First Course in the Theory of Equations
Constructions with Ruler and Compasses
Excerpts
Equations
Constructions with Ruler and Compasses
x^2 - ax + b = 0The quadratic whose real roots are constructed as the abscissas where a circle cuts the x-axis.
Constructions with Ruler and Compasses
\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{b+1}{2}\right)^2 = \frac{a^2 + (b-1)^2}{4}Equation of the circle having the segment BQ, with B=(0,1) and Q=(a,b), as a diameter.
Constructions with Ruler and Compasses
y = mx + bEquation of a straight line in coordinates, with slope m and intercept b.
Constructions with Ruler and Compasses
x = \frac{b' - b}{m - m'}Abscissa of the intersection of the lines y = mx + b and y = m'x + b'; a rational function of the coefficients.
Constructions with Ruler and Compasses
y = \frac{mb' - m'b}{m - m'}Ordinate of the intersection of the two lines y = mx + b and y = m'x + b'; a rational function of the coefficients.
Constructions with Ruler and Compasses
(x - c)^2 + (y - d)^2 = r^2Equation of a circle with centre (c, d) and radius r.
Constructions with Ruler and Compasses
\tfrac{1}{2}(a ± \sqrt{a^2 -4b)}Lengths of the segments constructed in the Remark after the criterion, the two roots of x^2 - ax + b = 0 halved-sum form. The printed brace is unbalanced in the source (sqrt{a^2 -4b) ), a typesetting point left as printed; the intended form is (a ± sqrt(a^2 - 4b))/2.
Constructions with Ruler and Compasses
s = \tfrac{1}{2}\sqrt{10 - 2\sqrt{5}}Side of a regular pentagon inscribed in a circle of unit radius, a number constructible by ruler and compasses.
Constructions with Ruler and Compasses
x^3 + \alpha x^2 + \beta x + \gamma = 0General cubic equation with rational coefficients, whose roots are examined for constructibility.
Constructions with Ruler and Compasses
x_1 = \frac{a + b \sqrt{k}}{c + d \sqrt{k}}A root of a superimposed-radical expression written as a fraction with the highest-order radical sqrt(k) in numerator and denominator.
Constructions with Ruler and Compasses
x_1 = e + f\sqrt{k}The root x_1 written as e + f·sqrt(k), with e and f free of sqrt(k).
Constructions with Ruler and Compasses
(e + f \sqrt{k})^3 + \alpha(e + f \sqrt{k})^2 + \beta(e + f \sqrt{k}) + \gamma = A + B\sqrt{k}Substituting x = e + f·sqrt(k) into the cubic and reducing gives A + B·sqrt(k), with A and B polynomials not involving sqrt(k).
Constructions with Ruler and Compasses
x_2 = e - f \sqrt{k}The conjugate e - f·sqrt(k) is a second root of the cubic.
Constructions with Ruler and Compasses
x_3 = -\alpha - x_1 - x_2 = -\alpha - 2eThe third root equals minus alpha minus the sum of the other two roots, which is -alpha - 2e (sum of roots equals -alpha).
Constructions with Ruler and Compasses
x_3 = g + h \sqrt{s}If e is irrational, the third root x_3 takes the form g + h·sqrt(s) with h not zero.
Constructions with Ruler and Compasses
g - h \sqrt{s} = e ± f \sqrt{k}The conjugate g - h·sqrt(s) of x_3 is one of the other two roots, so it equals e ± f·sqrt(k); the sign ± is not resolved by the relation.
Constructions with Ruler and Compasses
\cos A = 4 \cos^3 \frac{A}{3} - 3 \cos \frac{A}{3}The cosine of an angle expressed through the cosine of one third of it.
Constructions with Ruler and Compasses
x^3 - 3x = 2\cos AWith x = 2 cos(A/3), the angle A gives the cubic x^3 - 3x = 2 cos A.
Constructions with Ruler and Compasses
\cos A = -\frac{1}{2}For the angle A = 120 degrees, the cosine of A is -1/2.
Constructions with Ruler and Compasses
x^3 - 3x + 1 = 0The cubic whose root 2 cos(A/3) gives the trisection of A = 120 degrees; it has no rational root.
Constructions with Ruler and Compasses
x = 2 \cos BDefines x as twice the cosine of an angle B that would be constructible by ruler and compasses.
