Math 101 A Bit of Algebra
This is a quick review of some of the ideas we use when solving equations in Calculus typically
when nding zeros of the derivatives of a function in optimization and graphing problems, but
sometimes also when dealing with regions in integration problems.
If
AB = 0
where
A
and
B
are real numbers, then
A=0
or
B=0 :
A very popular mistake is to cancel expressions involving unknowns multiplicatively from both
sides of an equation. Consider this:
Both steps in the following are incorrect! Do not imitate them!
x3 = x Ô⇒ x2 = 1 Ô⇒ x = 1
Both steps in the preceding are incorrect! Do not imitate them!
In the rst step above the factor x was cancelled from both sides ignoring the possibility that
it can be 0 and in the second step the negative square root of 1 was completely ignored.
The correct reasoning should go as follows:
x3 = x Ô⇒ x = 0 or x2 = 1 Ô⇒ x = 0 or x = 1 or x = −1
The equation x3 = x has three real solutions, 1, −1 and 0; not just one. An even cleaner way
of doing this is as follows:
x3 = x Ô⇒ x3 − x = 0 Ô⇒ x(x − 1)(x + 1) = 0
Ô⇒ x = 0 or x − 1 = 0 or x + 1 = 0 Ô⇒ x = 0 or x = 1 or x = −1
Second Degree Equations
:
We can solve quadratic equations by factorization, like:
3x2 + 14x − 5 = 0 Ô⇒ (3x − 1)(x + 5) = 0 Ô⇒ x = 1/3 or x = −5
Or we can use
The Quadratic
Formula: The roots of the polynomial ax2 + bx + c with a =/ 0
√
are (−b ±
b2 − 4ac)/(2a).
like:
3x2 + 14x − 5 = 0 Ô⇒ x =
−14 ±
√
142 − 4 ⋅ 3 ⋅ (−5)
Ô⇒ x = 1/3 or x = −5
2⋅3
Sometimes we have to deal with situations where there are two unknowns, and we are trying
to solve one in terms of the other. In these cases factorization works rarely and we must rely
on the quadratic formula. For instance,
√
√
−1 ± 12 − 4 ⋅ y ⋅ (−y 2 )
−1 ± 1 + 4y 3
2
2
Ô⇒ x =
yx + x − y = 0 Ô⇒ x =
2⋅y
2y
if y =/ 0. On the other hand, if y = 0, then from yx2 + x − y 2 = 0 one obtains x = 0.
Some equations become quadratic after a change of variable. For instance, consider the fourth
2
2
degree equation x4 − 10x2 + 8 = 0. Letting
√ z = x this can be written as z √− 10z + 8√= 0 for which
x =√ z or − z√
, we obtain
the quadratic formula gives z = 5 ± 17. Now√using the fact
√ that
√
√
√
√
the four roots of the original equation as x = 5 + 17, − 5 + 17, 5 − 17, − 5 − 17.
As another application of the quadratic formula let us nd the inverse of the hyperbolic sine
function sinh x = (ex − e−x )/2. If x = sinh y , then x = (ey − e−y )/2. We can rewrite this in the
√
form√(ey )2 − 2xey − 1 = 0. This is quadratic in ey and we have ey = x ± x2 + 1. Now note that
x − x2 √
+ 1 is negative for all values√of x whereas ey is always positive. √Hence we must have
y
e = x + x2 + 1 giving us y = ln(x + x2 + 1). That is, sinh−1 x = ln(x + x2 + 1).
Higher Degree Equations
:
There are formulas for solving the third and fourth degree polynomial equations, but we will
not need them. There are no general formulas for polynomial equations of degree higher than
four, and in fact, it is proven that no such formulas can exist. If any such equation appears in
a Calculus problem, you may expect that it has a root which can be spotted easily, perhaps
using
The Rational Root Theorem: If a is a rational root of a polynomial f (x) =
cn xn + cn−1 xn−1 + ⋯ + c1 x + c0 with integer coecients where cn =/ 0, then a has the
form a = p/q where p is an integer dividing c0 and q is an integer dividing cn .
and then we can use
The Factor Theorem: If a is a root of a polynomial f (x) = cn xn + cn−1 xn−1 + ⋯ +
c1 x + c0 , then x − a is a factor of f (x).
to obtain a lower degree equation.
Consider x3 − 15x + 4 = 0. If this equation has a rational root, it must be among ±1, ±2, ±4.
By inspection we see that x = −4 is a root, and hence x + 4 is a factor. Polynomial division
3
2
gives
√ x − 15x + 4 = (x + 4)(x − 4x + 1). Using the quadratic formula for the second factor gives
2 ± 3 as the other two roots.
Consider the equation 2x1/3 − 3 + x−2/3 = 0. We rst use the substitution z = x1/3 to convert
this to the polynomial equation 2z 3 − 3z 2 + 1 = 0. Next we observe that z = 1 is a root of this
equation and 2z 3 − 3z 2 + 1 = (z − 1)(2z 2 − z − 1). The roots of the quadratic factor are z = 1
and z = −1/2. (So z = 1 is a double root.) Then x = z 3 gives x = 1 and x = −1/8 as the roots of
the original equation.
Finally, for the sake of completeness, we state
The Fundamental Theorem of Algebra: A polynomial f (x) = cn xn +cn−1 xn−1 +
⋯ + c1 x + c0 with real coecients can be written in a unique way as a product of
linear factors x−r where r is a real number, irreducible quadratic factors x2 +px+q
where p and q are real numbers such that p2 − 4q < 0 and a real constant.
whose proof, unlike those of the above which are easy exercises, is quite dicult.
