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JEE Main 8 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 26
Total: 90
The number of distinct real roots of the equation
|
x
+
1
|
|
x
+
3
|
−
4
|
x
+
2
|
+
5
=
0
, is ______.
[8 Apr 2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
|
x
+
1
|
|
x
+
3
|
−
4
|
x
+
2
|
+
5
=
0
case-1
x
≤
−
3
(
x
+
1
)
(
x
+
3
)
+
4
(
x
+
2
)
+
5
=
0
x
2
+
4
x
+
3
+
4
x
+
8
+
5
=
0
x
2
+
8
x
+
16
=
0
(
x
+
4
)
2
=
0
x
=
−
4
case-2
−
3
≤
x
≤
−
2
−
x
2
−
4
x
−
3
+
4
x
+
8
+
5
=
0
−
x
2
+
10
=
0
x
=
±
√
10
case- 3
−
2
≤
x
≤
−
1
−
x
2
−
4
x
−
3
−
4
x
−
8
+
5
=
0
−
x
2
−
8
x
−
6
=
0
x
2
+
8
x
+
6
=
0
x
=
−
8
±
2
√
10
2
=
−
4
±
√
10
case-4
x
≥
−
1
x
2
+
4
x
+
3
−
4
x
−
8
+
5
=
0
x
2
=
0
x
=
0
No. of solution
=
2
© examsnet.com
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