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Mathematics Programs 15–21
b
0
= a
0
(4a
2
a
3
2
) – a
1
2
.
Let
y
0
be the largest real root of the above cubic. Then the fourth–order polynomial
is reduced to two quadratic polynomials:
x
2
+ (J + L)x + (K + M) = 0
x
2
+ (JL)x + (KM) = 0
where
J = a
3
/2
K = y
0
/2
L =
02
2
yaJ +
×
(the sign of
JK – a
1
/2)
M =
0
2
aK
Roots of the fourth degree polynomial are found by solving these two quadratic
polynomials.
A quadratic equation
x
2
+ a
1
x + a
0
= 0 is solved by the formula
0
2
11
2,1
)
2
(
2
a
aa
x ±=
If the discriminant d = (a
1
/2)
2
a
o
0, the roots are real; if
d
<
0, the roots are
complex, being
diaivu ±=± )2(
1
.