Solve P^2-5=-4

$$P ^ { 2 } - 5 = - 4$$

Solve for P

$$P=1$$

$$P=-1$$

Steps Using Difference of Squares

Add 4 to both sides.

$$P^{2}-5+4=0$$

Add -5 and 4 to get -1.

$$P^{2}-1=0$$

Consider P^{2}-1. Rewrite P^{2}-1 as P^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(P-1\right)\left(P+1\right)=0$$

To find equation solutions, solve P-1=0 and P+1=0.

$$P=1 P=-1$$

Steps by Finding Square Root

Add 5 to both sides.

$$P^{2}=-4+5$$

Add -4 and 5 to get 1.

$$P^{2}=1$$

Take the square root of both sides of the equation.

$$P=1 P=-1$$

Steps Using the Quadratic Formula

Add 4 to both sides.

$$P^{2}-5+4=0$$

Add -5 and 4 to get -1.

$$P^{2}-1=0$$

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

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

Square 0.

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

Multiply -4 times -1.

$$P=\frac{0±\sqrt{4}}{2}$$

Take the square root of 4.

$$P=\frac{0±2}{2}$$

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

$$P=1$$

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

$$P=-1$$

The equation is now solved.

$$P=1 P=-1$$