Download Cubic Equations with Problems

THEORY OF EQUATIONS – Notes
p.7
C. Cubic and Quartic Equations
ex/ Transform the cubic equation, x3 + bx2 + cx + d = 0 into the reduced cubic
with the 'x2' term missing. (see the example on page 6) y 3  py  q  0
Since b/1 = − (r1 + r2 + r3) , we'll need r1 + r2 + r3 = −b
So we will diminish all three roots by −b/3 , obtaining: P (x − b/3) = 0
We are actually increasing each root by b/3.

  c  x  b3   d
2
3
 x3  3x 2 b3  3x  b3    b3   b  x 2  2 b3 x  b9   cx  bc
d
3
 x  b3

3

 b x  b3
2
2
The x2-terms cancel out (or they should!) and to distinguish this new function
from the old one, we'll use 'y' as the independent variable:
2 3 =0
P( y)  y3  c  13 b2 y  d  13 bc  27
b

 

2 3
with coefficients, p = c  13 b2 and q = d  13 bc  27
b
So we can take any cubic equation, a3x3 + a2x2 + a1x + a0 = 0 , and divide by a3
to get, x3 + bx2 + cx + d = 0, and then eliminate the x2-term to get: y 3  py  q  0
(Can you guess where we're going with all this?!?)
Now introduce 2 unknowns, u and v, whose sum is a root of this 'reduced cubic'.
Substituting we get: (u + v)3 + p(u + v) + q = 0
u3 + 3u2v + 3uv2 + v3 + p(u + v) + q = 0
u3 + v3 + 3uv(u + v) + p(u + v) + q = 0
u3 + v3 + (3uv + p)(u + v) + q = 0
Condition 1 - u + v was a root of the reduced cubic
Condition 2 - 3uv + p = 0 (v = −p/3u)
Now we have: u3 + v3 + q = 0 and substituting with v = −p/3u
p3
p3
3
6
3
u 
 q  0 or u  qu 
 0 (which is a quadratic in u3, whoa!)
3
27
27u
By the quadratic formula: u 
3
q  q 2  4(1)
  
 p3
27
1
q
2
1 3
p
R where R = 14 q 2  27
2(1)
Let u3 = A =  12 q  R and let v3 = B =  12 q  R (since u3 + v3 + q = 0 above)
From our study of complex numbers (DeMoivre's Theorem) the 3 cube roots of A
are: u = 3 A ,  3 A ,  2 3 A and the 3 cube roots of B are: v = 3 B ,  3 B ,  2 3 B
where    12  23 i  cis120 and  2   12  23 i  cis 240
So if u+v is a root of the reduced equation, then u+v − b/3 is a root of the original,
but which u+v combination? There are 9 combination!
With the condition that 3uv + p = 0 (uv = −p/3) , the only pairs of u and v satisfying
this condition yield the following 3 solutions to the original equation.
THEORY OF EQUATIONS – Notes
p.8
C. Cubic and Quartic Equations
Wait a second! Who checked to see if u = 3 A and v = 3 B satisfied uv = −p/3 ?
p
q2
q2
q2
p3
p3
uv = 3 AB  3 4  R  3 4  4  27  3  27 
okay, this pair worked, so
3
x1 = 3 A + 3 B − b/3 , so what are the other two solutions?
x2 =  3 A   2 3 B − b/3 Well since (1cis 120)(1cis 240) = 1cis 360 = 1 + 0i …
x3 =  2 3 A   3 B − b/3 I can see how the 'omegas' go away! Okay…
Cardan's Formulas – Girolamo Cardano (1501-1576) Ars Magna 1545
http://www.geocities.com/doracy/alunos/Cardano.html


credit for solving: x3 + px = q appears to belong to Scipione del Ferro (1465-1526)
credit for solving: x3 + px2 = q to Nicolo Tartaglia (1500-1557)
Public contests, prestige, and monetary awards kept many math discoveries secret.
An algebraic solution to a polynomial equation must be expressed in terms of its
coefficients by means of formulas involving a finite number of operations of
addition, subtraction, multiplication, division and extraction of roots.
Norwegian Niels Abel (1802-1829) first proved in 1824 that polynomial
equations, quintic (5th degree) and higher cannot be solved algebraically.
ex/ Solve x3 + 9x2 + 18x + 28 = 0 (Recall p = c − b2/3 and q = d − bc/3 + 2b3/27)
And R = q2/4 + p3/27 with u3 = A =  12 q  R and v3 = B =  12 q  R
So we get: p = 18 − 81/3 = −9 and q = 28 − 54 + 54 = 28
and R = −27 + 196 = 169
A = u3 = −14 + 13 = −1 and B = v3 = −14 − 13 = −27
So we get: 3 A  1 and 3 B  3
Cardan's Formulas:

