Chapter 3
Solving Equations
Equivalent Equations
You boil it in sawdust, you salt it in glue,
You condense it with locusts and tape,
Still keeping one principal object in view –
To preserve its symmetrical shape.
- Lewis Carroll, The Hunting of the Snark, Fit the Fifth
So far, every equation we’ve considered has been a declarative statement: a blunt assertion that two different-looking expressions represent the same thing. Other equations, however, have an interrogative flavor. For example, we can read $ 2x + 3 = 11 $ as asking us, “Which values of x make $ 2x + 3 $ equal 11?” Such values are called the equation’s solutions. [Here, 4 is a solution, since $ 2(4) + 3 = 11 $ .]
To solve an equation, we algebraically twist it into an equivalent equation – a new equation with the same solutions as the original. If need be, we twist this second equation into a third, then a fourth, and so on, always keeping one principal object in view: to preserve the original solutions. Eventually, if we are skillful algebraists, we’ll reach an equation with obvious solutions. But since this equation is equivalent to the original one, its solutions are also the original equation’s solutions. Oh frabous day!
The algebraic twists come in two basic varieties:
Two Basic Moves for Solving Equations.
These two “moves” turn an equation into an equivalent one (i.e. one with the same solutions):
Replace an expression in the equal by an equal expression.
‘Do the same thing’ to both sides.*
The first move is perfectly clear. It tells us, for example, that $ 5x + 3x = 16 $ is equivalent to 8x = 16. Since the sole solution to this last equation is obviously 2, the original equation’s only solution is also 2.
We usually use the second move to isolate the unknown on one side of the equation. For example, we can use it twice in succession to solve $ 5x + 2 = 9 $ . First, we subtract 2 from both sides, yielding the equivalent equation 5x = 7. Next, we divide both sides by 5, yielding x = 7/5. This last equation’s only solution is 7/5 (obviously!), so the original equation’s only solution must also be 7/5.
How to Write and How Not to Write
When solving an equation, it’s customary to write up one’s work vertically: Equivalent equations should be written in a column, without any symbol linking them. In rare circumstances (when space is tight, as in a footnote), one might employ a horizontal format, where the equivalent equations are written in a row, linked by arrows. The vertical format, however, is much preferred.
Here’s an example of how Student A (who gets an A on every test) uses the vertical format.
Example. Solve $ 5x + 2 = 9 $ .
Solution (A).
\[ \begin{aligned}5x+2&=9\\5x&=7\quad&(subtracting2frombothsides)\\x&=7/5.\quad&(dividingbothsidesby5)\end{aligned} \]
Note that the first line of Student A’s solution restates the original equation - a good practice to follow. Her parenthetical statements justifying each step are nice, but they aren’t always necessary: You can omit them when the nature of a step is clear - as it usually is. $ ^{*} $
Student D, alas, commits two terrible sins of mathematical writing: equals abuse and track covering.
Example. Solve $ 5x + 2 = 9 $ .
Solution (D). $ 5x + 2 - 2 = 9 - 2 $
\[ \begin{aligned}5x+2-2&=9-2\\=\frac{5x}{5}&=\boxed{\frac{7}{5}}\end{aligned} \]
Perhaps Student D began by restating the original equation, but if so, he then obscured it by scrawling –2 on both sides. Perhaps he wrote 5x = 7 on the next line, but the evidence vanished when he divided both sides by 5 and then recorded the result on the very same line, thus covering up his tracks a second time. Such track covering produces an unreadable mess. Don’t do it. Never forget that writing is for a reader. Equals abuse is, if anything, even worse. It isn’t just messy. It actually produces mathematical nonsense. For instance, Student D abused the equals sign by inappropriately placing one before 5x/x, thus equating that expression to 9 - 2. Note the absurd consequence: Student D has written that 9 - 2 = 7/5. Not so.
Equals abuse, incidentally, takes two forms: inserting the equals sign in inappropriate places, and omitting the equals sign where it is necessary (equals neglect). The rule for using this symbol is simple:
The equals signs goes between equal expressions. It never goes between equivalent equations.
