Consider $\left(1+P\right)\left(1-P\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.
$$1-P^{2}=1$$
Subtract $1$ from both sides.
$$-P^{2}=1-1$$
Subtract $1$ from $1$ to get $0$.
$$-P^{2}=0$$
Divide both sides by $-1$. Zero divided by any non-zero number gives zero.
$$P^{2}=0$$
Take the square root of both sides of the equation.
$$P=0$$ $$P=0$$
The equation is now solved. Solutions are the same.
$$P=0$$
Steps Using the Quadratic Formula
Consider $\left(1+P\right)\left(1-P\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$. Square $1$.
$$1-P^{2}=1$$
Subtract $1$ from both sides.
$$1-P^{2}-1=0$$
Subtract $1$ from $1$ to get $0$.
$$-P^{2}=0$$
Divide both sides by $-1$. Zero divided by any non-zero number gives zero.
$$P^{2}=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $0$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
