Solve pir^2=17

$$\pi r ^ { 2 } = 17$$

Solve for r

$$r=\sqrt{\frac{17}{\pi }}\approx 2.326213246$$

$$r=-\sqrt{\frac{17}{\pi }}\approx -2.326213246$$

Steps by Finding Square Root

Divide both sides by \pi .

$$\frac{\pi r^{2}}{\pi }=\frac{17}{\pi }$$

Dividing by \pi undoes the multiplication by \pi .

$$r^{2}=\frac{17}{\pi }$$

Take the square root of both sides of the equation.

$$r=\frac{17}{\sqrt{17\pi }} r=-\frac{17}{\sqrt{17\pi }}$$

Steps Using the Quadratic Formula

Subtract 17 from both sides.

$$\pi r^{2}-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \pi for a, 0 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{0±\sqrt{0^{2}-4\pi \left(-17\right)}}{2\pi }$$

Square 0.

$$r=\frac{0±\sqrt{-4\pi \left(-17\right)}}{2\pi }$$

Multiply -4 times \pi .

$$r=\frac{0±\sqrt{\left(-4\pi \right)\left(-17\right)}}{2\pi }$$

Multiply -4\pi times -17.

$$r=\frac{0±\sqrt{68\pi }}{2\pi }$$

Take the square root of 68\pi .

$$r=\frac{0±2\sqrt{17\pi }}{2\pi }$$

Now solve the equation r=\frac{0±2\sqrt{17\pi }}{2\pi } when ± is plus.

$$r=\frac{17}{\sqrt{17\pi }}$$

Now solve the equation r=\frac{0±2\sqrt{17\pi }}{2\pi } when ± is minus.

$$r=-\frac{17}{\sqrt{17\pi }}$$

The equation is now solved.

$$r=\frac{17}{\sqrt{17\pi }} r=-\frac{17}{\sqrt{17\pi }}$$