Solve {left(sqrt{left(3-2right)^2+(5-k)}right)}^2={left(sqrt{left(4-2right)^2+left(2-kright)^2}right)}^2

$${ \left( \sqrt{ { \left(3-2 \right) }^{ 2 } +(5-k) } \right) }^{ 2 } = { \left( \sqrt{ { \left(4-2 \right) }^{ 2 } + { \left(2-k \right) }^{ 2 } } \right) }^{ 2 }$$

Solve for k

$$k=1$$

$$k=2$$

Steps Using Factoring By Grouping

Subtract 2 from 3 to get 1.

$$\left(\sqrt{1^{2}+5-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\left(\sqrt{1+5-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Add 1 and 5 to get 6.

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

Calculate \sqrt{6-k} to the power of 2 and get 6-k.

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

Subtract 2 from 4 to get 2.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Add 4 and 4 to get 8.

$$6-k=\left(\sqrt{8-4k+k^{2}}\right)^{2}$$

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

$$6-k=8-4k+k^{2}$$

Subtract 8 from both sides.

$$6-k-8=-4k+k^{2}$$

Subtract 8 from 6 to get -2.

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

Add 4k to both sides.

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

Combine -k and 4k to get 3k.

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

Subtract k^{2} from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

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

$$a+b=3 ab=-\left(-2\right)=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=2 b=1$$

Rewrite -k^{2}+3k-2 as \left(-k^{2}+2k\right)+\left(k-2\right).

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

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

$$-k\left(k-2\right)+k-2$$

Factor out common term k-2 by using distributive property.

$$\left(k-2\right)\left(-k+1\right)$$

To find equation solutions, solve k-2=0 and -k+1=0.

$$k=2 k=1$$

Steps Using the Quadratic Formula

Subtract 2 from 3 to get 1.

$$\left(\sqrt{1^{2}+5-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\left(\sqrt{1+5-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Add 1 and 5 to get 6.

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

Calculate \sqrt{6-k} to the power of 2 and get 6-k.

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

Subtract 2 from 4 to get 2.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Add 4 and 4 to get 8.

$$6-k=\left(\sqrt{8-4k+k^{2}}\right)^{2}$$

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

$$6-k=8-4k+k^{2}$$

Subtract 8 from both sides.

$$6-k-8=-4k+k^{2}$$

Subtract 8 from 6 to get -2.

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

Add 4k to both sides.

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

Combine -k and 4k to get 3k.

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

Subtract k^{2} from both sides.

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

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.

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

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

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

Square 3.

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

Multiply -4 times -1.

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

Multiply 4 times -2.

$$k=\frac{-3±\sqrt{9-8}}{2\left(-1\right)}$$

Add 9 to -8.

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

Take the square root of 1.

$$k=\frac{-3±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{-3±1}{-2}$$

Now solve the equation k=\frac{-3±1}{-2} when ± is plus. Add -3 to 1.

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

Divide -2 by -2.

$$k=1$$

Now solve the equation k=\frac{-3±1}{-2} when ± is minus. Subtract 1 from -3.

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

Divide -4 by -2.

$$k=2$$

The equation is now solved.

$$k=1 k=2$$

Steps for Completing the Square

Subtract 2 from 3 to get 1.

$$\left(\sqrt{1^{2}+5-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\left(\sqrt{1+5-k}\right)^{2}=\left(\sqrt{\left(4-2\right)^{2}+\left(2-k\right)^{2}}\right)^{2}$$

Add 1 and 5 to get 6.

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

Calculate \sqrt{6-k} to the power of 2 and get 6-k.

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

Subtract 2 from 4 to get 2.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Add 4 and 4 to get 8.

$$6-k=\left(\sqrt{8-4k+k^{2}}\right)^{2}$$

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

$$6-k=8-4k+k^{2}$$

Add 4k to both sides.

$$6-k+4k=8+k^{2}$$

Combine -k and 4k to get 3k.

$$6+3k=8+k^{2}$$

Subtract k^{2} from both sides.

$$6+3k-k^{2}=8$$

Subtract 6 from both sides.

$$3k-k^{2}=8-6$$

Subtract 6 from 8 to get 2.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 3 by -1.

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

Divide 2 by -1.

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

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.

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

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

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

Add -2 to \frac{9}{4}.

$$k^{2}-3k+\frac{9}{4}=\frac{1}{4}$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

$$k=2 k=1$$