with b/3 = 3 and  2   12 
3
i
2

2
 1cis120  1cis 240   12 
2
x1 = −1 + −3 − b/3 = −7

x =  (1)    3  3  1 
x2 =  (1)   2  3  3  1  12 
2
3
1
2

  3 
i   3 
  3  1 
i   3  1 
3
i
2
3
i
2
1
2

3
i
2
3i
3
2
1
2

3
2
3i
(Oh yeah, the complex conjugate. I guess we didn't have to… )
* When x + bx2 + cx + d = 0 with real coefficients, has:
(i) 3 distinct roots, R < 0 (This is called the irreducible case. )
(ii) 1 real root, R > 0 (iii) at least 2 real roots, R = 0
3
THEORY OF EQUATIONS – Notes
C. Cubic and Quartic Equations
p.9
Irreducible Case of the Cubic (when R < 0) where there is no algebraic process to
find the exact cube roots of complex numbers. Hmm…
Recall: u3 = A =  12 q  R =   cos  i sin     cis
and v3 = B =  12 q  R =   cos  i sin  
Adding these two we get: −q = 2 cos or cos   q 2 
Multiplying these two to get:
q2
4


 R   2 cos2   sin 2    2
R = q2/4 + p3/27 now substitute for 'R' to get: −p3/27 =  2 or
 p3
27
=
Since R < 0 (Irreducible Case), p < 0 and hence  is real, so we may write:
 3  120  , u =  cis  3  240 
 cos 3  i sin 3  , v =   cos  3  120  i sin  3  120
v =   cos  3  240  i sin  3  240 
u1 =  3 cis 3 , u2 =  3 cis
1
v1 = 
1
3
1
3
2
3
1
3
1
3
1
3
Condition: uv = −p/3 results in only 3 pairs: u1 and v1 , u2 and v2 , u3 and v3
General Solution:
x1 = 2 3  cos 3  b3
 3  120  b3
= 2 3  cos  3  240   b3
x2 = 2 3  cos
x3
ex/ Solve x3 + 6x2 + 9x + 1 = 0 , p = −3 , q = −1 , R = −3/4 < 0 (3 distinct real roots)
q 1
 p3
cos 
    27  1 so we have  = 60 degrees with −b/3 = −2
2 2
x1 = 2cos 20 −2  −0.1206
x2 = 2cos 140 − 2  −3.5320
x3 = 2cos 260 − 2  −2.3472
Hey, you didn't have to type this all up!
One more page on 'quartic equations'!
THEORY OF EQUATIONS – Notes
C. Cubic and Quartic Equations
p.10
Solution of the Quartic Equation
P(x) = x4 + bx3 + cx2 + dx + e = 0
Transform to P(x − b/4) yielding the 'reduced quartic' by increasing each of
the four roots by b/4: y 4  qy 2  ry  s  0
Assuming the left side can be expressed as: (y2 + 2ky + )(y2 − 2ky + m)
where k, , m are to be determined.
y4 + qy2 + ry + s = y4 + ( + m − 4k2)y2 + 2k(m − )y + m
then: + m − 4k2 = q or
+m = q + 4k2
2k (m − ) = r
or
m − = r/2k
m=s
Adding and subtracting the equations above:
m = ½ (q + 4k2 + r/2k)
and
= ½ (q + 4k2 − r/2k)
Whew! Now substitute and m into that m = s equation…
¼ (q2 + 4k2q + (rq)/(2k) + 4k2q + 16k4 + 2kr − (rq)/(2k) − 2kr − r2/4k2) = s
or 64k6 + 32qk4 + 4(q2 − 4s)k2 − r2 = 0 (the resolvent cubic)
a solvable cubic in k2 !
Any nonzero k2 root of this equation can be used to obtain and m.
