Solve 1/2left(1-k/2right)left(2-kright)=left(k+2right)left(1-k/2right)

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

Solve for k

$$k=2$$

$$k=-\frac{2}{3}\approx -0.666666667$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

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

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

Apply the distributive property by multiplying each term of 1-\frac{k}{2} by each term of 2-k.

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

Express 2\left(-\frac{k}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

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

Combine -k and -k to get -2k.

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

Multiply -1 and -1 to get 1.

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

Express \frac{k}{2}k as a single fraction.

$$2-2k+\frac{kk}{2}=2\left(k+2\right)\left(1-\frac{k}{2}\right)$$

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

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

Use the distributive property to multiply 2 by k+2.

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

Apply the distributive property by multiplying each term of 2k+4 by each term of 1-\frac{k}{2}.

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

Express 2\left(-\frac{k}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

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

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

$$2-2k+\frac{k^{2}}{2}=2k-kk+4-2k$$

Combine 2k and -2k to get 0.

$$2-2k+\frac{k^{2}}{2}=-kk+4$$

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

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

Add k^{2} to both sides.

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

Combine \frac{k^{2}}{2} and k^{2} to get \frac{3}{2}k^{2}.

$$2-2k+\frac{3}{2}k^{2}=4$$

Subtract 4 from both sides.

$$2-2k+\frac{3}{2}k^{2}-4=0$$

Subtract 4 from 2 to get -2.

$$-2-2k+\frac{3}{2}k^{2}=0$$

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.

$$\frac{3}{2}k^{2}-2k-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2} for a, -2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times \frac{3}{2}\left(-2\right)}}{2\times \frac{3}{2}}$$

Square -2.

$$k=\frac{-\left(-2\right)±\sqrt{4-4\times \frac{3}{2}\left(-2\right)}}{2\times \frac{3}{2}}$$

Multiply -4 times \frac{3}{2}.

$$k=\frac{-\left(-2\right)±\sqrt{4-6\left(-2\right)}}{2\times \frac{3}{2}}$$

Multiply -6 times -2.

$$k=\frac{-\left(-2\right)±\sqrt{4+12}}{2\times \frac{3}{2}}$$

Add 4 to 12.

$$k=\frac{-\left(-2\right)±\sqrt{16}}{2\times \frac{3}{2}}$$

Take the square root of 16.

$$k=\frac{-\left(-2\right)±4}{2\times \frac{3}{2}}$$

The opposite of -2 is 2.

$$k=\frac{2±4}{2\times \frac{3}{2}}$$

Multiply 2 times \frac{3}{2}.

$$k=\frac{2±4}{3}$$

Now solve the equation k=\frac{2±4}{3} when ± is plus. Add 2 to 4.

$$k=\frac{6}{3}$$

Divide 6 by 3.

$$k=2$$

Now solve the equation k=\frac{2±4}{3} when ± is minus. Subtract 4 from 2.

$$k=-\frac{2}{3}$$

The equation is now solved.

$$k=2 k=-\frac{2}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

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

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

Apply the distributive property by multiplying each term of 1-\frac{k}{2} by each term of 2-k.

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

Express 2\left(-\frac{k}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

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

Combine -k and -k to get -2k.

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

Multiply -1 and -1 to get 1.

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

Express \frac{k}{2}k as a single fraction.

$$2-2k+\frac{kk}{2}=2\left(k+2\right)\left(1-\frac{k}{2}\right)$$

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

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

Use the distributive property to multiply 2 by k+2.

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

Apply the distributive property by multiplying each term of 2k+4 by each term of 1-\frac{k}{2}.

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

Express 2\left(-\frac{k}{2}\right) as a single fraction.

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

Cancel out 2 and 2.

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

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

$$2-2k+\frac{k^{2}}{2}=2k-kk+4-2k$$

Combine 2k and -2k to get 0.

$$2-2k+\frac{k^{2}}{2}=-kk+4$$

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

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

Add k^{2} to both sides.

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

Combine \frac{k^{2}}{2} and k^{2} to get \frac{3}{2}k^{2}.

$$2-2k+\frac{3}{2}k^{2}=4$$

Subtract 2 from both sides.

$$-2k+\frac{3}{2}k^{2}=4-2$$

Subtract 2 from 4 to get 2.

$$-2k+\frac{3}{2}k^{2}=2$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{3}{2}k^{2}-2k=2$$

Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{2}k^{2}-2k}{\frac{3}{2}}=\frac{2}{\frac{3}{2}}$$

Dividing by \frac{3}{2} undoes the multiplication by \frac{3}{2}.

$$k^{2}+\left(-\frac{2}{\frac{3}{2}}\right)k=\frac{2}{\frac{3}{2}}$$

Divide -2 by \frac{3}{2} by multiplying -2 by the reciprocal of \frac{3}{2}.

$$k^{2}-\frac{4}{3}k=\frac{2}{\frac{3}{2}}$$

Divide 2 by \frac{3}{2} by multiplying 2 by the reciprocal of \frac{3}{2}.

$$k^{2}-\frac{4}{3}k=\frac{4}{3}$$

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{4}{3}k+\left(-\frac{2}{3}\right)^{2}=\frac{4}{3}+\left(-\frac{2}{3}\right)^{2}$$

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{4}{3}k+\frac{4}{9}=\frac{4}{3}+\frac{4}{9}$$

Add \frac{4}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{4}{3}k+\frac{4}{9}=\frac{16}{9}$$

Factor k^{2}-\frac{4}{3}k+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{2}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{2}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$k-\frac{2}{3}=\frac{4}{3} k-\frac{2}{3}=-\frac{4}{3}$$

Add \frac{2}{3} to both sides of the equation.

$$k=2 k=-\frac{2}{3}$$