Algebra is a programming language for geometry.
When they teach you how to solve such equations at schools, they usually give you only a template formula, where you insert the coefficients , , from the equation 1 to compute the solution:
But I guess that when you look at this complicated formula, it probably doesn't tell you much :-P That's because at schools they don't explain where does this formula come from and what does it mean. So your only option is to memoize this formula and apply it mindlessly
Today it's gonna change! :-> Not only I'll show you how to get to this formula step-by-step, by investigating the geometry hidden in the quadratic equation, but you'll also see that in most cases this formula is totally superfluous and it unnecessarily complicates calculations, which you can easily perform in your head, with much less of work! It's enough if you understand how it all works.
Just imagine how much you'll impress your friends, when by mere looking at the equation you instantly tell them the solution: , before they even manage to recall the equation and compute the infamous "delta"! :-> I'm sure they'll be very shocked :-D
But before we try to solve the full quadratic equation shown above, let's first warm up and try to solve some simplier forms of it, deprived of some of its lower-degree terms. Thanks to that you'll have opportunity to observe what each of them contributes to the quadratic equation and how it affects the solution. This way we will build up the full solution step-by-step, so it will be easier for you to understand where it comes from.
The simplest quadratic equation has only the second-degree term (that is, quadratic term) equated to some constant number:
How to understand this equation? On the left we have , that is some mysterious square. We don't know its side length . But we know its area! :-) Notice that this whole square have to be equal to the right-hand side which tells us exactly the area: it's equal to , that is some number of known value. This situation is shown in the picture below:
From the article about squares and square roots you probably know already that the side of a square (ie. our unknown ) is the square root of its area (here the which is known). And such problems you already know how to solve The solution of such an equation is trivial: You simply take the square root of both sides of equation 3 and you get:
Since we have just plain on the left, and there is no trace of it on the right, our equation is solved! :-D See? It wasn't so hard
Notice that we've got actually two roots: positive and negative. That's because both these values, when squared (or multiplied with themselves), give the same positive square. Negative solutions were depreciated and ignored for hundreds of years, because mathematicians couldn't imagine what could such "less than zero" (negative number) mean. But it's possible to imagine such a "negative side of a square" and it's easy when you already know how to do it. I'll show you that in a separate article.
But what if there's some coefficient with the ?:
No problem. It just means that we have not one, but several such squares: there are of them. All of them taken together have an area equal to . For instance, when is 3, it means that we have three squares with total area . This example is shown in the animation below:
To figure out the side of one such square (that is our ), we need to know its area. Since squares has a total area , one such square has an area . So you just divide both sides of the equation 5 by to get the equation for one single square:
As you can see, the area of the square could be described not only by a single number, but also by a more compound expression. But as long as this expression doesn't contain any unknowns ('s), it's still some known area. You can wrap this under one single constant, let's say , if you like:
and from now on you can solve it the same way as before :-) So let's now find out the side of our square, by taking the square root of both sides of equation 7:
remembering to "unwrap" the letter at the end of the day by substituting it in the final formula by the whole expression hidden behind it
And this way we have our equation 5 finally solved :-) Notice that it doesn't differ much from the solution 4: simply the value under the square root has been divided (scaled down proportionaly) by , because we're interested in the side of one single square instead of all the squares taken together. Equation 9 could be understood as: " is the side (or square root) of one of squares with total area ".
