Solution of Cubic and Quartic Equations; Their Discriminants

Excerpts

Solution of Cubic and Quartic Equations; Their Discriminants
The product of the squares of the differences of the roots of any equation in which the coefficient of the highest power of the unknown is unity shall be called the *discriminant* of the equation.
Solution of Cubic and Quartic Equations; Their Discriminants
A cubic equation with real coefficients has three distinct real roots if its discriminant $\Delta$ is positive, a single real root and two conjugate imaginary roots if $\Delta$ is negative, and at least two equal real roots if $\Delta$ is zero.
Solution of Cubic and Quartic Equations; Their Discriminants
This is called the irreducible case since it may be shown that a cube root of a general complex number cannot be expressed in the form $a + bi$, where $a$ and $b$ involve only real radicals.
Solution of Cubic and Quartic Equations; Their Discriminants
The expression $A + B$ for a root was first published by Cardan in his *Ars Magna* of 1545, although he had obtained it from Tartaglia under promise of secrecy.
Solution of Cubic and Quartic Equations; Their Discriminants
The pairs of values of $z$ whose product is $5$ are $1$ and $5$, $\omega$ and $5\omega^2$, $\omega^2$ and $5\omega$.
Solution of Cubic and Quartic Equations; Their Discriminants
This expression (and not $P$ itself) is called the discriminant of $(13)$.
Solution of Cubic and Quartic Equations; Their Discriminants
Choose any root $y$ of this *resolvent cubic equation $(17)$*. Then the right member of $(16)$ is the square of a linear function, say $mx+n$.
Solution of Cubic and Quartic Equations; Their Discriminants
The value $k^2 = 1$ gives the factors $z^2 + 2z - 1$, $z^2 - 2z + 2$.