Clear writing is essential in mathematics. Pay attention to how mathematics is written in textbooks, and learn to imitate it. You must learn to write up your solutions like a civilized human being. Don’t be like Student D, who actually can solve equations, but whose sloppy writing makes him look poorly educated. Don’t cover your tracks and don’t commit equals abuse. Or as Student D might put that same advice,
\[ \begin{array}{r}{D o n^{\prime}t r a c k s_{y o u r t r a c k s}=a n d d o n^{\prime}t c o m m i t=e q u a l s=a b u s e.}\end{array} \]
Exercises. Which of the following solutions involve equals abuse (or neglect)? Explain your answers.
Solve: $ 2x + 3 = 8 $ .
Simplify: $ 3x^{2} - (-x^{2} + x) $
Simplify: $ 3x^{2} - (-x^{2} + x) $
Solution: $ 2x + 3 = 8 $
\[ = 2x = 5 \]
\[ = x = 5/2 \]
\[ \begin{aligned}Solution:3x^{2}-(-x^{2}+x)\\=3x^{2}+x^{2}-x\\=4x^{2}-x.\end{aligned} \]
\[ \begin{aligned}Solution:3x^{2}-(-x^{2}+x)\\3x^{2}+x^{2}-x\\4x^{2}-x.\end{aligned} \]
Solving Linear and Quadratic Equations
Any algebraic expression of the form $ ax^{n} $ (where n is any whole number or 0) is called a monomial in x. The constant a is called the monomial’s coefficient. Note that constants count as monomials, because we can always think of a constant such as 8 as $ 8x^{0} $ .
A polynomial in x is a sum of monomials in x. A polynomial’s degree is the largest exponent of its terms. For example, $ 5x^{3}-x^{2}+1 $ is a polynomial in x of degree 3. Polynomials of degree 1, 2, and 3 are called linear, quadratic, and cubic polynomials, respectively. $ ^{*} $
Any equation that has the form $ ax + b = 0 $ (or can be put into that form) is called a linear equation. Solving a linear equation is easy: we just isolate x on one side of the equation. Nothing could be simpler.
Example 1. Solve $ 4x + 3 = 0 $ .
Solution. $ 4x+3=0 $
Any equation that has (or can be put into) the form $ ax^{2} + bx + c = 0 $ is called a quadratic equation. We’ll start with a particularly simple example: $ x^{2} = 4 $ . Clearly, 2 is this equation’s only positive solution, but there’s also a negative solution: -2. From this trivial observation, we glean an important lesson:
Important Lesson. For any k > 0, the equation $ x^{2} = k $ has two solutions: $ $ and $ - $ . (We often express them with the shorthand notation $ $ .)
Having learned this much, you can already solve any quadratic that lacks an x-term. We just isolate x:
Example 2. Solve $ 2x^{2}-5=0 $
Solution. $ 2x^{2}-5=0 $
\[ \begin{aligned}2x^{2}&=5\quad&(Adding5tobothsides)\\x^{2}&=5/2\quad&(Dividingbothsidesby2)\\x&=\pm\sqrt{5/2}\quad&(Bytheimportantlesson!)\\\end{aligned} \]
When a quadratic equation has an x-term, isolating x is much harder. Don’t believe me? As a challenge, take a few minutes and try isolating x in the equation $ 3x^{2} + 10x - 1 = 0 $ . Success requires real ingenuity, so don’t feel bad when you fail. Please do try it, though, because trying and failing will help you appreciate the ingenious solution to this problem that I’ll present at the end of this chapter.
Before I can present that ingenious solution, however, we’ll need to discuss an important technique for solving factorable quadratic equations. The technique hinges upon the following important theorem.
The Zero Product Theorem. The values that make a product equal zero are the values that make any of its factors equal zero.
The Zero Product Theorem tells us that the solutions of $ (2x+1)(x-1)=0 $ are those of $ (2x+1)=0 $ and $ (x-1)=0 $ : Thus, -1/2 and 1 are the solutions we seek. Substitute them back into the original product, and you’ll see why the theorem holds.
We can solve certain “nice” (i.e. factorable) quadratics by pushing all of their terms to one side of the equals sign, factoring the resulting polynomial, and using the Zero Product Theorem. Here’s an example of this technique in action.
Example 3. Solve the equation $ 3x^{2}=8x-4 $
Solution. $ 3x^{2}=8x-4 $
\[ 3x^{2}-8x+4=0 \]
(pushing all terms to one side)
\[ (3x-2)(x-2)=0 \]
(factoring)
\[ x=2/3,\ x=2. \]
(Zero Product Theorem)
That is, the original equation’s solutions are 2/3 and 2.
