- Figure out a way to solve the following equations:
\[ a){\sqrt{{\sqrt{x+16}}-{\sqrt{x}}}}=2 \]
- $ -=1 $
29. (A paradox! A paradox! A most ingenious paradox!)
Criticize the following argument:
Let x = 1.
Then $ x^{2}=x $ .
Hence $ x^{2}-1=x-1 $ .
That is, $ (x-1)(x+1)=x-1 $
Consequently, $ x + 1 = 1 $ .
Thus, x = 0.
But x = 1 by definition, so it follows that 0 = 1.
How did this happen? Where is the flaw in the reasoning?
Reproduce this argument for a friend who knows algebra – or even better, for a friend who knows calculus – and see if he or she can identify the problem.
Deriving the Quadratic Formula
Now if a 6 turned out to be 9, I don’t mind, I don’t mind.
- Jimi Hendrix
Solving a linear equation is simple: We just isolate its variable. Bringing this “isolationist” strategy to quadratic equations, however, seems hopeless, because in a quadratic equation, the variable generally appears in two places at once. And yet, amazingly, we can isolate x under these circumstances, but doing so requires some real algebraic artistry.
Let’s begin with a specific quadratic: $ x^{2} + 6x + 6 = 0 $ . First, try solving it on your own without the quadratic formula. You won’t succeed, but your failed attempts will deepen your own understanding and appreciation of the solution when you see it.
Had you spent hours or days trying to solve that equation, you might eventually have found yourself, exhausted and desperate, wishing “If only the constant term were 9! For then we could rewrite the quadratic polynomial as the square of a linear polynomial, and solve the equation by factoring”:
\[ x^{2}+6x+\mathbf{9}=0\quad\Rightarrow\quad(x+3)^{2}=0\quad\Rightarrow\quad x=-3 \]
If only… And yet, desperate wishing can be the mother of invention. We don’t have a 9 in our equation, but… we can put one there if we take care to keep the equation “balanced”. Watch carefully:
\[ \begin{aligned}x^{2}+6x+6&=0\\x^{2}+6x+\mathbf{9}+6&=\mathbf{9}\ $ x+3)^{2}+6&=9\ $ x+3)^{2}&=3\\x+3&=\pm\sqrt{3}\\x&=-3\pm\sqrt{3}\end{aligned} \]
(adding 9 to both sides)
Huzzah! We’ve solved our equation! Now let’s consider our solution carefully to see what makes it tick; doing so will show us how to generalize it, so that it will work for all quadratic equations.
Our method was to introduce a constant $ (9) $ that allowed us to replace our unruly quadratic with something much more manageable – a squared linear polynomial – in which the symbol x appeared only once. After that, the equation was easy to solve; we simply isolated the one appearance of x.
Let us think about what makes certain quadratics (such as $ x^{2} + 6x + 9 $ ) capable of being folded up into the form $ (x + a)^{2} $ . Expanding the “folded” form, we have
\[ (x+a)^{2}=x^{2}+2\boldsymbol{a}x+\boldsymbol{a}^{2}. \]
The coefficients on the right give us the answer we seek. Reading this last equation backwards, we see that a quadratic polynomial will fold up into a perfect square of the form $ (x + a)^{2} $ if…
Its leading coefficient (i.e. coefficient of $ x^{2} $ ) is 1,
Its constant term is… the square of half its x-coefficient.
Thus, we can’t fold $ x^{2} + 6x + 6 $ up into a perfect square of the specified form, since its constant term, 6, isn’t the square of half its x-coefficient, which is 9. But as we saw above, once we know the magic constant
that we need (9), it’s easy enough to inject it into an equation. This insight yields a general method so important that it has its own name.
Completing the Square. To rewrite x² + bx so that x appears just once, add and subtract a certain “magic constant” that we get as follows: Cut the x-coefficient in half, then square the result. Note well: This recipe for the magic constant works only when the leading coefficient of the quadratic is 1.
Let’s pause for a moment and practice completing the square. Here are three examples.
Example 1. Rewrite the expression $ x^{2} + 5x $ so that the symbol x occurs only once.
Solution. Since this quadratic polynomial has a leading coefficient of one, we can complete the square by the process described above. Cutting the x-coefficient in half gives us 5/2; squaring this gives us our magic constant, 25/4. Let us use it now to complete the square:
\[ \begin{aligned}x^{2}+5x&=\underbrace{x^{2}+5x+\frac{25}{4}}_{a perfect square}-\frac{25}{4}\\&=\left(x+\frac{5}{2}\right)^{2}-\frac{25}{4}.\end{aligned} \]
We’ve rewritten our original quadratic in a form in which x appears only once.
The next example will lead us almost to the doorstep of the quadratic formula.
Example 2. Solve $ x^{2}-8x+11=0 $ without using the quadratic formula.