Constructions with Ruler and Compasses
\cos 3B = \cos 4BSince 7B = 360 degrees, the angles 3B and 4B have equal cosines.
Constructions with Ruler and Compasses
2(4 \cos^3 B - 3 \cos B) = x^3 - 3xTwice the cosine of 3B, expressed in x = 2 cos B, equals x^3 - 3x.
Constructions with Ruler and Compasses
4(2 \cos^2 B - 1)^2 - 2 = (x^2 - 2)^2 - 2Twice the cosine of 4B, expressed in x = 2 cos B, equals (x^2 - 2)^2 - 2.
Constructions with Ruler and Compasses
0 = x^4 - 4x^2 + 2 - (x^3 - 3x) = (x - 2)(x^3 + x^2 - 2x - 1)Equating the two expressions for 2 cos 3B and 2 cos 4B gives a quartic which factors as (x - 2)(x^3 + x^2 - 2x - 1).
Constructions with Ruler and Compasses
x^3 + x^2 - 2x - 1 = 0The cubic satisfied by x = 2 cos B for the regular polygon of 7 sides; it has no rational root, so the heptagon cannot be constructed.
Constructions with Ruler and Compasses
R = \cos\frac{2\pi}{7} + i \sin\frac{2\pi}{7}R is the complex seventh root of unity with argument 2 pi / 7.
Constructions with Ruler and Compasses
\frac{1}{R} = \cos\frac{2\pi}{7} -i \sin\frac{2\pi}{7}The reciprocal of R is its complex conjugate.
Constructions with Ruler and Compasses
R + \frac{1}{R} = 2\cos\frac{2\pi}{7}The sum of R and its reciprocal is twice the cosine of 2 pi / 7.
Constructions with Ruler and Compasses
y^6 + y^5 + y^4 + y^3 + y^2 + y + 1 = 0The equation with roots R, R^2, ..., R^6, obtained by removing the factor y - 1 from y^7 - 1.
Constructions with Ruler and Compasses
y+ \frac{1}{y} = xThe substitution that converts the reciprocal equation in y into an equation in x.
Constructions with Ruler and Compasses
\left(y^3 + \frac{1}{y^3}\right) + \left(y^2 + \frac{1}{y^2}\right) + \left(y + \frac{1}{y }\right) + 1=0Dividing the equation in y by y^3 gives this form, whose brackets are the powers y^k + 1/y^k.
Constructions with Ruler and Compasses
y^2 + \frac{1}{y^2} = x^2 - 2Squaring y + 1/y = x gives y^2 + 1/y^2 in terms of x.
Constructions with Ruler and Compasses
y^3 + \frac{1}{y^3} = x^3 - 3xCubing y + 1/y = x gives y^3 + 1/y^3 in terms of x.
Constructions with Ruler and Compasses
x_1 = R + \frac{1}{R} = R + R^6The first of the three sums of pairs of seventh roots of unity equals 2 cos(2 pi / 7).
Constructions with Ruler and Compasses
x_2 = R^2 + \frac{1}{R^2} = R^2 + R^5The second of the three sums of pairs of seventh roots of unity.
Constructions with Ruler and Compasses
x_3 = R^3 + \frac{1}{R^3} = R^3 + R^4The third of the three sums of pairs of seventh roots of unity.
Constructions with Ruler and Compasses
x_1 + x_2 + x_3 = R + R^2 + \dotsb + R^6 = -1The sum of the three values x_1, x_2, x_3 equals -1, since R, ..., R^6 are the roots of the equation (13).
Constructions with Ruler and Compasses
x_1 x_2 + x_1 x_3 + x_2 x_3 = 2(R + R^2 + \dotsb +R^6) = -2The sum of pairwise products of x_1, x_2, x_3 equals -2.
Constructions with Ruler and Compasses
x_1 x_2 x_3 = 2 + R + R^2 + \dotsb + R^6 = 1The product x_1 x_2 x_3 equals 1.
Constructions with Ruler and Compasses
y^n f\left(\frac{1}{y}\right) \equiv ±f(y)For a reciprocal equation f(y)=0 of degree n with constant term c, the reversed polynomial equals plus or minus f(y).
