Look at the constant term
If the roots of the polynomial equation 4 x 4 + 6 x 3 + 2 x 2 + 2 0 0 3 x − 4 0 1 2 0 0 9 = 0 are of the form c − a ± b , c 2 − a ± i ⋅ 7 ( a + c ) ⋅ d , where b , c and d are primes. Find a + b + c + d .
The answer is 4119.
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2 solutions
4 x 4 + 6 x 3 + 2 x 2 + 2 0 0 3 x - 4012009 = 0 4 x 4 can be written as 2 x 2 × 2 x 2 . 6 x 3 can be written as ( 2 x 2 × 2 x ) + ( x × 2 x 2 ) . 2 x 2 can be written as ( x × 2 x ) + ( 2 x 2 × 2 0 0 3 ) + ( 2 x 2 × − 2 0 0 3 ) . 2 0 0 3 x can be written as ( 2 0 0 3 × 2 x ) + ( − 2 0 0 3 × x ) . -4012009 = 2 0 1 3 × − 2 0 1 3 .
So, 4 x 4 + 6 x 3 + 2 x 2 + 2 0 0 3 x - 4012009 can be written as ( 2 x 2 + x + 2 0 0 3 ) ( 2 x 2 + 2 x − 2 0 0 3 ) .
So,
(
2
x
2
+
x
+
2
0
0
3
)
(
2
x
2
+
2
x
−
2
0
0
3
)
=
0
Either
(
2
x
2
+
x
+
2
0
0
3
)
=
0
or
(
2
x
2
+
2
x
−
2
0
0
3
)
=
0
Applying quadratic formula on both equations we get:-
x
=
2
−
1
±
4
0
0
7
or
x
=
4
−
1
±
7
i
3
2
7
=
4
−
1
±
7
i
(
1
+
2
)
×
1
0
9
Here
4
0
0
7
,
2
and
1
0
9
are primes. So,
a
=
1
;
b
=
4
0
0
7
;
c
=
2
;
d
=
1
0
9
∴
a
+
b
+
c
+
d
=
1
+
2
+
4
0
0
7
+
1
0
9
=
4
1
1
9
.
The Equation 4x^4 + 6x^3 + 2x^2 + 2003x - 4012009 = (2x^2 + x + 2003)(2x^2 + 2x - 2003) =0, so we have 2 quadratic equations to solve:The solutions are: for the real pair, x = [-1 +/- sqrt(4007)]/2 , so b = 4007, a = 1, c = 2. For the complex pair, x = [ -1 +/- 7i sqrt(3 109)]/4, so d = 109, and a + b + c + d = 4119. Ed Gray