Quadratic Equations

Do you remember the quadratic formula? Once you have memorized it, you can use a part of it, called the discriminant, and it helps you quickly visualize a parabola. 

You don't have to memorize the discriminant formula if you know the quadratic formula:

$\dfrac{-b\pm \sqrt{\bf{b^2-4ac}}}{2a}$

The bolded part of the quadratic formula is the discriminant:

$b^2-4ac=D$

When we encounter a polynomial, the first action we take is to simplify that polynomial by factoring. Trinomials, in particular, can be factored into binomials.  We have written out the process for factoring a trinomial into binomials, but there is a set of special trinomials (special products) that have memorizable patterns. 

Trinomials, in the form of $x^2 +bx+c$ and $ax^2+bx+c$, can often be factored into binomials.  Factoring trinomials into binomials can make them much easier to solve (see lesson on solving quadratic equations by factoring).

There are many processes for factoring trinomials into binomials.  We teach the one below because it eliminates guesswork and makes factoring trinomials in the form $ax^2+bx+c$ just as easy as factoring trinomials in the form $x^2 +bx+c$.

Polynomials can often be simplified by factoring.  Factoring means breaking a polynomial down into its "factors": variables or terms that are multiplied together to create the polynomial.

The most basic form of factoring is to "factor out a variable," which means to divide each term in the polynomial by a common factor. You can think of this as a sort of "reverse distributive property" process.  Rather than multiplying each term by something, you're dividing each term by something.

There are several ways to solve quadratic equations.  The first methods that come to mind are factoring (either factoring out a constant or factoring into binomials) and using the quadratic formula. 

But, many quadratics that cannot be solved by factoring into binomials can be solved by a process called completing the square. 

The process of completing the square relies on your knowledge of special products.

Do you remember that $x^2+6x+9=(x+3)^2$?

Do you remember that $x^2-8x+16=(x-4)^2$?

We usually try to solve quadratics by factoring. But, some quadratics can't  be factored. Do you remember how to solve those?  The quadratic formula!  Have it memorized!

On a standardized test, you'll know you need to use the quadratic formula to find the solutions to a quadratic equation when you see that the multiple choice answers include a $\pm$ and a square root! If you see that, don't even try to factor; just use the quadratic formula. 

To remind you, the quadratic formula is:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Factoring polynomials is an important step in solving polynomial equations.  Getting a polynomial into binomials is one of the easiest ways to solve for x-intercepts as you can set each binomial equal to zero and find multiple answers. 

We have learned how to factor out a constant and how to factor different types of trinomials into binomials, but what about when you have four terms in a polynomial?

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: 

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

You can graph a quadratic function by plugging in x and y values and plotting out the parabola, just like you can use an x-y table to graph a line.