This method works well when quadratics factor nicely. However, when factoring fails (as it often does), we use the big gun: the quadratic formula, which is powerful enough to solve any quadratic equation. I’ll state the formula here, and I’ll derive it at the end of the chapter, once you’ve grown accustomed to using it. The derivation at the end of the chapter will answer the riddle I posed earlier: How can we isolate x in a quadratic equation? All this in good time. For now, I just want you to become comfortable using the quadratic formula, which you should commit to memory right away.
The Quadratic Formula.
The solutions to the equation $ ax^{2} + bx + c = 0 $ are
\[ \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. \]
Example 4. Solve the equation $ 2x^{2}-7x-1=0 $ .
Solution. This quadratic doesn’t factor, so we’ll solve it with the quadratic formula, obtaining
\[ x=\frac{7\pm\sqrt{49-4(2)(-1)}}{2(2)}=\frac{7\pm\sqrt{57}}{4}. \]
Thus, this equation has two solutions, $ (7 + )/4 $ and $ (7 - )/4 $ . In an applied problem, we might run these through a calculator and note that they are approximately 3.637 and -0.137. In a purely mathematical problem, we would leave these solutions in their exact form.
The quadratic formula will also tell us – in its own way – when a quadratic equation has no solution, as the next example shows.
Example 5. Solve $ 3x^{2}-2x+1=0 $ .
Solution. By the quadratic formula, the “solutions” are
\[ x=\frac{2\pm\sqrt{4-4(3)(1)}}{6}=\frac{2\pm\sqrt{-8}}{6}, \]
but… these aren’t real solutions, since -8 has no square root. In fact, we should read that square root of a negative number as an “error message” indicating that the equation has no solution.
Before we can apply the quadratic formula, we must first put the quadratic equation that we wish to solve in the appropriate form: $ ax^{2} + bx + c = 0 $ . The following example shows how to do that when necessary.
Example 6. Solve $ -3x{2}+5x(x-2)-2x-7=x{2}-10x $ .
\[ \begin{aligned}-3x^{2}+5x(x-2)-2x-7&=x^{2}-10x\\-3x^{2}+5x^{2}-10x-2x-7&=x^{2}-10x\\x^{2}-2x-7&=0\\x&=\frac{2\pm\sqrt{32}}{2}\\&=\frac{2\pm4\sqrt{2}}{2}\\&=1\pm2\sqrt{2}.\end{aligned} \]
(distributing the 5x)
(pushing all terms to one side)
(quadratic formula).
Thus, we conclude that the solutions to the original messy equation are $ 1 + 2 $ and $ 1 - 2 $ .
Exercises
- (Revisiting Example 6.) Carefully explain why $ = 1 $ .
[Hint. The left-hand side represents two numbers. Write them out separately and play with them.]
Explain why $ $ is the same thing as $ $ . [Hint. Recall the old “multiply-by-1 trick” from Chapter 1.]
Solve each of the following equations:
3x - 5 = 11
$ -u+17-u= $
$ 7 + 3(-x - 5) = -x + $
$ 16x^{2}=121 $
$ x^{2}-4x-5=0 $
$ z^{2}-z-12=0 $
$ 2x^{2} + 15x - 7 = 0 $
$ -t^{2} + t + 1 = 0 $
$ 2x^{2}-3x-5=0 $
\[ j)w(-2w-5)=w^{2}+1 \]
- $ x^{2} + x - 2 = 0 $
\[ \left| Ⅰ \right)-\left(-x-\left(-2x-x^{2}\right)+x\right)=-1 \]
$ x^{2} + (x + 2)^{2} = 16 $
$ x^{2} + 1 = 0 $
$ x^{2}-2x=-1 $
- Dividing both sides of an equation by a constant yields an equivalent equation, so we can sometimes simplify our work at the outset by dividing out any constant factors common to all terms.
To understand this trick, solve $ 10x^{2}-15x-30=0 $ twice: First, by applying the quadratic formula directly; second, by dividing both sides by 5 and then applying the quadratic formula.
The answers you obtained by the two methods may look different, but they must, of course, represent the same numbers. Show that they are indeed equal (if you haven’t done this already).
