Solve sqrt2^2+1^2=c^2

$$\sqrt { 2 } ^ { 2 } + 1 ^ { 2 } = c ^ { 2 }$$

Solve for c

$$c=\sqrt{3}\approx 1.732050808$$

$$c=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

The square of \sqrt{2} is 2.

$$2+1^{2}=c^{2}$$

Calculate 1 to the power of 2 and get 1.

$$2+1=c^{2}$$

Add 2 and 1 to get 3.

$$3=c^{2}$$

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

$$c^{2}=3$$

Take the square root of both sides of the equation.

$$c=\sqrt{3} c=-\sqrt{3}$$

Steps Using the Quadratic Formula

The square of \sqrt{2} is 2.

$$2+1^{2}=c^{2}$$

Calculate 1 to the power of 2 and get 1.

$$2+1=c^{2}$$

Add 2 and 1 to get 3.

$$3=c^{2}$$

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

$$c^{2}=3$$

Subtract 3 from both sides.

$$c^{2}-3=0$$

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

$$c=\frac{0±\sqrt{0^{2}-4\left(-3\right)}}{2}$$

Square 0.

$$c=\frac{0±\sqrt{-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$c=\frac{0±\sqrt{12}}{2}$$

Take the square root of 12.

$$c=\frac{0±2\sqrt{3}}{2}$$

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

$$c=\sqrt{3}$$

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

$$c=-\sqrt{3}$$

The equation is now solved.

$$c=\sqrt{3} c=-\sqrt{3}$$