Constructions with Ruler and Compasses
c^2 = 1Equating constant terms in the reversed polynomial gives c^2 = 1, so c = ±1.
Constructions with Ruler and Compasses
f(y) \equiv y^n ± 1 + p_1(y^{n-1} ± y) + p_2 (y^{n-2} ± y^2) + \dotsbForm of a reciprocal polynomial: the coefficients of terms equidistant from the ends are equal up to sign.
Constructions with Ruler and Compasses
y^{n-1} Q \left(\frac{1}{y}\right) \equiv Q(y)The quotient Q(y) = f(y)/(y ± 1) is again reciprocal, with reversal degree n - 1.
Constructions with Ruler and Compasses
y^{2t} + 1 + c_1 (y^{2t-1} + y) + c_2 (y^{2t-2} + y^2) + \dotsb + c_{t-1} (y^{t+1} + y^{t-1}) + c_t y^t = 0The standard form to which any reciprocal equation of even degree 2t can be reduced.
Constructions with Ruler and Compasses
n = 2t+1Writing an odd degree n as 2t + 1.
Constructions with Ruler and Compasses
y^k + \frac{1}{y^k} = x \left(y^{k-1} + \frac{1}{y^{k-1}}\right) - \left(y^{k-2} + \frac{1}{y^{k-2}}\right)Recurrence giving y^k + 1/y^k in terms of x = y + 1/y and the two previous powers.
Constructions with Ruler and Compasses
x (x^3-3x) - (x^2-2) = x^4 - 4x^2 + 2Applying the recurrence gives y^4 + 1/y^4 = x^4 - 4x^2 + 2.
Constructions with Ruler and Compasses
R = \cos\frac{ 2\pi}{9} + i \sin\frac{ 2\pi}{9}R is the complex ninth root of unity with argument 2 pi / 9.
Constructions with Ruler and Compasses
\frac{y^9 - 1}{y^3 - 1} = y^6 + y^3 + 1 = 0The primitive ninth roots of unity are roots of y^6 + y^3 + 1 = 0.
Constructions with Ruler and Compasses
R^8 = RSince R^7 = 1 for the seventh root of unity, the fourth power of the ordering g = 2 returns to R, so g = 2 is rejected.
Constructions with Ruler and Compasses
z_1 = R + R^2 + R^4Period of three terms formed from alternate terms of the ordering R, R^3, R^2, R^6, R^4, R^5.
Constructions with Ruler and Compasses
z_2 = R^3 + R^6 + R^5The second period of three terms of the seventh roots of unity.
Constructions with Ruler and Compasses
z_1 z_2 = 3 + R + \dotsb + R^6 = 2The product of the two periods of three terms equals 2.
Constructions with Ruler and Compasses
z^2 + z + 2 = 0The periods z_1 and z_2 are the roots of this quadratic (their sum is -1 and product 2).
Constructions with Ruler and Compasses
w^3 - z_1w^2 + z_2w - 1 = 0The numbers R, R^2, R^4 are the roots of this cubic whose coefficients are the periods z_1 and z_2.
Constructions with Ruler and Compasses
\frac{R^{17} - 1}{R - 1} = R^{16} + R^{15} + \dotsb + R + 1 = 0The seventeenth roots of unity other than 1 are roots of this reciprocal polynomial.
Constructions with Ruler and Compasses
y_1 + y_2 = -1The two periods of eight terms of the 17th roots of unity sum to -1.
Constructions with Ruler and Compasses
y_1 y_2 = 4(R + \dotsb + R^{16}) = -4The product of the two eight-term periods equals -4.
Constructions with Ruler and Compasses
y^2 + y - 4 = 0The periods y_1 and y_2 of the seventeenth roots of unity are roots of this quadratic.
Constructions with Ruler and Compasses
R + R^9 + R^{13} + R^{15} + R^{16} + R^8 + R^4 + R^2The period y_1 of eight seventeenth roots of unity (the sum R + ... + R^2 taken over the even-position terms of the cube-ordering).
Constructions with Ruler and Compasses
R^3 + R^{10} + R^5 + R^{11} + R^{14} + R^7 + R^{12} + R^6The period y_2 of eight seventeenth roots of unity.
Constructions with Ruler and Compasses
z_1 = R + R^{13} + R^{16} + R^4Period of four seventeenth roots of unity, obtained from alternate terms of y_1.