Use this trick to solve the following equations:
\[ 300x^{2}+500x+100=0 \]
\[ ii)-28x^{2}+35x=42. \]
\[ iii)5x^{2}=50+15x \]
- Multiplying both sides of an equation by a constant yields an equivalent equation, so we can sometimes simplify our work by “clearing fractions” before solving an equation. The equation $ (2/3)x^{2}+(1/2)x-1=0 $ , for example, will be easier to handle if we first multiply both sides by 6.
Solve the equation in the previous paragraph both ways: First, by directly applying the quadratic formula; second, by clearing fractions and then applying the quadratic formula.
The answers you obtained by the two methods look different, but they must, of course, represent the same numbers. Show that they are indeed equal. (You may find this part challenging. Proving that two different forms of the same number are equal can be difficult when radicals are involved.)
Use this “clearing the fractions” trick to solve the following equations:
\[ i)\ -\frac{3}{7}x+\frac{1}{2}=3 \]
\[ ii)\frac{1}{10}x-\frac{7}{5}=\frac{3}{20} \]
\[ iii)\ \frac{3}{4}x^{2}-4=\frac{1}{9}x \]
Find the solutions of the following equation: $ bx^{2} + cx + a = 0 $ .
The “solution” of the following problem is incorrect. Point out precisely where the would-be solver’s reasoning goes astray. (Note well: I am not asking, What should the solver have done?, but rather, At what point did he make a logically invalid step, and why is that step invalid?)
Problem. Solve the equation $ x^{2}-x-6=1 $
“Solution”. $ x^{2}-x-6=1 $
\[ (x+2)(x-3)=1 \]
Thus, the solutions are -2 and 3.
Finally, now that you’ve pointed out the logical error, find the equation’s true solutions.
- The expression $ b^{2}-4ac $ is called the discriminant of a quadratic equation $ ax^{2}+bx+c=0 $ .
(Thus, for example, the discriminant of the equation $ 3x^{2} + 4x - 6 = 0 $ is 88, as you should verify.)
Quadratic equations “usually” have two solutions, but those whose discriminant is 0 have only one solution. Explain why this is true.
Come up with an example of a quadratic equation with only one solution.
What can we say about a quadratic equation with a negative discriminant?
- Sometimes higher degree polynomial equations can be solved by recognizing that they are actually quadratics in disguise. Consider, for example, $ x^{6} + 2x^{3} - 3 = 0 $ . This terrifying $ 6^{th} $ -degree polynomial in x turns out to be, if we look at it in the right way, a perfectly harmless quadratic polynomial… in $ x^{3} $ . Once we see this, we can solve it by making a substitution as follows:
\[ \begin{aligned}&x^{6}+2x^{3}-3=0\\&u^{2}+2u-3=0.\\&(u-1)(u+3)=0\\&u=1,\ u=-3\\&x^{3}=1,\ x^{3}=-3.\\&x=1,\ x=\sqrt[3]{-3}\\ \end{aligned} \]
\[ \left(\mathbf{Letting}u=x^{3}\right) \]
(Translating back to x)
Thus 1 and $ $ are the solutions to the original $ 6^{th} $ -degree polynomial equation.
Explain why $ $ is a real number (and why $ $ isn’t).
Solve the following equations:
\[ \begin{aligned}&i)x^{8}-x^{4}-2=0\quad&ii)3x^{4}-x^{2}-4=0\quad&iii)-x^{62}+2x^{31}+7=0\end{aligned} \]
How to Write, Part 2 (Story Problems)
Quadratic equations are easy to solve. If you take care not to cover your tracks, abuse the equals sign, or make arithmetic errors, little can go wrong. The procedures involved are purely algorithmic; they can be carried out by brainless bundles of plastic and silicon. In contrast, story problems require thought and the ability to express it in writing. As regards the latter, you should always follow two cardinal rules:
Two Rules for Writing.
Define any symbols you introduce, either in words (“let x be such and such…”), or, if appropriate, on a clearly labeled figure.
If the solution involves several parts, write phrases or sentences explaining how they are related.
You may find it helpful to do your exploratory scratchwork on one page, and then write up your polished solution on another. In this spirit, I’ll pose a problem and then I will give two solutions: First, a “stream of consciousness solution,” which attempts to show how one might think about solving the problem; Second, a formal solution, as one would actually write it up for a homework assignment or test.