And what will happen if instead of a known area we have another unknown of the first degree, namely the linear term?:
Although at the first glance this problem seems harder than the former one, it turns out that it's actually simplier! :-D Why? Well, let's try to decipher its meaning:
On the left-hand side we have some square with an unknown side . Its area must be equal to whatever the right-hand side tells us. And on the right-hand side we have a product of two factors: and , which we can consider as sides of a rectangle. But wait! The right-hand side tells us that one side of this rectangle is the same as the side of our square! If that's true, then this rectangle must be a square too, because only then its area could still be equal to the area of the square from the left-hand side. Therefore both sides of the equation describe the same square! It is shown in the picture below:
Let's recap this in somewhat simpler words: We have a square of unknown area; one of its sides is unknown and equals , and its second side is known and equals . But we know that squares have equal sides! :-D Therefore the unknown side must be also equal to . And that's exactly the solution:
If it's still unclear to you, then recall the definition of a square. A square is a number multiplied with itself:
If you write again equation 10 below:
and compare it with equation 12 above, then you'll surely notice that one of the 's on the right-hand side has hidden itself under the letter , so . It's like you had two curtains with the same object hidden behind each of them. If you discover one of them, then you'll already know what's behind the other
But it's not the only solution to equation 10. Every 2nd-degree equation has always to have two solutions. Where's the other one? Notice that equation 10 contains a common factor on both sides. It's easier to see if you write it this way:
If you remember from the article about equations, such a factor, which repeat in the same form on both sides of the equation, is just like the same weighting piece placed on both plates of a balance, and it is then equivalent to zero. That's why our lost solution is:
This time the coefficient of will also stand no problem for us:
Simply instead of one square with known one of its sides, we now have several such squares: as many, as the coefficient of tells us. So, same as before, you simply divide both sides by to get the equation describing one square:
and the further solution method goes exactly the same way as for the equation 10, resulting in:
That's why this solution is so easy to find by mere looking at the equation: will be always equal to the coefficient of , and of course zero as the other possible solution. And if you have any coefficient of , then you quickly divide in your head by this coefficient and tell the answer before your friend will compute it from the formulas ;-)
As you see, in these several starting simple cases you can easily deal without computing the "delta" and memoizing complex formulas at all, which are like using a sledgehammer to crack a nut :-P But even with the subsequent examles, which will be full-fledged quadratic equations with all terms included, you can deal with ease, if you just understand how they work :-)
OK, we know already how to solve a quadratic equation when on one side of the equal sign we have some constant , and on the other side we have just plain square of . But what if there's something else there? For example, a term linear in ?
Well… then we don't have simply a square which we could just take a square root of. Instead we have a square () and a rectangle (), as in the animated picture below:
We cannot take the square root of a rectangle, only of a square. So we have somehow transform this rectangle into an equivalent square. How can we achieve it?
We can divide the rectangle into two halves, as you can see in the animation, and obtain two smaller rectangles with sides and .
Next we can move one of these halves somewhere else. This maneuvre doesn't change the area of our figure from the left side of the equation at all, but… it will get us closer to the shape we already know from the section about short multiplication formulas!
Well, almost: there is one little square piece in the corner which is missing :-P Its side is , so its area is . So when we complete the square by adding this little missing piece, we will get the short multiplication formula, whic we can then "wrap" as a single square – the big square in the animation. Of course, to keep the balance, we have to add this small little square to both sides of our original equation! This way:
And we get on the left side of the equation the short multiplication formula. If you don't see it yet, I'll try to say it more verbose:
You can also read it out from the picture: the side of the big square consists of the lengths and , which together gives you the full length of the big square, that is . The area of the big square is . Therefore we can wrap the whole left side (big square) as:
and this way we obtained on the left side of the equation a normal square! So now we can take the square root of it without any problems. On the right side we have just a bunch of constants, which we can join into one. This resulted in that our equation now has exactly the same form as the one we dealt with in the previous section! And we know already how to solve such a creature :-) Just take the plain old square root of both sides to get:
To solve for , you just get rid with that from the left side, by subtracting it from both sides:
and in principle we have the equation solved – because we have just single on one side, and on the other side we have an expression in terms of known quantities, so we can easily compute its value. But before we do that, it is a good idea to simplify it a bit to avoid repeating work…
For those concerned:
Before we proceed with that, notice one thing: subtracted on the left side we can also write as:
which enlighten us that under both radicals there is a common term (but under one radical there's additionaly subtracted from it). Moreover, we gain some extra knowledge about the solution of the quadratic equation: If it's not a simple square root (a side of a square), that is, a square root of some perfect square, then it will always be a sum or a difference of two such square roots (the difference of sides of two squares). Euclid wrote on this in the 10th book of his "Elements". Such sum of square roots he called a "binomium" (lat. "two names"), and the difference of square roots he called a "residuum" (lat. "remainder", "left-over"). We'll get back to this case when we'll try to solve the cubic equation. Then you'll see that its solutions have very similar form.
The meaning of this sum/difference of square roots you can easily understand if you remember that square roots are the sides of some squares. So the difference of square root represents the difference between the sides of some two such squares. What are these squares? Is it possible that one of them is our square from the original equation, and the second one is the closest perfect square? Think about that for a while
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