Two Calculus Examples
:
Now we look at two Calculus examples involving solving equations or systems of equations.
The rst one is about nding horizontal tangents to a curve. The second one is about
expressing the volume of a solid of revolution as an integral. In each solution the parts
where the algebra occurs are enclosed between ⌜ and ⌟.
Example: Find all points (x, y) =/ (0, 0) on the cardioid (x2 + y 2 − x)2 = x2 + y 2 where the
tangent line is horizontal.
Implicit dierentiation gives:
2(x2 + y 2 − x)(2x + 2y
dy
dy
− 1) = 2x + 2y
dx
dx
(e)
At a point where the tangent is horizontal we have dy/dx = 0 and this becomes:
(x2 + y 2 − x)(2x − 1) = x
⌜So we are looking for the points (x, y) whose coordinates satisfy both
(x2 + y 2 − x)2 = x2 + y 2 and (x2 + y 2 − x)(2x − 1) = x
(ee)
simultaneously. Note that x cannot be 1/2 as substituting x = 1/2 in the second equation gives
0 = 1/2, a contradiction. Therefore from the second equation we obtain x2 + y 2 − x = x/(2x − 1)
and x2 + y 2 = x/(2x − 1) + x = 2x2 /(2x − 1). Substituting these in the rst equation we obtain
(x/(2x − 1))2 = 2x2 /(2x − 1). This can be rearranged as x2 (4x − 3) = 0. We deduce that x = 0
or x = 3/4. √
Substituting x = 0 in the second equation gives y = 0, √
and substituting x√= 3/4
gives y = ±3 3/4. Hence we obtain the points (x, y) = (0, 0), (3/4, 3 3/4) and (3/4, −3 3/4)
as the set of solutions of the system of equations (ee). (Always verify!)
⌟
√
√
We are not quite done. Although the points (3/4, 3 3/4) and (3/4, −3 3/4) do satisfy (ee),
we still have to verify that dy/dx is indeed zero at these points. Note that (e) can be written
in the form:
dy
y (2(x2 + y 2 − x) − 1)
= x − (x2 + y 2 − x)(2x − 1)
(eee)
dx
For the right hand side of this equality the second equation
of the system (ee) gives x −
√
(x2 + y 2 − x)(2x − 1) = 0 for both of the points (3/4, ±3 3/4). On the other hand,
√ for the
2
2
factor multiplying dy/dx on the left hand side we have y (2(x + y − x) − 1) = ±3 3/2 =/ 0 at
√
(3/4, ±3 3/4), respectively. So in both cases we can conclude that dy/dx = 0.
√
Therefore
the
cardioid
has
horizontal
tangent
lines
at
the
points
(x,
y)
=
(3/4,
3
3/4) and
√
(3/4, −3 3/4) dierent from the origin.
Remark: The cardioid has a cusp at the origin. Note that (eee) gives 0 = 0 at (0,0) and
we cannot conclude that dy/dx = 0 as we did above for the other two points. However, the
negative x-axis is still a left-tangent to both the upper and the lower halves of the cardioid at
(0, 0) as can be shown using other means.
Example: The region bounded by the curve y 2 = x2 − x4 is revolved about the x-axis to
generate a solid. Express the volume V of the solid as an integral using the cylindrical shells
method.
By symmetry, V is 2 times the volume obtained by revolving the portion of the region lying
in the rst quadrant. The green horizontal rectangles generate the cylindrical shells that are
used in the computation of V .
⌜To nd the heights of these shells we have to solve (x ) −x +y
2 2
solutions are:
x=
√
(1 +
√
1 − 4y 2 )/2
and x =
√
2
(1 −
2
√
= 0 for x. The two nonnegative
1 − 4y 2 )/2
⌟
⌜Finally we nd the largest value of y, which will be the upper limit of the integral. Implicit
The dierence of these gives the height of the shell, and y is the radius of the shell.
dierentiation gives y dy/dx = x−2x3 , and therefore at
√ the points
√ with the largest y -coordinates
on the curve we must have 0 = x − 2x3 = −2x(x − 1/ 2)(x√+ 1/ 2). We conclude that the one
in the rst quadrant has y = 1/2, corresponding to x = 1/ 2.
⌟
So we have:
V = 2 ⋅ 2π ∫
0
1/2
(radius of shell)(height of shell) dy
¿
¿
Á 1 + √1 − 4y 2 Á 1 − √1 − 4y 2
1/2
Á
Á
À
À
= 2 ⋅ 2π ∫
y(
−
) dy
2
2
0
Remark: This
√ integral can be computed. For instance, the substitution cos 2θ =
√
1 − 4y 2 ,
sin 2θ dθ = 2y/ 1 − 4y 2 dy gives
V = 2π ∫
π/4
0
(cos θ − sin θ) sin 2θ cos 2θ dθ
after using the half-angle formulas. With a little more trigonometric manipulation this turns
into
V = 8π ∫
π/4
0
cos4 θ sin θ dθ − 4π ∫
π/4
cos2 θ sin θ dθ
0
− 4π ∫
π/4
0
sin2 θ cos θ dθ + 8π ∫
π/4
sin4 θ cos θ dθ
0
which eventually leads to V = 4π/15. Of course, this volume can be computed much more
easily using the disk method.
Last revision: September 18, 2015