Example 1. A right triangle’s hypotenuse is 4 units. Its long leg is 2 units longer than its short leg. How long are its legs?
“Stream of Consciousness Solution.” Well, there’s a right triangle in this problem, so I’d better draw one. Its hypotenuse is 4 units, so I’ll scribble a 4 on there.
What am I looking for in this problem? The lengths of the legs. I guess I’d better give them names. How about x and y. I’ll put them on my figure, too. Now what? Well, I know the sides of any right triangle are related by the Pythagorean Theorem. If I apply that to my triangle, I’ll get $ x^{2} + y^{2} = 16 $ . Hmmm… If there were only one variable in that equation, I could solve for it… but there are two. Well, can I get rid of one? Maybe. What was that business about the long leg and short leg in the problem? The long leg is 2 units longer than the short leg. Ah, I don’t need y at all:
I can just call the long leg $ x + 2 $ instead. Let’s redraw that figure.
Now the Pythagorean theorem says that $ x^{2} + (x + 2)^{2} = 16 $ . OK… now that’s much better. Now I have a quadratic equation in x, and I can solve those. Once I’ve solved it, I’ll know x, the short leg’s length. To get the long leg’s length I’ll just add 2 to the short leg. So basically the problem is solved, apart from carrying out all the computational details. OK – let’s carry out those details… [Et cetera. The rest is straightforward.] ♦
Now that we’ve discovered a path to the solution, we must write it up the solution in a polished form. We’ll do this next, in our actual polished solution, which will follow the “two rules for writing” listed above, as well as the earlier advice on how to write a readable column of equivalent equations.
Polished Solution to Example 1.
Let x be the short leg’s length.
Then the long leg’s length must be $ x + 2 $ .
By the Pythagorean Theorem,
\[ x^{2}+(x+2)^{2}=16. \]
We can solve for x, the short leg’s length, as follows:
\[ x^{2}+(x+2)^{2}=16 \]
\[ x^{2}+(x^{2}+4x+4)=16 \]
\[ 2x^{2}+4x-12=0 \]
\[ x^{2}+2x-6=0 \]
\[ x=\frac{-2\pm\sqrt{28}}{2}=\frac{-2\pm2\sqrt{7}}{2}=-1\pm\sqrt{7}\;. \]
Since x is a length, it must be positive. Thus, we can throw out the equation’s negative solution. Hence, the short leg’s length is the remaining solution, $ -1 + $ units.
The long leg is 2 units longer than this, so the long leg’s length is $ 1 + $ units.
Observe how that followed the rules of writing we’ve discussed. The solution has a definite narrative arc: The characters are introduced (i.e. the one new symbol x is explicitly defined), the scene is set (the figure), a relationship is established (via the Pythagorean Theorem, cited by name), a problem is resolved (the quadratic equation is solved – without any equals abuse or track-covering), an interpretation is presented (the concluding lines), and they all live happily ever after.
Let’s put this problem to rest, and look at a few more story problems. Try to solve the problems yourself before you read the solutions. When you do read the solutions, pay attention not only to the method by which they are solved, but also the style in which the solutions are written.
Story Problems – More Examples
Example 2. The hypotenuse of a right isosceles triangle is 2 units long. Find the triangle’s area.
Solution. Let x be the length of the triangle’s two legs. By the Pythagorean Theorem, we have $ x^{2} + x^{2} = 4 $ . Solving for x yields $ x = $ . Since x is a length, it must be positive, so the triangle’s legs are each $ $ units long. Finally, a triangle’s area is half its base times its height, this triangle’s area must be $ (1/2)()() = 1 $ unit $ ^{2} $ .
We can solve many problems by expressing the same thing two different ways; this gives us one equation, which we can then solve. In the next example, we’ll find a cylinder’s surface area twice. One way will be easy – in fact, its numerical value will be given to us. The other way will require some ingenuity, which I hope you’ll appreciate.
Example 3. Suppose that the height of a cylindrical can (closed at both ends) is twice the diameter of its base. If the can’s surface area is $ 1000 , cm^2 $ , what must its radius and height be?
Solution
Let r be the radius of the can’s base (in cm).