Solution. The polynomial doesn’t factor, but we can complete the square and rewrite it so that x appears just once; we can then solve the equation by isolating x. Since the leading coefficient is 1, we can follow the procedure above. Half of the x-coefficient is -4. Squaring this, we find that our magic constant is 16. Using this to complete the square, we obtain:
\[ \begin{array}{r}x^{2}-8x+11=0\\\underbrace{x^{2}-8x+\mathbf{16}}_{a perfect square}-\mathbf{16}+11=0\end{array} \]
(completing the square)
\[ \begin{aligned}(x-4)^{2}-5&=0\ $ x-4)^{2}&=5\\x-4&=\pm\sqrt{5}\\x&=4\pm\sqrt{5}.\end{aligned} \]
The method we used in this last example can be used to solve any quadratic equation whose leading coefficient is 1. And even if the leading coefficient is not 1, we can divide both sides of the equation by the leading coefficient to make it so. Watch carefully:
Example 3. Solve $ 3x^{2} + 12x + 5 = 0 $ .
Solution. $ 3x^{2}+12x+5=0 $
$ x^{2} + 4x + = 0 $ (dividing both sides by 3 to make the leading coefficient 1)
\[ \begin{aligned}x^{2}+4x+\frac{5}{3}&=0\quad&(dividing both sides by3\\x^{2}+4x+\mathbf{4}-\mathbf{4}+\frac{5}{3}&=0\quad&(completing the square)\ $ x+2)^{2}+\frac{5}{3}&=4\ $ x+2)^{2}&=\frac{7}{3}\\x+2&=\pm\sqrt{7/3}\\x&=-2\pm\sqrt{7/3}.\end{aligned} \]
We can use this method to solve any quadratic equation. However, rather than completing the square every time we need to solve a quadratic equation, we can automate the process by deriving a formula (“the quadratic formula”) that takes us directly from a quadratic to its solutions, keeping the square-completing details under the hood, where we need not think about them. If you understood this last example, then deriving the quadratic formula will be simple, even trivial.
Problem. Derive the quadratic formula. That is, extract a formula for the solutions of the general quadratic equation $ ax^{2} + bx + c = 0 $ from out of that equation itself.
Solution. The method we use to derive the quadratic formula can be summarized in three words: Complete the square. A fair amount of algebra is involved, but it’s all routine. Here are the details.
\[ ax^{2}+bx+c=0 \]
\[ x^{2}+\frac{b}{a}x+\frac{c}{a}=0 \]
(dividing both sides by a to make the leading coefficient 1)
\[ x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}+\frac{c}{a}=\frac{b^{2}}{4a^{2}} \]
(completing the square)
\[ \left(x+\frac{b}{2a}\right)^{2}+\frac{c}{a}=\frac{b^{2}}{4a^{2}} \]
\[ \left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}}{4a^{2}}-\frac{c}{a} \]
\[ \left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}-4ac}{4a^{2}} \]
\[ x+\frac{b}{2a}=\pm\sqrt{\frac{b^{2}-4ac}{4a^{2}}} \]
\[ x=-\frac{b}{2a}\pm\sqrt{\frac{b^{2}-4ac}{4a^{2}}} \]
\[ x=-\frac{b}{2a}\pm\frac{\sqrt{b^{2}-4ac}}{2a} \]
\[ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. \]
We’ve done it! Simply by completing the square, we’ve derived a remarkable formula for the exact solutions of any quadratic equation whatsoever.
To sum up our discussion of quadratics, let us note that, apart from trivial equations in which x can be isolated (such as $ 4x^{2} = 9 $ ), there are only two ways that anyone ever solves a quadratic equation in practice: by factoring (with help from the zero product theorem), and, when factoring fails, by the quadratic formula. $ ^{*} $
Exercises
- Rewrite the following expressions in equivalent forms in which the symbol x appears only once:
$ x^{2} + 8x $
$ x^{2} + 9x $
$ x^{2} - 6x $
$ x^{2} - x $
$ 4x^{2} + 8x $
$ -3x^{2} + x $
$ x^{2} + 3x + 1 $
$ 2x^{2} - 4x + 4 $
Solve the equation $ 2x^{2}-5x-1=0 $ without using the quadratic formula.
Derive the quadratic formula (without using any notes, of course).
Here’s an appealing geometric interpretation of completing the square.
Given an expression of the form $ x^{2} + bx $ , we can complete the square by thinking of the two terms as the areas of a square and a rectangle. If we split the rectangle in half (along its “b” side) and paste the halves to two sides of the square as shown below, the resulting figure will be a square… with a bite out of one corner.
To literally complete this square, we must fill in the missing corner.
What is the area of the missing piece that will complete our square?
Observe that this is indeed the “magic constant” described in the section above.
This geometric demonstration only works when b > 0. Explain why.
Can you come up with a related geometric demonstration that will works when b < 0?
Why Negative Numbers?
Whole numbers are for counting, fractions are for measuring, but what are negative numbers really for? When, outside of a classroom, do we ever need to add, subtract, multiply, and divide negative numbers? Despite all evidence to the contrary, authors of arithmetic textbooks often assert that we need negative number arithmetic to make sense of debts, subzero temperatures, and the like. This is utter nonsense, as anyone with a brain in his head should be able to recognize after a few minutes of thought.
