solve the equation x^4-2x^3-21x^2+22x+40=0 whose roots are in arithmetical progression
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Answer:
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Step-by-step explanation:
x
4
−2x
3
−21x
2
+22x+40=0
Let the roots be a−3d,a−d,a+d,a+3d
S
1
=a−3d+a−d+a+d+a+3d=−
1
−2
4a=2
a=
2
1
S
4
=(a−3d)(a+3d)(a−d)(a+d)=
1
40
(a
2
−9d
2
)(a
2
−d
2
)=40
(
4
1
−9d
2
)(
4
1
−d
2
)=40
(1−36d
2
)(1−4d
2
)=640
1−4d
2
−36d
2
+144d
4
−640=0
144d
4
−40d
2
−639=0
144d
4
+284d
2
−324d
2
−639=0
(4d
2
−9)(36d
2
+71)=0
4d
2
−9=0
⇒d=±
2
3
So the roots are −4,−1,2,5
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