Since the can’s height is twice its diameter, it must be 4 times its radius, as indicated in the figure:
To find an algebraic expression for the can’s surface area, we’ll do some surgery: First, we’ll remove the can’s circular top and bottom, then we’ll slice the remaining tube along the dashed line, and finally, we’ll roll it out flat, transforming it into a rectangle. Thus, the can’s total surface area, A, turns out to be the sum of areas of the two circles and the rectangle. (Note that the circular base, when rolled out, becomes one side of the rectangle; the rectangle’s width is therefore equal to the circle’s circumference.) Thus,
\[ A=\pi r^{2}+\pi r^{2}+(2\pi r)(4r)=10\pi r^{2}. \]
Since the surface area is known to be $ 1000 , cm^{2} $ , we can substitute this for A in the previous equation, giving us $ 1000 = 10r^{2} $ . Solving this for r, we find that
\[ \begin{aligned}1000&=10\pi r^{2}\\\frac{100}{\pi}&=r^{2}\\r&=\frac{10}{\sqrt{\pi}}.\end{aligned} \]
That is, the can’s radius is $ 10/ $ cm.
Finally, the can’s height is 4 times its radius, so its height is $ 40/ $ cm.
Another common strategy for solving story problems is to find two equations involving two unknowns. We can then find the unknowns’ values as follows. First, we isolate one variable in one equation. Second, we substitute the result into the other equation, which will then have only a single variable. From there, the rest is easy: Solve the resulting equation in one variable, use the result to find the other variable, and finally, interpret the result. Here are a couple of examples in this style.
Example 4. Suppose that a certain tiny rectangle has a perimeter of $ 35 , $ (micrometers) and an area of $ 49 , ^2 $ . What are the rectangle’s dimensions?
Solution. Let b and h be the base and height of the rectangle (in $ $ ).
Translating the conditions of the problem into equations, we have
\[ 2b+2h=35\quad\text{and}\quad b h=49. \]
The second equation is equivalent to h = 49/b. By substituting 49/b for h in the first equation, we transform it into $ 2b + 2(49/b) = 35 $ , which contains only one unknown. Solving this for b yields
\[ \begin{aligned}2b+2(49/b)&=35\\2b+(98/b)&=35\\2b^{2}+98&=35b\\2b^{2}-35b+98&=0\end{aligned} \]
(multiplying both sides by b to clear fractions)
\[ \begin{aligned}b&=\frac{35\pm\sqrt{441}}{4}\quad&(the quadratic formula)\\ &=\frac{35\pm21}{4}=14and7/2.\\ \end{aligned} \]
Thus b could be either 14 or 7/2. We can use the relationship h = 49/b that we established earlier to find the corresponding values of h. When we do so, we find that if b = 14, then h = 7/2, and vice-versa. Thus, we reach the same conclusion either way: The rectangle must be 14 by 7/2 $ $ m.
Example 5. Today, Rosencrantz is 1.2 times as old as Guildenstern.
Twenty years ago, he was twice as old as Guildenstern was. How old are they today?
Solution. Let R be Rosencrantz’s age today (in years).
Let G be Guildenstern’s age today (in years).
Translating the problem’s conditions into equations, we obtain
\[ R=1.2G\quad\text{and}\quad R-20=2(G-20). \]
Substituting the expression for R (from the first equation) into the second equation yields
\[ 1.2G-20=2(G-20). \]
Solving this linear equation for G, we find that
\[ \begin{aligned}1.2G-20&=2(G-20)\\1.2G-20&=2G-40\\20&=0.8G\\G&=20/0.8=25.\end{aligned} \]
Thus, Guildenstern is 25 years old today.
Finally, since \(R = 1.2G\) (as noted earlier), we have \(R = 1.2(25) = 30\).
That is, Rosencrantz is 30 years old today.
When you define a symbol, be precise. For example, in the solution above, I defined R as Rosencrantz’s age (in years) today. Don’t settle for “R = Rosencrantz”, which is both sloppy and counterproductive. Sloppy because variables in equations represent numbers, not people; counterproductive because vague
definitions make your job of interpretation harder. He who writes “R = Rosencrantz” may have trouble translating this problem’s second sentence into an equation. In contrast, he who precisely defines R as Rosencrantz’s age in years today recognizes that Rosencrantz’s age twenty years ago must be $ (R - 20) $ .
