Solve (k-2/2)^2=(k-1)(k-3/3)

$$( \frac { k - 2 } { 2 } ) ^ { 2 } = ( k - 1 ) ( \frac { k - 3 } { 3 } )$$

Solve for k

$$k=4$$

$$k=0$$

Steps Using Factoring

Multiply both sides of the equation by 3.

$$3\times \left(\frac{k-2}{2}\right)^{2}=\left(k-1\right)\left(k-3\right)$$

To raise \frac{k-2}{2} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(k-2\right)^{2}}{2^{2}}=\left(k-1\right)\left(k-3\right)$$

Express 3\times \frac{\left(k-2\right)^{2}}{2^{2}} as a single fraction.

$$\frac{3\left(k-2\right)^{2}}{2^{2}}=\left(k-1\right)\left(k-3\right)$$

Use the distributive property to multiply k-1 by k-3 and combine like terms.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-2\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract k^{2} from both sides.

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

Use the distributive property to multiply 3 by k^{2}-4k+4.

$$\frac{3k^{2}-12k+12}{4}-k^{2}=-4k+3$$

Divide each term of 3k^{2}-12k+12 by 4 to get \frac{3}{4}k^{2}-3k+3.

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

Combine \frac{3}{4}k^{2} and -k^{2} to get -\frac{1}{4}k^{2}.

$$-\frac{1}{4}k^{2}-3k+3=-4k+3$$

Add 4k to both sides.

$$-\frac{1}{4}k^{2}-3k+3+4k=3$$

Combine -3k and 4k to get k.

$$-\frac{1}{4}k^{2}+k+3=3$$

Subtract 3 from both sides.

$$-\frac{1}{4}k^{2}+k+3-3=0$$

Subtract 3 from 3 to get 0.

$$-\frac{1}{4}k^{2}+k=0$$

Factor out k.

$$k\left(-\frac{1}{4}k+1\right)=0$$

To find equation solutions, solve k=0 and -\frac{k}{4}+1=0.

$$k=0 k=4$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

$$3\times \left(\frac{k-2}{2}\right)^{2}=\left(k-1\right)\left(k-3\right)$$

To raise \frac{k-2}{2} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(k-2\right)^{2}}{2^{2}}=\left(k-1\right)\left(k-3\right)$$

Express 3\times \frac{\left(k-2\right)^{2}}{2^{2}} as a single fraction.

$$\frac{3\left(k-2\right)^{2}}{2^{2}}=\left(k-1\right)\left(k-3\right)$$

Use the distributive property to multiply k-1 by k-3 and combine like terms.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-2\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract k^{2} from both sides.

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

Use the distributive property to multiply 3 by k^{2}-4k+4.

$$\frac{3k^{2}-12k+12}{4}-k^{2}=-4k+3$$

Divide each term of 3k^{2}-12k+12 by 4 to get \frac{3}{4}k^{2}-3k+3.

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

Combine \frac{3}{4}k^{2} and -k^{2} to get -\frac{1}{4}k^{2}.

$$-\frac{1}{4}k^{2}-3k+3=-4k+3$$

Add 4k to both sides.

$$-\frac{1}{4}k^{2}-3k+3+4k=3$$

Combine -3k and 4k to get k.

$$-\frac{1}{4}k^{2}+k+3=3$$

Subtract 3 from both sides.

$$-\frac{1}{4}k^{2}+k+3-3=0$$

Subtract 3 from 3 to get 0.

$$-\frac{1}{4}k^{2}+k=0$$

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

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

Take the square root of 1^{2}.

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

Multiply 2 times -\frac{1}{4}.

$$k=\frac{-1±1}{-\frac{1}{2}}$$

Now solve the equation k=\frac{-1±1}{-\frac{1}{2}} when ± is plus. Add -1 to 1.

$$k=\frac{0}{-\frac{1}{2}}$$

Divide 0 by -\frac{1}{2} by multiplying 0 by the reciprocal of -\frac{1}{2}.

$$k=0$$

Now solve the equation k=\frac{-1±1}{-\frac{1}{2}} when ± is minus. Subtract 1 from -1.

$$k=-\frac{2}{-\frac{1}{2}}$$

Divide -2 by -\frac{1}{2} by multiplying -2 by the reciprocal of -\frac{1}{2}.

$$k=4$$

The equation is now solved.

$$k=0 k=4$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

$$3\times \left(\frac{k-2}{2}\right)^{2}=\left(k-1\right)\left(k-3\right)$$

To raise \frac{k-2}{2} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(k-2\right)^{2}}{2^{2}}=\left(k-1\right)\left(k-3\right)$$

Express 3\times \frac{\left(k-2\right)^{2}}{2^{2}} as a single fraction.

$$\frac{3\left(k-2\right)^{2}}{2^{2}}=\left(k-1\right)\left(k-3\right)$$

Use the distributive property to multiply k-1 by k-3 and combine like terms.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-2\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Subtract k^{2} from both sides.

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

Use the distributive property to multiply 3 by k^{2}-4k+4.

$$\frac{3k^{2}-12k+12}{4}-k^{2}=-4k+3$$

Divide each term of 3k^{2}-12k+12 by 4 to get \frac{3}{4}k^{2}-3k+3.

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

Combine \frac{3}{4}k^{2} and -k^{2} to get -\frac{1}{4}k^{2}.

$$-\frac{1}{4}k^{2}-3k+3=-4k+3$$

Add 4k to both sides.

$$-\frac{1}{4}k^{2}-3k+3+4k=3$$

Combine -3k and 4k to get k.

$$-\frac{1}{4}k^{2}+k+3=3$$

Subtract 3 from both sides.

$$-\frac{1}{4}k^{2}+k=3-3$$

Subtract 3 from 3 to get 0.

$$-\frac{1}{4}k^{2}+k=0$$

Multiply both sides by -4.

$$\frac{-\frac{1}{4}k^{2}+k}{-\frac{1}{4}}=\frac{0}{-\frac{1}{4}}$$

Dividing by -\frac{1}{4} undoes the multiplication by -\frac{1}{4}.

$$k^{2}+\frac{1}{-\frac{1}{4}}k=\frac{0}{-\frac{1}{4}}$$

Divide 1 by -\frac{1}{4} by multiplying 1 by the reciprocal of -\frac{1}{4}.

$$k^{2}-4k=\frac{0}{-\frac{1}{4}}$$

Divide 0 by -\frac{1}{4} by multiplying 0 by the reciprocal of -\frac{1}{4}.

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

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

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

Square -2.

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

Factor k^{2}-4k+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add 2 to both sides of the equation.

$$k=4 k=0$$