Solve 2^2+(2sqrt{2})^2=k^2

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

Solve for k

$$k=2\sqrt{3}\approx 3.464101615$$

$$k=-2\sqrt{3}\approx -3.464101615$$

Steps by Finding Square Root

Calculate 2 to the power of 2 and get 4.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

$$4+4\times 2=k^{2}$$

Multiply 4 and 2 to get 8.

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

Add 4 and 8 to get 12.

$$12=k^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}=12$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Calculate 2 to the power of 2 and get 4.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

$$4+4\times 2=k^{2}$$

Multiply 4 and 2 to get 8.

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

Add 4 and 8 to get 12.

$$12=k^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$k^{2}=12$$

Subtract 12 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{0±\sqrt{0^{2}-4\left(-12\right)}}{2}$$

Square 0.

$$k=\frac{0±\sqrt{-4\left(-12\right)}}{2}$$

Multiply -4 times -12.

$$k=\frac{0±\sqrt{48}}{2}$$

Take the square root of 48.

$$k=\frac{0±4\sqrt{3}}{2}$$

Now solve the equation k=\frac{0±4\sqrt{3}}{2} when ± is plus.

$$k=2\sqrt{3}$$

Now solve the equation k=\frac{0±4\sqrt{3}}{2} when ± is minus.

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

The equation is now solved.

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