Finally, since many story problems involve formulas for geometric magnitudes, be sure that you have memorized some of the basic ones. Apart from the Pythagorean Theorem, you should obviously know how to find the areas and perimeters of rectangles, triangles, and circles. And in three dimensions, cylinders are particularly simple: The volume is just the base area times its height; the surface area can be found by the “surgery” method explained in example 3 above.
Exercises
The product of two consecutive whole numbers is 6592056. Find the numbers.
Find the area of an isosceles triangle whose equal sides are 9 inches, and whose base is 4 inches.
The area of an equilateral triangle is $ 1 , mile^{2} $ . Find its side length.
[Hint: Find a second expression for the triangle’s area – this one should be in terms of its side length.]
A rectangle has an area of 4 furlongs $ ^{2} $ and a perimeter of 40/3 furlongs. Find its dimensions.
Suppose that the height of a cylindrical can (closed at both ends) is half its base’s radius. Given that the can’s volume is $ 500 , cm^{3} $ , what is its surface area?
At noon, Vladimir and Estragon depart from the same point. Vladimir walks east at a constant rate of 3 miles per hour, while Estragon walks north at a constant rate of 2 miles per hour. At what time (to the nearest minute) will they be exactly 20 miles apart?
Consider a picture frame of uniform width. Its inner edges form a rectangular window through which the picture is seen. This window has an area of $ 100 , in^{2} $ . Its outer edges constitute a $ 15^{} $ by $ 18^{} $ rectangle. What is the frame’s width? (Give an exact expression and an approximation.)
Suppose that we form two circles as follows: we begin with a line segment whose length is 5 units; we then cut this segment into two unequal pieces and bend each piece into a circle. Is it possible to do this in such a way that the total area enclosed by the two circles will be $ 17/4$ units $ ^{2} $ ? If not, why not? If so, how should we cut the segment to accomplish this?
Mr. Punch was three times as old as Judy was 32 years ago. Three years after that, Mr. Punch was only twice as old as Judy. How old are they now?
Suppose we must construct a box with a square bottom (and no lid) as follows: we start with a square piece of cardboard, cut a 3-inch by 3-inch square from each of its corners, and then we fold up the resulting “flaps”. If we want the box to have a volume of $ 42 , in^{3} $ , how large should the original square piece of cardboard be?
Suppose a object is launched upwards from a 180’ platform at a speed of 60 ft/sec. The formula for the object’s height h (in feet) after t seconds is $ h = -16t^{2} + 60t + 180 $ . (You’ll learn why in physics.) How much time elapses before the object strikes the ground?
A mother is 21 years older than her son. In 6 years, she’ll be 5 times as old as he will be. Where is the father?
Solving Equations Containing Rational Expressions
A ratio of polynomials is called a rational expression. When rational expressions turn up in equations, we can always chase them away by “clearing the fractions.” $ ^{*} $ Here’s a handy trick for doing this:
Cross-Multiplication. We can rewrite $ = $ in the equivalent form ad = bc.
Proof. $ =bd()=bd()ad=bc, $ as claimed.
The following example shows the technique in action.
Example. Solve the equation $ = $ .
Solution. $ = $
\[ \begin{aligned}(3x+2)(-2x+1)&=(5x-7)5\\-6x^{2}-x+3&=25x-35\\-6x^{2}-26x+38&=0\\3x^{2}+13x-19&=0\\x&=\frac{-13\pm\sqrt{397}}{6}\end{aligned} \]
(clearing the fractions by cross multiplying)
(dividing both sides by $ -2^{} $ )
(Quadratic Formula)
Solving Equations Containing Radical Expressions
So-called radical expressions can assume a variety of forms, from the relative simplicity of $ + 6x + 1 $ to the nested nastiness of
\[ \sqrt[3]{2+\sqrt{1+2x+\sqrt[5]{x}}}. \]
You can solve many equations containing radicals just by following your nose. For example,
Example 1. Solve the equation $ 2-1=5 $ .
