Solving Quadratic Equations: In Factored Form

Once you get a quadratic equation into factored form, it's easy to find the solutions, roots, or x-intercepts of the equation (where the parabola crosses the x-axis). 

The mathematical premise at play here is the Zero-Product Property, which states that when $ab=0$, then either $a$ or $b$ (or both) must equal $0$.

Examples: 

$\eqalign{3x&=0\\\div 3 &\quad \div 3\\x&=0}$

$\eqalign{xy&=0\\\div x \qquad &\text{ OR }\qquad \div y\\x=0 \qquad &\text{ OR } \qquad y=0}$

As you can see in the second example, when there are two variables that multiply to make $0$, we know that at least one of them must equal $0$, but we don't know which one it is. 

So, once you get a quadratic equation into factored form, and you know that those factors multiply to $0$ (you are trying to find the x-intercepts, so $y$, by definition, will equal $0$), you know that at least one of those terms equals 0 (and you don't know how many, so there can be more than one right answer).

Examples: 

$\eqalign{(x+4)(x-5)&=0\\x+4=0 &\quad\text{ OR }\quad x-5=0\\x=-4 &\qquad \text{  }\qquad x=5}$

Because this is a parabola, it makes sense that there are two answers: the parabola crosses the x-axis at two places, $-4$ and $5$. 

When you factor out a variable, that constant can also equal $0$ (think about it, if $abc=0$ then $a$, $b$, or $c$ could equal $0$.

$\eqalign{x(2x-3)(x+7)&=0\\x=0 &\quad\text{ OR }\qquad 2x-3=0 \qquad\text{ OR }\qquad x+7=0\\\quad &\quad \text{     }\qquad \qquad x=\dfrac{3}{2} \qquad \text{     }\qquad \qquad \qquad x=-7}$

Once you get a quadratic into factored form, it's pretty easy to find the solutions. Just take each term on the non-zero side of the equal, and set it equal to zero, and solve if necessary.  You may get 1 solution, 2 solutions, or no real solutions (see the lesson on graphing quadratics for more on different solution sets).