Constructions with Ruler and Compasses
z_2 = R^9 + R^{15} + R^8 + R^2Period of four seventeenth roots of unity, the other half of y_1.
Constructions with Ruler and Compasses
w_1 = R^3 + R^5 + R^{14} + R^{12}Period of four seventeenth roots of unity, obtained from alternate terms of y_2.
Constructions with Ruler and Compasses
w_2 = R^{10} + R^{11} + R^7 + R^6Period of four seventeenth roots of unity, the other half of y_2.
Constructions with Ruler and Compasses
z_1 + z_2 = y_1The two four-term periods z_1, z_2 add to the eight-term period y_1.
Constructions with Ruler and Compasses
z_1 z_2 = w_1 w_2 = -1The products of the paired four-term periods both equal -1.
Constructions with Ruler and Compasses
z^2 - y_1 z - 1 = 0The periods z_1 and z_2 are the roots of this quadratic.
Constructions with Ruler and Compasses
w^2 - y_2 w - 1 = 0The periods w_1 and w_2 are the roots of this quadratic.
Constructions with Ruler and Compasses
v_1 = R + R^{16}Period of two seventeenth roots of unity, from alternate terms of z_1.
Constructions with Ruler and Compasses
v_2 = R^{13} + R^4Second period of two seventeenth roots of unity, from alternate terms of z_1.
Constructions with Ruler and Compasses
v_1 + v_2 = z_1The two periods v_1, v_2 add to z_1.
Constructions with Ruler and Compasses
v_1v_2 = w_1The product of v_1 and v_2 equals the four-term period w_1.
Constructions with Ruler and Compasses
v^2 - z_1v + w_1 = 0The periods v_1 and v_2 are the roots of this quadratic.
Constructions with Ruler and Compasses
\rho^2 - v_1\rho + 1 = 0R and R^16 are the roots of this quadratic, so R can be found by solving quadratics.
Constructions with Ruler and Compasses
v_1 = 2 \cos\frac{2\pi}{17}The larger period v_1 equals twice the cosine of 2 pi / 17, so the angle 2 pi / 17 is constructible.
Constructions with Ruler and Compasses
2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} = 2 \cos \frac{6\pi}{17} - 2 \cos \frac{7\pi}{17}Since cos(10 pi/17) = -cos(7 pi/17), the sum of the two cosines equals the expression with -2 cos(7 pi/17), which shows w_1 > 0.
Constructions with Ruler and Compasses
2 \cos \frac{6\pi}{17} + 2 \cos \frac{10\pi}{17} + 2 \cos \frac{12\pi}{17} + 2 \cos \frac{14\pi}{17} < 0The period y_2 is negative, since only the first cosine is positive and it is numerically less than the third.
Constructions with Ruler and Compasses
OE = \tfrac{1}{4} \sqrt{17}Length OE, the radius of the circle through which the point of AS is located for the 17-gon construction; equals sqrt(17)/4 for unit radius.
Constructions with Ruler and Compasses
\cos LOP = OL = \cos\frac{2\pi}{17}The perpendicular bisector of OM meets the unit circle at P so that angle LOP equals 2 pi / 17, which gives the side of the 17-gon.
Constructions with Ruler and Compasses
\tfrac{1}{2}\sqrt{10 - 2\sqrt{5}}placeholder not used
Constructions with Ruler and Compasses
st = \sqrt{5}The product of s (side of the pentagon) and t equals sqrt(5).
Constructions with Ruler and Compasses
t = \tfrac{1}{2} \sqrt{10 + 2\sqrt{5}}Definition of t as half the square root of 10 + 2 sqrt(5), a number of order 2.
Problems
Exercise Page30
Exercise Page30, problem 1, p. 30
$x^2 - 5x + 4 = 0$.
Printed answer:- $1$, $4$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[1, 4]
Exercise Page30, problem 2, p. 30
$x^2 + 5x + 4 = 0$.
Printed answer:- $-1$, $-4$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[-1, -4]
Exercise Page30, problem 3, p. 30
$x^2 + 5x - 4 = 0$.
Printed answer:- $0.7$, $-5.7$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[0.7, -5.7]
Exercise Page30, problem 4, p. 30
$x^2 - 5x - 4 = 0$.
