3.1 (OPTIONAL) The General Quadratic Formula

Lesson materials located below the video overview.

Be sure to have looked at PART 2 of this QUADRATICS course so that this section makes sense to you! 


Many students are taught to solve quadratics by memorizing the famous “quadratic formula.” My students don’t! THE BOX METHOD WILL NEVER LET YOU DOWN! And the reason I say this is because the box method really is the quadratic formula that everyone else does in disguise! Here’s why:


Students who use the quadratic formula are taught to bring all the terms to one side so that it is an equation with zero on one side. We can do this too if we want (but the box method does not require it!)


If we use the box method on this, we multiply through by \(a\) to make the first term a perfect square and through by \(4\) to avoid fractions. This gives:


Now apply the box method:


The box shows that we need the number \(b^{2}\), not \(4ac\), in the equation \(4a^{2}x^{2}+4abx+4ac=0\). Let’s subtract \(4ac\):


and add \(b^{2}\):



Now looking at the box we are set to go.




\(2ax+b=\sqrt{b^{2}-4ac}\)  or  \(-\sqrt{b^{2}-4ac}\)


Now add \(-b\) throughout:

\(2ax=-b+\sqrt{b^{2}-4ac}\)  or  \(-b-\sqrt{b^{2}-4ac}\)


And divide through by \(2a\):

\(x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\)  or   \(x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}\)


This statement can be combined into a single expression:  \(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\).


IF \(ax^{2}+bx+c=0\) THEN  \(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\).


COMMENT: Many curricula have students solve all quadratic equations by plugging numbers into this memorized formula. It has the advantage of being quick. I will never stop a student from following this route too, but I personally prefer following a method I feel I understand and can just do without any memorization. It stays with me in the long run and makes sense. There is no need to do math fast!



The box method shows that the general quadratic equation \(ax^{2}+bx+c=0\) has general solution: \(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\).

Many curricula have students not only memorise this formula, but also study different features of the formula, and give parts of it names. (AS WE SHALL SEE IN THE NEXT PART OF THIS COURSE, NONE OF THIS IS NECESSARY!)


The quantity under the square root sign, \(b^{2}-4ac\), is called the discriminant of the quadratic. Folk like to give this quantity a name because its sign determines the types of solutions one obtains:

If \(b^{2}-4ac\)   is negative, then the quadratic has no real solutions. (One cannot compute the square root of a negative value.)

If \(b^{2}-4ac\)   equals 0, then the quadratic has precisely one solution. (Zero is the only number with precisely one square root.)

If \(b^{2}-4ac\)   is positive, then the quadratic has two real solutions. (There are two square roots to a positive quantity.)

 For example, without any effort we can see:

\(2w^{2}-3w+4=0\) has no solutions because is \(b^{2}-4ac =3^{2}-4 \cdot 2 \cdot 4=-23\)  is negative.

\(x^{2}-4x+4=0\) has precisely one solution because \(b^{2}-4ac =16-16=0\).

\(3y^{2}-y-1\) has precisely two solutions because \(b^{2}-4ac =1+12=13\) is positive.

 (REMINDER: We shall see in PART 4 of this course there is no need for any of this. One determine the number of solutions to a quadratic just as quickly by a better, non-memorised, technique.)


People also note that the two solutions to a quadratic can be written:

\(x=-\frac {b}{2a} + \frac {\sqrt{b^{2}-4ac}} {2a} \)  and  \(x=-\frac {b}{2a} – \frac {\sqrt{b^{2}-4ac}}{2a}\).

 This shows that the two solutions lie at symmetrical positions about the value:

\(x=-\frac {b} {2a}\).

Some folk consider this important to hold in mind.

(REMINDER: I don’t! As we shall see in the Part 4 of this course one can determine all this information with ease without one whit of memorization.)


COMMENT: If you know the shape of a quadratic graph (see PART 4), then solutions to the quadratic equation \(ax^{2}+bx+c=0\)  correspond to the locations where \(y=ax^{2}+bx+c\) crosses the \(x\)-axis.  And because the graph is symmetrical we have that \(x=-\frac {b}{2a}\) is the location of the vertex of the graph. [But there are better ways to see this than from analyising the quadratic formula! See PART 4.  Ignore this comment!]



If speed matters to you … maybe.

If a curriculum  insists that you and your students know it and that you must use the word “discriminant,”  then yes, I suppose.

But if understanding has a higher priority in your mind, then the box method will not let you down.

To solve \(ax^{2}+bx+c=0\) just multiply through by \(a\) to make \(ax^{2}\) a square and also multiply through by \(4\)  to avoid fractions. Then let the box guide you!

Please join the conversation on Facebook and Twitter and kindly share this page using the buttons below.
Share on Facebook
Tweet about this on Twitter




Take your understanding to the next level with easy to understand books by James Tanton.



Guides & Solutions

Dive deeper into key topics through detailed, easy to follow guides and solution sets.


light bulb


Consider supporting G'Day Math! with a donation, of any amount.

Your support is so much appreciated and enables the continued creation of great course content. Thanks!


Ready to Help?

Donations can be made via PayPal and major credit cards. A PayPal account is not required. Many thanks!