If I owe Groucho 3, Chico 5, and Harpo 10, how much do I owe the Marx brothers altogether? Obviously, I owe them 18, a figure that I reached, just as you did, by adding three positive numbers. Granted, someone could reach the same conclusion by computing $ (-3) + (-5) + (-10) = -18 $ , but why bother? This example, moreover, is typical. Every so-called “real world application” of negative number arithmetic can be solved in a natural way with positive numbers alone. If you don’t believe me, try to come up with an example that proves me wrong. Examine an arithmetic textbook sometime, and try to find among its “applied” examples and exercises even a single instance in which negative number arithmetic is indispensable. You will search in vain. History bears out my claim, too. In traditional bookkeeping, negative numbers were not used. Instead, debits were written in red ink. (This is why a business in debt is said to be “in the red.”) Chronology works similarly: We say that Archimedes died in 212 B.C. We do not say that he died in -212.
Having cleared the air of disingenuous answers, let’s repeat the question. Why do we need negative number arithmetic? The honest answer is that negative number arithmetic exists to simplify algebra. Without algebra, negative number arithmetic would not exist. And yet, not many centuries ago, even algebra was done without negatives. To give you a feel for algebra in the “pre-negative age,” let’s reconsider the problem of solving quadratic equations.
To take a specific example, the equation $ -3x^{2}+5x+1=0 $ would have been utter gibberish to those in the pre-negative age, on account of that mysterious leading coefficient. However, they would have found the equivalent (to our eyes) equation $ 3x^{2}=5x+1 $ satisfactory. To solve this equation, we, of course, would use the quadratic formula, which yields two solutions, one negative and one positive:
\[ \frac{-5+\sqrt{37}}{-6}\text{and}\frac{-5-\sqrt{37}}{-6}. \]
But because the solution on the left is negative, pre-negative minds would say the equation has one solution: the one on the right, which they would have expressed in the equivalent form
\[ \frac{5+\sqrt{37}}{6}. \]
How did they get this? We, of course, could obtain it from “our” form of the positive solution as follows:
\[ \frac{-5-\sqrt{37}}{-6}=\frac{-5-\sqrt{37}}{-6}\Big(\frac{-1}{-1}\Big)=\frac{5+\sqrt{37}}{6}. \]
But this would have been so much bollocks to our pre-negative friends. The original expression on the left-hand side would have been viewed as a heap of nonsense; our subsequent attempt to multiply it by another such heap, -1/-1, would have seemed the ravings of a lunatic.
Nonetheless, our pre-negative friends could easily have produced the positive solution directly, with no intervening negative numbers! How did they solve the problem? Exactly as most people would solve it today – by appealing to a formula that they had memorized, perhaps without entirely understanding it. Their formula (or rather formulas) would have looked different than ours, though. This is how, in the pre-negative age, a textbook discussion of quadratic equations could have been summarized:
A “Pre-Negative” Approach to Quadratic Equations.
Quadratic equations come in four basic forms:
$ Ax^{2} + Bx + C = 0 $ . These obviously have no solution (a sum of positives can never be zero!)
$ Ax^{2} + Bx = C $ . These always have one solution. Namely, $ x = $ .
$ Ax^{2} = Bx + C $ . These always have one solution. Namely, $ x = $ .
$ Ax^{2} + C = Bx $ . These are the oddballs.
If $ B^{2} < 4AC $ , then there is no solution.
If $ B^{2} = 4AC $ , there is one solution. Namely, $ x = $ .
If $ B^{2} > 4AC $ , then there are two solutions. Namely, $ x = $ .
Thus, to solve $ 3x^{2}=5x+1 $ in the pre-negative era, one would have used formula number 3 in the box above, which directly produces the solution
\[ x=\frac{\sqrt{5^{2}+4(3)(1)}+5}{2(3)}=\frac{\sqrt{37}+5}{6} \]
in such a way that negative numbers never intervene!
The moral of this story is that it is entirely possible to do algebra without negative numbers. However, once we admit them, algebra becomes ever so much simpler. Algebra students sometimes grumble about having to memorize the quadratic formula. Be thankful at least that we have negatives! Memorizing the quadratic formula is a trifle compared to memorizing the contents of the box above. Negative number arithmetic is a tremendous laborsaving device, which is precisely why it was invented.
Exercises
- Pretend you are living in the pre-negative age. Solve the following quadratic equations while playing on “period instruments” (the boxed formulae in this section). Then solve them as you would today.
$ 8x^{2} + 4 = 5x $
$ 3x^{2}+4x=1 $
\[ 2x^{2}=x+5 \]
- Ponder the following words of Frances Maseres, an 18 $ ^{th} $ -century mathematician:
They [negative numbers] are useful only… to darken the very whole doctrines of the equations and to make dark of the things which are in their nature excessively obvious and simple.
The pre-negative days were surprisingly recent.