Then the 4 roots can be extracted from the quadratics above, viz.,
y2 + 2ky + = 0 and y2 − 2ky + m = 0
ex/ Try x4 − 8x3 + 21x2 − 14x − 10 = 0
Reduced Quartic: y4 − 3y2 + 6y − 2 = 0
Resolvent Cubic: 16k6 − 24k4 + 17k2 − 9 = 0 (k2 = 1 works! so =−1 and m = 2
Quadratics: y2 + 2y + 1 = 0 and y2 − 2y + 2 = 0 yield
x1 and x2 = 1  2
x3 and x4 = 3  i
The first solution was given by Lodovico Ferrari (1522-1560), a pupil of Cardan.
The above solution is basically the same as the one found by Descartes (1596-1650).
Below some cubic equations to solve!
Cubic Equations – Problems & Solutions
p.11
3
3
2
5. 4x − 9x + 14 = 0
9. x + 12x + 30x − 43 = 0
1. x − 9x + 28 = 0
3
3
6. 3x + 9x + 8 = 0
10. x3 − 3x2 + 27x − 9 = 0
2. x + 9x − 26 = 0
7. x3 + 3x2 + 27x − 31 = 0 11. 6x3 − 18x2 + 36x − 19 = 0
3. x3 + 6x + 2 = 0
4. x3 + 108x + 180 = 0 8. 3x3 + 9x2 + 4x + 12 = 0 12. y3 − 6  y + 2 = 0
1. 4,2  3 i
3
2. p = 9 and q = −26 so R =
 262
4
3
9
 27
 132  27  196  142
So A = u = −q/2 + R = 13 + 14 = 27 and B = v3 = −q/2 − R = 13 − 14 = −1
The equation is already 'reduced' so b/3 = 0
x1 = 3 A  3 B  b3 = 3 − 1 − 0 = 2, Now we'll short cut our work by using the factor
3
theorem: (x − 2)(x2 + 2x + 13) = 0 and the quadratic formula to get: x = 2, 1  2 3 i
3.
3
2  3 4 , 12
 3 4  3 2   23  3 4  3 2 
4. p = 108 and q = 180 so R =
1802
4
2
 108
 54756  2342 (calculator, anybody?)
27
So A = u3 = −q/2 + R = −90 + 234 = 144 and B = v3 = −q/2 − R = −90 − 234 = −324
x1 =
3
A  3 B  b3 =
5. 2,1 
3
144  3 324  −1.62680,  12
 3 144  3 324   23  3 144  3 324  i
3
i
2
6. For the 'reduced' equation, p = 3 and q = 8/3, R = 25/9 and x1 =
with x2 and x3 =  12

3 1
3
 
33 
3 3 1
2
3

3 1
3
 3 3  −.74889
33 i
7. 1,  2  3 3 i
8. For x3 + 3x2 + 43 x + 4 = 0, we need to get to the 'reduced' equation where
p = c − b2/3 = 4/3 − 3 = −5/3 and q = d − bc/3 + 2b3/27 = 4 − 4/3 + 2 = 14/3
R = q2/4 + p3/27 = 3844/729 = 622/272 (of course!) Here b/3 = 1 so subtract 1 later.
A = u3 = −q/2 + R = −7/3 + 62/27 = 20/27, B = v3 = −q/2 − R = −7/3 − 62/27 = −104/27
x1 =−1 +
3
20  3 104 , x2, x3 = −1  12
 3 20  3 104   23  3 20  3 104  i
9. 1,  13
 23 i
2
10. p = 27 − 9/3 = 24 and q = −9 − (−27) − 2 = 16 with b/3 = −1 so we'll add 1 later.
R = 64 + 512 = 576 and A = u3 = −8 + 24 = 16, B = v3 = −8 −24 = −32
 3 16  3 32   23  3 16  3 32  i
11. 1  13 3 18  12 3 12 , 1  16 3 18  14 3 12  23  13 3 18  12 3 12  i
x1 = 1 + 3 16  3 32 , x2 and x3 = 1  12
12.  = 1cis120 , p = −6  , q = 2, R = 1 − 8 = −7(irreducible case),   8 , cos 
x1 = 2 2 cos 3 , x2 = 2 2 cos
 3  120 , x
3
= 2 2 cos
 3  240
1
8