HISTORY OF MATH HOMEWORK 5 SOLUTIONS 9.30, 32, 35, 37 ...

HISTORY OF MATH
HOMEWORK 5 SOLUTIONS
9.30, 32, 35, 37, 38
9.30. Use Cardano’s formula to solve x 3 = 6x + 6.
We have p = −2, q = 3, so q 2 + p 3 = 1. The positive real solution is
x = [3 + 1] 1/3 + [3 − 1] 1/3 = 4 1/3 + 2 1/3 .
The general solution is
w + 2 w ,
where w is any cube root of 4.
9.32. Solve x 3 + 21x = 9x 2 + 5.
Making the substitution x = y + 3 leads to the equation
y 3 − 6y + 4 = 0.
Trying divisors of 4, we find that y = 2 is a root, and that
y 3 − 6y + 4 = (y − 2)(y 2 + 2y − 2).
The latter quadratic has roots −1 ± √ 3, so the original equation has roots
x = 5, 2 ± √ 3.
We don’t need Cardano for this, since there was a rational root. In fact, Cardano gives
x = [−2 + 2 √ −1] 1/3 + [−2 − 2 √ −1] 1/3 .
The Bombelli procedure gives the cube root 1 ± √ −1 of −2 ± 2 √ −1, and indeed (1 + √ −1) +
(1 − √ −1) = 2.
1

2 HISTORY OF MATH HOMEWORK 5 SOLUTIONS
9.35 Solve the system of equations
x 3 + 3xy 2 = x + y + 64
y 3 + 3yx 2 = x + y
Subtract the two equations and get (x − y) 3 = 64, so x − y = 4. Substitute y = x − 4 into the
second equation and get
y 3 + 6y 2 + 23
2 y − 1 = 0.
Let y = z − 2, to get
z 3 − 1z = 8. 2
So p = −1/6 and q = 4 and q 2 + p 3 = 16 − 1/216 = 3455/216. Cardano’s (actually Tartaglia’s
in this case!) formula gives
y = z − 2 = −2 +
and x = y + 4 as shown above.
[
4 +
√ ] 1/3 [
3455
+ 4 −
216
√ ] 1/3
3455
,
216
9.37 The equation x 3 + 3x = 36 has the root r = 3, but Cardano’s formula gives
Explain this using Bombelli’s methods.
r = [18 + √ 325] 1/3 + [18 − √ 325] 1/3 .
The numbers
α = [18 + √ 325] 1/3 , β = [18 − √ 325] 1/3
Satisfy
α + β = r, αβ = −p,
so they are the roots of the quadratic Y 2 − rY − p = 0, where r = 3 and p = 1. The quadratic
formula then gives
α, β = 1[3 ± √ 13].
2
You can check that
( 1 [3 + √ 13] ) 3 √
2 = 18 + 325,
( 1 [3 − √ 13] ) 3 √
2 = 18 − 325,
so
α = 1 2 [3 + √ 13], β = 1 2 [3 − √ 13].

HISTORY OF MATH HOMEWORK 5 SOLUTIONS 3
9.38 Express [52 + √ −2209] 1/3 in the form a + b √ −1.
Note that 2209 = 47 2 . So we must find a, b such that
Working out the cube gives two equations
(a + b √ −1) 3 = 52 + 47 √ −1.
a 3 − 3ab 2 = 52, 3a 2 b − b 3 = 47.
Adding these gives
(a − b)(a 2 + 4ab + b 2 ) = 99.
Subtracting gives
(a + b)(a 2 − 4ab + b 2 ) = 5.
Now let’s be optimistic and look for integer solutions for a, b. The latter equation says that a + b
divides 5, so a + b = ±1, ±5. We find that a + b = 5 leads to
a 2 − 4a(5 − a) + (5 − a) 2 = 1,
which has solutions a = 1, 4, hence b = 4, 1. But the first equation says that a − b > 0, so we
take a = 4, b = 1, and note that this satisfies both equations. Hence
[52 + √ −2209] 1/3 = 4 + √ −1.
Another way to do this, without guessing, is to reconstruct the cubic for which [52+ √ −2209] 1/3
arises in the Cardano formula. We have q = 52 and p 3 = −2209 − q 2 = −17 3 so p = −17. The
cubic
x 3 − 51x = 104
has the root r = 8. So
8 = [52 + √ −2209] 1/3 + [52 − √ −2209] 1/3 .
The quadratic
Y 2 − rY − p = Y 2 − 8Y + 17
has roots 4 ± √ −1. Cubing these, we find that (4 + √ −1) 3 = 52 + √ −2209.