Equations

Solution of Cubic and Quartic Equations; Their Discriminants
x^3 + bx^2 + cx + d = 0
The general cubic equation, with the coefficient of x^3 equal to one.
Solution of Cubic and Quartic Equations; Their Discriminants
y^3 + py + q = 0
The reduced cubic equation, which lacks the square of the unknown.
Solution of Cubic and Quartic Equations; Their Discriminants
p = c - \frac{b^2}{3}
The coefficient p of the reduced cubic is c minus one third of b squared.
Solution of Cubic and Quartic Equations; Their Discriminants
q = d - \frac{bc}{3} + \frac{2b^3}{27}
The constant q of the reduced cubic is expressed through the coefficients of the general cubic.
Solution of Cubic and Quartic Equations; Their Discriminants
x_1 = y_1 - \frac{b}{3}
Each root of the general cubic is the corresponding root of the reduced cubic shifted by minus b/3.
Solution of Cubic and Quartic Equations; Their Discriminants
y = z - \frac{p}{3z}
Vieta's substitution that turns the reduced cubic into an equation in z.
Solution of Cubic and Quartic Equations; Their Discriminants
z^6 + qz^3 - \frac{p^3}{27} = 0
Substituting the auxiliary form into the reduced cubic gives a sextic that is quadratic in z^3.
Solution of Cubic and Quartic Equations; Their Discriminants
R = \left(\frac{p}{3}\right)^3 + \left(\frac{q}{2}\right)^2
R is defined as the sum of the cube of p/3 and the square of q/2; its sign decides the number of real roots.
Solution of Cubic and Quartic Equations; Their Discriminants
\omega = -\tfrac{1}{2} + \tfrac{1}{2} \sqrt{3}i
omega is one of the imaginary cube roots of unity.
Solution of Cubic and Quartic Equations; Their Discriminants
\omega^2 = -\tfrac{1}{2} - \tfrac{1}{2} \sqrt{3}i
omega squared is the other imaginary cube root of unity, the conjugate of omega.
Solution of Cubic and Quartic Equations; Their Discriminants
A = \sqrt[3]{-\frac{q}{2} + \sqrt{R}}
A is a chosen cube root of minus q/2 plus the square root of R.
Solution of Cubic and Quartic Equations; Their Discriminants
B = \sqrt[3]{-\frac{q}{2} - \sqrt{R}}
B is a chosen cube root of minus q/2 minus the square root of R.
Solution of Cubic and Quartic Equations; Their Discriminants
AB = -\frac{p}{3}
The chosen cube roots A and B have product minus p/3.
Solution of Cubic and Quartic Equations; Their Discriminants
y_1 = A + B
The first root of the reduced cubic is the sum A + B.
Solution of Cubic and Quartic Equations; Their Discriminants
y_2 = \omega A + \omega^2 B
The second root of the reduced cubic is omega times A plus omega squared times B.
Solution of Cubic and Quartic Equations; Their Discriminants
y_3 = \omega^2 A + \omega B
The third root of the reduced cubic is omega squared times A plus omega times B.
Solution of Cubic and Quartic Equations; Their Discriminants
(y_1 - y_2)^2 (y_1 - y_3)^2 (y_2 - y_3)^2 = -4p^3 - 27q^2
The product of the squares of the differences of the roots of the reduced cubic equals minus four p cubed minus 27 q squared.
Solution of Cubic and Quartic Equations; Their Discriminants
(x-1)(x-\omega)(x-\omega^2) \equiv x^3 - 1
The cube roots of unity 1, omega, omega squared are the roots of x^3 - 1, identically in x.
Solution of Cubic and Quartic Equations; Their Discriminants
(A-B)(A-\omega B)(A-\omega^2 B) = A^3 - B^3 = 2 \sqrt{R}
The product (A-B)(A-omega B)(A-omega^2 B) equals A cubed minus B cubed, which equals 2 times the square root of R.
Solution of Cubic and Quartic Equations; Their Discriminants
(1-\omega)(1-\omega^2) = 3
The product of one minus omega and one minus omega squared equals 3.
Solution of Cubic and Quartic Equations; Their Discriminants
\omega - \omega^2 = \sqrt{3}i
The difference of the two imaginary cube roots of unity is root 3 times i.
Solution of Cubic and Quartic Equations; Their Discriminants
(y_1-y_2)(y_1-y_3)(y_2-y_3) = 6\sqrt{3}\sqrt{R}i
The product of the differences of the three roots of the reduced cubic equals 6 root 3 times root R times i.
Solution of Cubic and Quartic Equations; Their Discriminants
-108R = -4p^3 - 27q^2
