Solve 1^2+(-2)^2-r^2=1

$$1 ^ { 2 } + ( - 2 ) ^ { 2 } - r ^ { 2 } = 1$$

Solve for r

$$r=-2$$

$$r=2$$

Steps by Finding Square Root

Calculate 1 to the power of 2 and get 1.

$$1+\left(-2\right)^{2}-r^{2}=1$$

Calculate -2 to the power of 2 and get 4.

$$1+4-r^{2}=1$$

Add 1 and 4 to get 5.

$$5-r^{2}=1$$

Subtract 5 from both sides.

$$-r^{2}=1-5$$

Subtract 5 from 1 to get -4.

$$-r^{2}=-4$$

Divide both sides by -1.

$$r^{2}=\frac{-4}{-1}$$

Fraction \frac{-4}{-1} can be simplified to 4 by removing the negative sign from both the numerator and the denominator.

$$r^{2}=4$$

Take the square root of both sides of the equation.

$$r=2 r=-2$$

Steps Using the Quadratic Formula

Calculate 1 to the power of 2 and get 1.

$$1+\left(-2\right)^{2}-r^{2}=1$$

Calculate -2 to the power of 2 and get 4.

$$1+4-r^{2}=1$$

Add 1 and 4 to get 5.

$$5-r^{2}=1$$

Subtract 1 from both sides.

$$5-r^{2}-1=0$$

Subtract 1 from 5 to get 4.

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

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-r^{2}+4=0$$

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

$$r=\frac{0±\sqrt{0^{2}-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Square 0.

$$r=\frac{0±\sqrt{-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$r=\frac{0±\sqrt{4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$r=\frac{0±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

$$r=\frac{0±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$r=\frac{0±4}{-2}$$

Now solve the equation r=\frac{0±4}{-2} when ± is plus. Divide 4 by -2.

$$r=-2$$

Now solve the equation r=\frac{0±4}{-2} when ± is minus. Divide -4 by -2.

$$r=2$$

The equation is now solved.

$$r=-2 r=2$$