Solve (sqrt{(3-2)^{2}+(3-k)})^2=(sqrt{(4-2)^2+(2-k)^2})^2

$$( \sqrt { ( 3 - 2 ) ^ { 2 } + ( 3 - k ) } ) ^ { 2 } = ( \sqrt { ( 4 - 2 ) ^ { 2 } + ( 2 - k ) ^ { 2 } } ) ^ { 2 }$$

Solve for k

$$k=\frac{3+\sqrt{7}i}{2}\approx 1.5+1.322875656i$$

$$k=\frac{-\sqrt{7}i+3}{2}\approx 1.5-1.322875656i$$

Steps Using the Quadratic Formula

Subtract 2 from 3 to get 1.

$$\left(\sqrt{1^{2}+3-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\left(\sqrt{1+3-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Add 1 and 3 to get 4.

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

Calculate \sqrt{4-k} to the power of 2 and get 4-k.

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

Subtract 2 from 4 to get 2.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Add 4 and 4 to get 8.

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

Calculate \sqrt{8-4k+k^{2}} to the power of 2 and get 8-4k+k^{2}.

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

Subtract 8 from both sides.

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

Subtract 8 from 4 to get -4.

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

Add 4k to both sides.

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

Combine -k and 4k to get 3k.

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

Subtract k^{2} from both sides.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-3±\sqrt{3^{2}-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Square 3.

$$k=\frac{-3±\sqrt{9-4\left(-1\right)\left(-4\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$k=\frac{-3±\sqrt{9+4\left(-4\right)}}{2\left(-1\right)}$$

Multiply 4 times -4.

$$k=\frac{-3±\sqrt{9-16}}{2\left(-1\right)}$$

Add 9 to -16.

$$k=\frac{-3±\sqrt{-7}}{2\left(-1\right)}$$

Take the square root of -7.

$$k=\frac{-3±\sqrt{7}i}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{-3±\sqrt{7}i}{-2}$$

Now solve the equation k=\frac{-3±\sqrt{7}i}{-2} when ± is plus. Add -3 to i\sqrt{7}.

$$k=\frac{-3+\sqrt{7}i}{-2}$$

Divide -3+i\sqrt{7} by -2.

$$k=\frac{-\sqrt{7}i+3}{2}$$

Now solve the equation k=\frac{-3±\sqrt{7}i}{-2} when ± is minus. Subtract i\sqrt{7} from -3.

$$k=\frac{-\sqrt{7}i-3}{-2}$$

Divide -3-i\sqrt{7} by -2.

$$k=\frac{3+\sqrt{7}i}{2}$$

The equation is now solved.

$$k=\frac{-\sqrt{7}i+3}{2} k=\frac{3+\sqrt{7}i}{2}$$

Steps for Completing the Square

Subtract 2 from 3 to get 1.

$$\left(\sqrt{1^{2}+3-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\left(\sqrt{1+3-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Add 1 and 3 to get 4.

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

Calculate \sqrt{4-k} to the power of 2 and get 4-k.

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

Subtract 2 from 4 to get 2.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Add 4 and 4 to get 8.

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

Calculate \sqrt{8-4k+k^{2}} to the power of 2 and get 8-4k+k^{2}.

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

Add 4k to both sides.

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

Combine -k and 4k to get 3k.

$$4+3k=8+k^{2}$$

Subtract k^{2} from both sides.

$$4+3k-k^{2}=8$$

Subtract 4 from both sides.

$$3k-k^{2}=8-4$$

Subtract 4 from 8 to get 4.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

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

Divide 4 by -1.

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

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

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

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

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

Add -4 to \frac{9}{4}.

$$k^{2}-3k+\frac{9}{4}=-\frac{7}{4}$$

Factor k^{2}-3k+\frac{9}{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}.

$$\left(k-\frac{3}{2}\right)^{2}=-\frac{7}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Simplify.

$$k-\frac{3}{2}=\frac{\sqrt{7}i}{2} k-\frac{3}{2}=-\frac{\sqrt{7}i}{2}$$

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

$$k=\frac{3+\sqrt{7}i}{2} k=\frac{-\sqrt{7}i+3}{2}$$