The discriminant of the reduced cubic equals -108 R.
Solution of Cubic and Quartic Equations; Their Discriminants
\Delta = 18bcd - 4b^3 d + b^2 c^2 - 4c^3 - 27d^2
The discriminant of the general cubic is expressed directly in its coefficients b, c, d.
Solution of Cubic and Quartic Equations; Their Discriminants
ax^3 + bx^2 + cx +d = 0 \quad (a \neq 0)
A cubic equation whose leading coefficient a is not required to be one.
Solution of Cubic and Quartic Equations; Their Discriminants
a^4 P = 18 abcd - 4b^3 d + b^2 c^2 - 4ac^3 - 27a^2 d^2
For a cubic with leading coefficient a, a to the fourth times the product P of squared root differences equals this expression in a, b, c, d; this expression is the discriminant of that cubic.
Solution of Cubic and Quartic Equations; Their Discriminants
x^4 +bx^3 +cx^2 +dx+e=0
The general quartic equation, with leading coefficient one.
Solution of Cubic and Quartic Equations; Their Discriminants
(x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y)^2 = (\tfrac{1}{4}b^2 - c + y)x^2 + (\tfrac{1}{2}by - d)x + \tfrac{1}{4}y^2 - e
Adding the same terms to both sides of the quartic makes the left side a perfect square, leaving a quadratic in x on the right, for any y.
Solution of Cubic and Quartic Equations; Their Discriminants
(\tfrac{1}{2}by - d)^2 - 4(\tfrac{1}{4}b^2 - c + y)(\tfrac{1}{4}y^2 - e) = 0
The right side of the previous identity is a perfect square exactly when its discriminant is zero.
Solution of Cubic and Quartic Equations; Their Discriminants
y^3 - cy^2 + (bd - 4e)y - b^{2}e + 4ce - d^2 = 0
The resolvent cubic of the quartic, whose roots y allow the quartic to be factored into quadratics.
Solution of Cubic and Quartic Equations; Their Discriminants
x^2 + \tfrac{1}{2}bx + \tfrac{1}{2}y = mx+n
One of the two quadratic equations obtained by taking the square root of the perfect square, with m and n the coefficients of the linear function.
Solution of Cubic and Quartic Equations; Their Discriminants
y_1 = x_1 x_2 + x_3 x_4
The first root of the resolvent cubic is the sum of the products of the roots of the quartic paired as (x1,x2) and (x3,x4).
Solution of Cubic and Quartic Equations; Their Discriminants
y_2 = x_1 x_3 + x_2 x_4
The second root of the resolvent cubic pairs the roots as (x1,x3) and (x2,x4).
Solution of Cubic and Quartic Equations; Their Discriminants
y_3 = x_1 x_4 + x_2 x_3
The third root of the resolvent cubic pairs the roots as (x1,x4) and (x2,x3).
Solution of Cubic and Quartic Equations; Their Discriminants
x_1 x_2 = \tfrac{1}{2} y_1 - n
The product of the first pair of roots of the quartic equals half of y1 minus n.
Solution of Cubic and Quartic Equations; Their Discriminants
x_3 x_4 = \tfrac{1}{2} y_1 + n
The product of the second pair of roots of the quartic equals half of y1 plus n.
Solution of Cubic and Quartic Equations; Their Discriminants
x_1 x_2 + x_3 x_4 = y_1
The sum of the two pair-products of the quartic roots equals y1.
Solution of Cubic and Quartic Equations; Their Discriminants
x_1 + x_2 + x_3 + x_4 = -b
The sum of the four roots of the quartic equals minus the coefficient b.
Solution of Cubic and Quartic Equations; Their Discriminants
x_1 x_2 x_3 + x_1 x_2 x_4 + x_1 x_3 x_4 + x_2 x_3 x_4 = -d
The sum of the triple products of the quartic roots equals minus d.
Solution of Cubic and Quartic Equations; Their Discriminants
x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = c
The sum of the pairwise products of the quartic roots equals c.
Solution of Cubic and Quartic Equations; Their Discriminants
x_1 x_2 x_3 x_4 = e
The product of the four roots of the quartic equals e.
Solution of Cubic and Quartic Equations; Their Discriminants
y_1 + y_2 + y_3 = c
The sum of the roots of the resolvent cubic equals c.
Solution of Cubic and Quartic Equations; Their Discriminants
\Delta = ( x_1 - x_2 )^2 ( x_1 - x_3 )^2 ( x_1 - x_4 )^2 ( x_2 - x_3 )^2 ( x_2 - x_4 )^2 ( x_3 - x_4 )^2
The discriminant of the quartic is defined as the product of the squares of the differences of its four roots.