Printed answer:- $-0.7$, $5.7$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[-0.7, 5.7]
Exercise Page30, problem 5, p. 30
$x^2 - 4x + 4 = 0$.
Printed answer:- $2$, $2$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[2, 2]
Exercise Page30, problem 6, p. 30
$x^2 - 3x + 4 = 0$.
Printed answer:- Imaginary.
verified: the printed answer passed a computed check
How it was checked
solve: passes[]
Exercise Page40
Exercise Page40, problem 1, p. 39
Show by $(16)$ that the roots of $(12)$ are $2\cos 2\pi/7$, $2\cos 4\pi/7$, $2\cos 6\pi/7$.
Printed answer:- (none printed)
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Exercise Page40, problem 10, p. 39
To construct a straight line representing the distance from the circular base of a hemisphere to the parallel plane which bisects the hemisphere.
Printed answer:- See $(11)$, §32.
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Exercise Page40, problem 11, p. 39
To construct lines representing the lengths of the edges of an existing rectangular parallelopiped having a diagonal of length $5$, surface area $24$, and volume $1$, $2$, $3$, or $5$.
Printed answer:- Edges roots of $x^3 - 7x^2 + 12x - v = 0$, all real (§45) and irrational.
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Exercise Page40, problem 12, p. 39
To trisect an angle whose cosine is $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, $\frac{1}{8}$ or $p/q$, where $p$ and $q$ ($q>1$) are integers without a common factor, and $q$ is not divisible by a cube.
Printed answer:- (none printed)
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Exercise Page40, problem 13, p. 39
To trisect an angle whose cosine is $(4a^3 - 3ab^2)/b^3$, where the integer $a$ is numerically less than the integer $b$; for example, $\cos^{-1} 11/16$ if $a = -1$, $b = 4$.
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Exercise Page40, problem 14, p. 39
To construct the legs of a right triangle, given its area and hypotenuse.
Printed answer:- $\Delta = \text{area}$, $c = \text{hypotenuse}$, squares of legs $\tfrac{1}{2}(c^2 ± \sqrt{c^4 - 16\Delta^2})$.
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Exercise Page40, problem 15, p. 39
To construct the third side of a triangle, given two sides and its area.
Printed answer:- $\Delta$ area, $a$, $b$ given sides, square third side is $a^2 + b^2 ± 2\sqrt{a^2b^2 - 4\Delta^2}$.
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Exercise Page40, problem 16, p. 39
To locate the point $P$ on the side $BC=1$ of a given square $ABCD$ such that the straight line $AP$ cuts $DC$ produced at a point $Q$ for which the length of $PQ$ is a given number $g$. Show that $y=BP$ is a root of a reciprocal quartic equation, and solve it when $g = 10$.
Printed answer:- $y^4 - 2y^3 + (2 - g^2)y^2 - 2y + 1 = 0$, pos. roots $0.09125$, $10.95862$.
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Exercise Page40, problem 2, p. 39
The imaginary fifth roots of unity satisfy $y^4 + y^3 + y^2 + y + 1 = 0$, which by the substitution $(14)$ becomes $x^2 + x - 1 = 0$. It has the root R + 1R = 2 25 = 12(5-1). In a circle of radius unity and center $O$ draw two perpendicular diameters $AOA'$, $BOB'$. With the middle point $M$ of $OA'$ as center and radius $MB$ draw a circle cutting $OA$ at $C$ (Fig. 10). Show that $OC$ and $BC$ are the sides $s_{10}$ and $s_5$ of the inscribed regular decagon and pentagon respectively. Hints:
Printed answer:- (none printed)
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Exercise Page40, problem 3, p. 39
If $R$ is a root of $(19)$ verify as at the end of §35 that $R+R^8$, $R^2+R^7$, and $R^4+R^5$ are the roots of $(11)$.
Printed answer:- (none printed)
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Exercise Page40, problem 4, p. 39
Hence show that the roots of $(11)$ are $2\cos 2\pi/9$, $2\cos 4\pi/9$, $2\cos 8\pi/9$.
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Exercise Page40, problem 5, p. 39
Reduce $y^{11} = 1$ to an equation of degree $5$ in $x$.
Printed answer:- $x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$.
