Solve p^2=4

$$p ^ { 2 } = 4$$

Solve for p

$$p=2$$

$$p=-2$$

Steps Using Difference of Squares

Subtract 4 from both sides.

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

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

$$\left(p-2\right)\left(p+2\right)=0$$

To find equation solutions, solve p-2=0 and p+2=0.

$$p=2 p=-2$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$p=2 p=-2$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

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

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}.

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

Square 0.

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

Multiply -4 times -4.

$$p=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$p=\frac{0±4}{2}$$

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

$$p=2$$

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

$$p=-2$$

The equation is now solved.

$$p=2 p=-2$$