$$

Combine 1k^{2} and -2k^{2} to get -k^{2}.

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

Add 8k to both sides.

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

Subtract 8 from both sides.

$$-k^{2}-4+8k-8=0$$

Subtract 8 from -4 to get -12.

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

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}.

$$k=\frac{-8±\sqrt{8^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Square 8.

$$k=\frac{-8±\sqrt{64-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$k=\frac{-8±\sqrt{64+4\left(-12\right)}}{2\left(-1\right)}$$

Multiply 4 times -12.

$$k=\frac{-8±\sqrt{64-48}}{2\left(-1\right)}$$

Add 64 to -48.

$$k=\frac{-8±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$k=\frac{-8±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{-8±4}{-2}$$

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

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

Divide -4 by -2.

$$k=2$$

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

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

Divide -12 by -2.

$$k=6$$

The equation is now solved.

$$k=2 k=6$$

Steps for Completing the Square

Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$1k^{2}-4=4\left(\frac{k}{2}-1\right)\left(k-2\right)$$

Use the distributive property to multiply 4 by \frac{k}{2}-1.

$$1k^{2}-4=\left(4\times \frac{k}{2}-4\right)\left(k-2\right)$$

Cancel out 2, the greatest common factor in 4 and 2.

$$1k^{2}-4=\left(2k-4\right)\left(k-2\right)$$

Use the distributive property to multiply 2k-4 by k-2 and combine like terms.

$$1k^{2}-4=2k^{2}-8k+8$$

Subtract 2k^{2} from both sides.

$$1k^{2}-4-2k^{2}=-8k+8$$

Combine 1k^{2} and -2k^{2} to get -k^{2}.

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

Add 8k to both sides.

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

Add 4 to both sides.

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

Add 8 and 4 to get 12.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 8 by -1.

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

Divide 12 by -1.

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

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.

$$k=6 k=2$$