Solve 9p^2=49

$$9 p ^ { 2 } = 49$$

Solve for p

$$p = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$p = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

Steps Using Difference of Squares

Divide both sides by 9.

$$p^{2}=\frac{49}{9}$$

Subtract \frac{49}{9} from both sides.

$$p^{2}-\frac{49}{9}=0$$

Multiply both sides by 9.

$$9p^{2}-49=0$$

Consider 9p^{2}-49. Rewrite 9p^{2}-49 as \left(3p\right)^{2}-7^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(3p-7\right)\left(3p+7\right)=0$$

To find equation solutions, solve 3p-7=0 and 3p+7=0.

$$p=\frac{7}{3} p=-\frac{7}{3}$$

Steps by Finding Square Root

Divide both sides by 9.

$$p^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$p=\frac{7}{3} p=-\frac{7}{3}$$

Steps Using the Quadratic Formula

Divide both sides by 9.

$$p^{2}=\frac{49}{9}$$

Subtract \frac{49}{9} from both sides.

$$p^{2}-\frac{49}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{49}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{0±\sqrt{0^{2}-4\left(-\frac{49}{9}\right)}}{2}$$

Square 0.

$$p=\frac{0±\sqrt{-4\left(-\frac{49}{9}\right)}}{2}$$

Multiply -4 times -\frac{49}{9}.

$$p=\frac{0±\sqrt{\frac{196}{9}}}{2}$$

Take the square root of \frac{196}{9}.

$$p=\frac{0±\frac{14}{3}}{2}$$

Now solve the equation p=\frac{0±\frac{14}{3}}{2} when ± is plus.

$$p=\frac{7}{3}$$

Now solve the equation p=\frac{0±\frac{14}{3}}{2} when ± is minus.

$$p=-\frac{7}{3}$$

The equation is now solved.

$$p=\frac{7}{3} p=-\frac{7}{3}$$