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KEAM 2015 Math Paper
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© examsnet.com
Question : 52
Total: 120
The roots of the equation
|
1
+
x
3
5
2
2
+
x
5
2
3
x
+
4
|
= 0 are
2, 1,-9
1, 1,-9
- 1 ,1,-9
-2 ,-1 ,-8
-2, 1, 1
Validate
Solution:
|
1
+
x
3
5
2
2
+
x
5
2
3
x
+
4
|
=
0
Applying
C
1
→
C
1
+
C
2
+
C
3
|
x
+
9
3
5
x
+
9
2
+
x
5
x
+
9
3
x
+
4
|
=
0
(
x
+
9
)
|
1
3
5
1
2
+
x
5
1
3
x
+
4
|
=
0
[taking common
(
x
+
9
)
from
C
1
]
Applying
R
1
→
R
1
−
R
2
and
R
2
→
R
2
−
R
3
(
x
+
9
)
|
0
1
−
x
0
0
x
−
1
1
−
x
1
3
x
+
4
|
=
0
Expanding along
C
1
, we get
(
x
+
9
)
[
(
1
−
x
)
2
−
0
]
=
0
⇒
(
x
+
9
)
(
1
−
x
)
2
=
0
⇒
x
=
−
9
,
1
,
1
© examsnet.com
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