Solution of Cubic and Quartic Equations; Their Discriminants
y_1 - y_2 = (x_1-x_4)(x_2-x_3)
The difference of two resolvent cubic roots factors as a product of root differences of the quartic.
Solution of Cubic and Quartic Equations; Their Discriminants
(y_1-y_2)^2 (y_1-y_3)^2 (y_2- y_3)^2 = \Delta
The discriminant of the quartic equals the discriminant of its resolvent cubic.
Solution of Cubic and Quartic Equations; Their Discriminants
p = bd - 4e - \tfrac{1}{3} c^2
The coefficient p of the reduced cubic obtained from the resolvent cubic of the quartic.
Solution of Cubic and Quartic Equations; Their Discriminants
q = -b^2 e + \tfrac{1}{3} bcd + \tfrac{8}{3} ce - d^2 - \tfrac{2}{27} c^3
The constant q of the reduced cubic obtained from the resolvent cubic of the quartic.
Solution of Cubic and Quartic Equations; Their Discriminants
z^4 + qz^2 + rz + s = 0
The quartic with its z-cubed term removed, the starting point for Descartes' solution.
Solution of Cubic and Quartic Equations; Their Discriminants
(z^2 + 2kz + l)(z^2 - 2kz + m) = z^4 + (l + m - 4k^2)z^2 + 2k(m - l)z + lm
The product of two quadratic factors expands to a quartic with the stated coefficients.
Solution of Cubic and Quartic Equations; Their Discriminants
64k^6 + 32qk^4 + 4(q^2 - 4s)k^2 - r^2 = 0
The cubic in k squared that yields a factorization of the reduced quartic into quadratics (Descartes' resolvent).
Solution of Cubic and Quartic Equations; Their Discriminants
k_1^2 + k_2^2 + k_3^2 = -\tfrac{1}{2}q
The sum of the squares of the three roots of the Descartes cubic equals minus q/2.
Solution of Cubic and Quartic Equations; Their Discriminants
k_1^2 k_2^2 k_3^2 = \frac{r^2}{64}
The product of the squares of the three k roots equals r squared over 64.
Solution of Cubic and Quartic Equations; Their Discriminants
k_1 k_2 k_3 = -\frac{r}{8}
The signs of the square roots k1, k2, k3 must be chosen so that their product equals minus r/8.
Solution of Cubic and Quartic Equations; Their Discriminants
k_1 + k_2 + k_3
One of the four roots of the reduced quartic in the symmetric form of Descartes' solution, with signs chosen as in the relation for k1 k2 k3.
Solution of Cubic and Quartic Equations; Their Discriminants
64y^3 + 32qy^2 + 4(q^2 - 4s)y - r^2 = 0
The cubic whose roots are k1^2, k2^2, k3^2, used in the symmetric form of Descartes' solution.
Solution of Cubic and Quartic Equations; Their Discriminants
z = \sqrt{y_1} + \sqrt{y_2} + \sqrt{y_3}
A root of the reduced quartic is the sum of the square roots of the three roots of the cubic in y, with suitable signs.
Solution of Cubic and Quartic Equations; Their Discriminants
\sqrt{y_1}·\sqrt{y_2}·\sqrt{y_3} = -\frac{r}{8}
The square roots in the sum for z must be chosen so that their product equals minus r/8.
Solution of Cubic and Quartic Equations; Their Discriminants
z^3 - \tfrac{3}{4}z - \tfrac{1}{4}\cos 3A = 0 \qquad (z = \cos A)
The triple-angle identity rewritten as a cubic in z = cos A, which is the form used for the trigonometric solution.
Solution of Cubic and Quartic Equations; Their Discriminants
\cos 3A = 4\cos^3 A - 3\cos A
The triple-angle identity for cosine, expressing cos 3A in terms of cos A.
Solution of Cubic and Quartic Equations; Their Discriminants
n = \sqrt{-\tfrac{4}{3}p}
The scale factor n that makes the substitution y = nz produce the cosine form of the cubic.
Solution of Cubic and Quartic Equations; Their Discriminants
\cos{3A} = -\tfrac{1}{2}q ÷ \sqrt{-p^{3}/27}
The angle A is defined by cos 3A equal to minus q/2 divided by the square root of minus p cubed over 27.
Solution of Cubic and Quartic Equations; Their Discriminants
z^3 + \frac{p}{n^2}z + \frac{q}{n^3} = 0
After the substitution y = nz the reduced cubic becomes this equation in z, to be matched with the triple-angle form.
Solution of Cubic and Quartic Equations; Their Discriminants
R = -\Delta/108
In the irreducible case the quantity R equals minus the discriminant divided by 108.