Solution. $ 2-1=5 $
\[ \begin{aligned}2\sqrt[3]{x+1}-1&=5\\2\sqrt[3]{x+1}&=6\\\sqrt[3]{x+1}&=3\\\left(\sqrt[3]{x+1}\right)^{3}&=3^{3}\\x+1&=27\\x&=26\end{aligned} \]
(cubing both sides to “undo” the cube root)
In principle, we can solve any equation whose unknown is wrapped up inside a radical by isolating the radical on one side of the equation, then eliminating it by raising both sides to the appropriate power. There is, however, a potential complication. I warned earlier that exceptions exist to the general rule that “doing the same thing to both sides of an equation produces an equivalent equation.” One big exception to this rule occurs when we raise both sides of an equation to an even power. $ ^{*} $ (There’s only one other exception you’ll need to worry about in this book; you’ll meet it shortly, in exercise 26. $ ^{} $ )
Raising both sides of an equation to an even power can produce spurious “solutions” that aren’t solutions of the original equation. (For example, if we square both sides of x = 3, we obtain $ x^{2} = 9 $ , whose negative solution does not satisfy the original equation, x = 3.) Consequently, if we ever raise both sides of an equation to an even power, we must remember to go back and check our alleged solutions by substituting them into the original equation. That way, we can catch (and throw away) any spurious ones that might have crept in.
Example 2. Solve the equation $ 3 + 5 = 2x $ .
Solution. $ 3 + 5 = 2x $
\[ 5\sqrt{x}=2x-3 \]
(isolating the radical)
\[ \left(5\sqrt{x}\right)^{2}=(2x-3)^{2} \]
(squaring both sides to eliminate the radical)
\[ 25x=4x^{2}+9-12x \]
\[ 0=4x^{2}-37x+9 \]
\[ x=\frac{37\pm\sqrt{1225}}{8} \]
(by the quadratic formula)
\[ x=9,x=\frac{1}{4} \]
\[ (since\sqrt{1225}=35) \]
Since we squared both sides of the equation, we must check our two prospective solutions to see if either is spurious. Substituting 9 into the original equation yields the true statement 18 = 18, so 9 is a genuine solution. But when we substitute 1/4 into the original equation, we obtain the false statement 11/2 = 1/2, so 1/4 is a spurious solution picked up as a byproduct of squaring. Thus, our original equation has just one solution, x = 9.
Exercises
- Solve each of the following equations for the unknown it contains:
$ = $
$ += $
$ +=3 $
$ =+t+3 $
$ ++=- $
$ = $
$ x^{2}=144 $
$ =3 $
$ (-)()= $
$ 5+3=-12 $
$ w + = 4w $
$ = x - 1 $
- We’ve seen that raising both sides of an equation to an even power can produce too many “solutions,” some of which do not actually satisfy the original equation. The opposite problem – too few solutions – can occur if we divide both sides by an expression that involves the equation’s variable. (For example, the equation $ x^{2} = x $ has two solutions, 0 and 1. Dividing both sides by x, however, yields x = 1, an equation with only one solution.)
It’s easy to see why this happens: To divide both sides by x, we must tacitly assume that $ x $ , because division by 0 is, of course, undefined. Thus, when we divide by x, the number 0 is effectively in our blind spot. Having 0 in our blind spot is a serious problem if 0 happens to be one of the solutions that we’re looking for. (Similarly, if we divide both sides of an equation by $ (x - 2) $ , we tacitly assume that $ x $ , which means that 2 is now in our algebraic blind spot.*)
Fortunately, this problem is easy to avoid. If you are ever tempted to divide both sides of an equation by an expression involving x, do this instead: Push everything to one side of the equation, then factor that expression out. You can then solve the equation with the Zero Product Theorem. (This doesn’t require division, so it keeps our blind spot empty.) For example, let’s apply this strategy to the equation we dealt with above:
\[ x^{2}=x\quad\Rightarrow\quad x^{2}-x=0\quad\Rightarrow\quad x(x-1)=0\quad\Rightarrow\quad x=0,1. \]
Your problem: Without using the quadratic formula, solve the following equations:
$ x^{2} = 5x $
$ 3x^{2}-4x=0 $
\[ -x^{2}=\frac{3}{2}x \]
- $ 9(x - 2) = 3x(x - 2) $ .
- To solve an equation containing two algebraic expressions under square roots, the best strategy is to isolate one of them, and square both sides. This will produce an equation with only one radical, which is much better, since you already know how to solve such equations.
Solve $ -=1 $ by adopting this strategy.
Now solve the same equation by ignoring this strategy and squaring both sides right away, without bothering to isolate one of the radicals first.
- 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.