Solve 16p^2=25

$$16 p ^ { 2 } = 25$$

Solve for p

$$p = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

$$p = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

Steps Using Difference of Squares

Divide both sides by 16.

$$p^{2}=\frac{25}{16}$$

Subtract \frac{25}{16} from both sides.

$$p^{2}-\frac{25}{16}=0$$

Multiply both sides by 16.

$$16p^{2}-25=0$$

Consider 16p^{2}-25. Rewrite 16p^{2}-25 as \left(4p\right)^{2}-5^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(4p-5\right)\left(4p+5\right)=0$$

To find equation solutions, solve 4p-5=0 and 4p+5=0.

$$p=\frac{5}{4} p=-\frac{5}{4}$$

Steps by Finding Square Root

Divide both sides by 16.

$$p^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$p=\frac{5}{4} p=-\frac{5}{4}$$

Steps Using the Quadratic Formula

Divide both sides by 16.

$$p^{2}=\frac{25}{16}$$

Subtract \frac{25}{16} from both sides.

$$p^{2}-\frac{25}{16}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{25}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$p=\frac{0±\sqrt{0^{2}-4\left(-\frac{25}{16}\right)}}{2}$$

Square 0.

$$p=\frac{0±\sqrt{-4\left(-\frac{25}{16}\right)}}{2}$$

Multiply -4 times -\frac{25}{16}.

$$p=\frac{0±\sqrt{\frac{25}{4}}}{2}$$

Take the square root of \frac{25}{4}.

$$p=\frac{0±\frac{5}{2}}{2}$$

Now solve the equation p=\frac{0±\frac{5}{2}}{2} when ± is plus.

$$p=\frac{5}{4}$$

Now solve the equation p=\frac{0±\frac{5}{2}}{2} when ± is minus.

$$p=-\frac{5}{4}$$

The equation is now solved.

$$p=\frac{5}{4} p=-\frac{5}{4}$$