Problems

Exercise Page46

Exercise Page46, problem 1, p. 46
$y^3 - 18y + 35 = 0$.
Printed answer:
- $-5$, $\tfrac{1}{2}(5±\sqrt{-3})$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [-5, (5 + sqrt(-3))/2, (5 - sqrt(-3))/2]
Exercise Page46, problem 2, p. 46
$x^3 + 6x^2 + 3x + 18 = 0$.
Printed answer:
- $-6$, $±\sqrt{-3}$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [-6, sqrt(-3), -sqrt(-3)]
Exercise Page46, problem 3, p. 46
$y^3 - 2y + 4 = 0$.
Printed answer:
- $-2$, $1± i$.
unverified: no computed check settled this one (yet)
How it was checked
- solve: the printed answer does not match the problem [-2, 1 + I, 1 - I]
Exercise Page46, problem 4, p. 46
$28x^3 + 9x^2 - 1 = 0$.
Printed answer:
- $\tfrac{1}{4}$, $\tfrac{1}{7}(-2±\sqrt{-3})$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [Rational(1, 4), (-2 + sqrt(-3))/7, (-2 - sqrt(-3))/7]

Exercise Page48

Exercise Page48, problem 1, p. 48
$y^3 - 2y - 4 = 0$.
Printed answer:
- $\Delta = -400$, one.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes -400
Exercise Page48, problem 2, p. 48
$y^3 - 15y + 4 = 0$.
Printed answer:
- $\Delta = 4 · 27 · 121$, three.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 4*27*121
Exercise Page48, problem 3, p. 48
$y^3 - 27y + 54 = 0$.
Printed answer:
- $\Delta = 0$, two.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 0
Exercise Page48, problem 4, p. 48
$x^3 + 4x^2 - 11x + 6 = 0$.
Printed answer:
- $\Delta = 0$, two.
verified: the printed answer passed a computed check
How it was checked
- evaluate: passes 0
Exercise Page48, problem 5, p. 48
Show by means of §21 that a double root of a real cubic is real.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles

Exercise Page49

Exercise Page49, problem 1, p. 49
Solve $y^3 -15y+4=0$.
Printed answer:
- $-4$, $2±\sqrt{3}$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [-4, 2 + sqrt(3), 2 - sqrt(3)]
Exercise Page49, problem 2, p. 49
Solve $y^3 -2y-1=0$.
Printed answer:
- See Ex. 1, §47.
unverified: no computed check settled this one (yet)
How it was checked
- solve: no printed answer to check
Exercise Page49, problem 3, p. 49
Solve $y^3 -7y+7=0$.
Printed answer:
- $1.3569$, $1.6920$, $-3.0489$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [1.3569, 1.6920, -3.0489]
Exercise Page49, problem 4, p. 49
Solve $x^3+ 3x^2 -2x-5=0$.
Printed answer:
- $-1.201639$, $1.330058$, $-3.128419$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [-1.201639, 1.330058, -3.128419]
Exercise Page49, problem 5, p. 49
Solve $x^3 +x^2 -2x-1=0$.
Printed answer:
- $1.24698$, $-1.80194$, $-0.44504$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [1.24698, -1.80194, -0.44504]
Exercise Page49, problem 6, p. 49
Solve $x^3 +4x^2 -7=0$.
Printed answer:
- $1.1642$, $-1.7729$, $-3.3914$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [1.1642, -1.7729, -3.3914]

Exercise Page49b

The data holds no problems for this exercise yet.