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Exercise Page40, problem 6, p. 39
Solve $y^5 - 7y^4 + y^3 - y^2 + 7y - 1 = 0$ by radicals. [One root is $1$.]
Printed answer:- $-\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(7±\sqrt{45})$.
verified: the printed answer passed a computed check
How it was checked
solve: passes["-1/2 + sqrt(-3)/2", "-1/2 - sqrt(-3)/2", "(7 + sqrt(45))/2", "(7 - sqrt(45))/2"]
Exercise Page40, problem 7, p. 39
After finding so easily in [chap:I]Chapter I the trigonometric forms of the complex roots of unity, why do we now go to so much additional trouble to find them algebraically?
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Exercise Page40, problem 8, p. 39
Prove that every real root of $x^4 + ax^2 + b = 0$ can be constructed with ruler and compasses, given lines of lengths $a$ and $b$.
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Exercise Page40, problem 9, p. 39
Show that the real roots of $x^3 - px - q = 0$ are the abscissas of the intersections of the parabola $y = x^2$ and the circle through the origin with the center $(\frac{1}{2}q, \frac{1}{2} + \frac{1}{2}p)$.
Printed answer:- (none printed)
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other: not a kind the checker handles
Exercise Page44
Exercise Page44, problem 1, p. 44
If $a$ and $b$ are relatively prime numbers, so that their greatest common divisor is unity, we can find integers $c$ and $d$ such that $ac + bd = 1$. Show that, if regular polygons of $a$ and $b$ sides can be constructed and hence angles $2\pi/a$ and $2\pi/b$, a regular polygon of $a·b$ sides can be derived.
Printed answer:- (none printed)
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Exercise Page44, problem 2, p. 44
If $p = 2^h + 1$ is a prime, $h$ is a power of $2$. For $h = 2^0$, $2^1$, $2^2$, $2^3$, the values of $p$ are $3$, $5$, $17$, $257$ and are primes. [Show that $h$ cannot have an odd factor other than unity.]
Printed answer:- (none printed)
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Exercise Page44, problem 3, p. 44
For $13$th roots of unity find the least $g$ (§38), write out the three periods each of four terms, and find the cubic equation having them as roots.
Printed answer:- $g=2$, $R + R^8 + R^{12} + R^5$, etc., $z^3 + z^2 - 4z + 1 = 0$.
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Exercise Page44, problem 4, p. 44
For the primitive ninth roots of unity find the least $g$ and write out the three periods each of two terms.
Printed answer:- $g=2$, $R+R^8$, $R^2+R^7$, $R^4+R^5$.
unverified: no computed check settled this one (yet)
How it was checked
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Exercise Page44, problem 5, p. 44
$y^4 + 4y^3 - 3y^2 + 4y + 1 = 0$.
Printed answer:- $\tfrac{1}{2}(1±\sqrt{-3})$, $\tfrac{1}{2}(-5±\sqrt{21})$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[(1+sqrt(-3))/2, (1-sqrt(-3))/2, (-5+sqrt(21))/2, (-5-sqrt(21))/2]
Exercise Page44, problem 6, p. 44
$y^5 - 4y^4 + y^3 + y^2 - 4y + 1 = 0$.
Printed answer:- $-1$, $2±\sqrt{3}$, $\tfrac{1}{2} ± \tfrac{1}{2}\sqrt{-3}$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[-1, 2+sqrt(3), 2-sqrt(3), (1+sqrt(-3))/2, (1-sqrt(-3))/2]
Exercise Page44, problem 7, p. 44
$2y^6 - 5y^5 + 4y^4 - 4y^2 + 5y - 2 = 0$.
Printed answer:- $1$, $1$, $1$, $-1$, $\tfrac{1}{4}(1±\sqrt{-15})$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[1, 1, 1, -1, (1+sqrt(-15))/4, (1-sqrt(-15))/4]
Exercise Page44, problem 8, p. 44
$y^5 + 1 = 31(y + 1)^5$.
Printed answer:- $-1$, $-2$, $-\tfrac{1}{2}$, $\tfrac{1}{6}(-5±\sqrt{-11})$.
verified: the printed answer passed a computed check
How it was checked
solve: passes[-1, -2, -1/2, (-5+sqrt(-11))/6, (-5-sqrt(-11))/6]