Exercise Page51

Exercise Page51, problem 1, p. 51
Solve $x^4 - 8x^3 + 9x^2 + 8x - 10 = 0$. Note that $(17)$ is $(y - 9) (y^2 - 24) = 0$.
Printed answer:
- $1$, $-1$, $4±\sqrt{6}$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [1, -1, 4 + sqrt(6), 4 - sqrt(6)]
Exercise Page51, problem 2, p. 51
Solve $x^4 - 2x^3 - 7x^2 + 8x + 12 = 0$. Since the right member of $(16)$ is $(8 + y) (x^2 - x) + \frac{1}{4} y^2 - 12$, use $y = -8$.
Printed answer:
- $-1$, $-2$, $2$, $3$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [-1, -2, 2, 3]
Exercise Page51, problem 3, p. 51
Solve $x^4 - 3x^2 + 6x - 2 = 0$.
Printed answer:
- $1± i$, $-1±\sqrt{2}$.
unverified: no computed check settled this one (yet)
How it was checked
- solve: the printed answer does not match the problem [1 + I, 1 - I, -1 + sqrt(2), -1 - sqrt(2)]
Exercise Page51, problem 4, p. 51
Solve $x^4 - 2x^2 - 8x - 3 = 0$.
Printed answer:
- $1±\sqrt{2}$, $-1±\sqrt{-2}$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes [1 + sqrt(2), 1 - sqrt(2), -1 + sqrt(-2), -1 - sqrt(-2)]
Exercise Page51, problem 5, p. 51
Solve $x^4 - 10x^2 - 20x - 16 = 0$.
Printed answer:
- $4$, $-2$, $-1± i$.
unverified: no computed check settled this one (yet)
How it was checked
- solve: the printed answer does not match the problem [4, -2, -1 + I, -1 - I]

Exercise Page52

The data holds no problems for this exercise yet.

Exercise Page53

The data holds no problems for this exercise yet.

Exercise Page54

Exercise Page54, problem 1, p. 54
Find the coordinates of the single real point of intersection of the parabola $y = x^2$ and the hyperbola $xy - 4x + y + 6 = 0$.
Printed answer:
- $(-3, 9)$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes {x: -3, y: 9}
Exercise Page54, problem 2, p. 54
Show that the abscissas of the points of intersection of $y=x^2$ and $ax^2 - xy + y^2 - x - (a+5)y - 6 = 0$ are the roots of $x^4 - x^3 - 5x^2 - x - 6 = 0$. Compute the discriminant of the latter and show that only two of the four points of intersection are real.
Printed answer:
- $\Delta=-250000$, $x=3$, $-2$, $±i$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page54, problem 3, p. 54
Find the coordinates of the two real points in Ex. 2.
Printed answer:
- $(3,9)$, $(-2,4)$.
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page54, problem 4, p. 54
A right prism of height $h$ has a square base whose side is $b$ and whose diagonal is therefore $b\sqrt{2}$. If $v$ denotes the volume and $d$ a diagonal of the prism, $v = hb^2$ and $d^2 = h^2 + (b\sqrt{2})^2$. Multiply the last equation by $h$ and replace $hb^2$ by $v$. Hence $h^3 - d^2h + 2v = 0$. Its discriminant is zero if $d = 3\sqrt{3}$, $v = 27$; find $h$.
Printed answer:
- $h=3$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes, with the problem read into an equation 3
Exercise Page54, problem 5, p. 54
Find the admissible values of $h$ in Ex. 4 when $d = 12$, $v = 332.5$.
Printed answer:
- $6.856$, $7$.
verified: the printed answer passed a computed check
How it was checked
- solve: passes, with the problem read into an equation [6.856, 7]
Exercise Page54, problem 6, p. 54
Find a necessary and sufficient condition that quartic equation $(15)$ shall have one root the negative of another root. Hint: $(x_1 + x_2)(x_3 + x_4) = q - y_1$. Hence substitute $q$ for $y$ in $(17)$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page54, problem 7, p. 54
In the study of parabolic orbits occurs the equation % [** PP: Displayed for better line breaking.] 12v + 13^3 12v = t. Prove that there is a single real root and that it has the same sign as $t$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles
Exercise Page54, problem 8, p. 54
In the problem of three astronomical bodies occurs the equation $x^3 + ax + 2 = 0$. Prove that it has three real roots if and only if $a\leqq{-3}$.
Printed answer:
- (none printed)
unverified: no computed check settled this one (yet)
How it was checked
- other: not a kind the checker handles