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Question
Mathematics
If x2+(a-b) x+(1-a-b)=0, a, b ∈ R then the value of 'a' for which both roots of the equation are real and unequal ∀ b ∈ R is
Q. If
x
2
+
(
a
−
b
)
x
+
(
1
−
a
−
b
)
=
0
,
a
,
b
∈
R
then the value of 'a' for which both roots of the equation are real and unequal
∀
b
∈
R
is
236
178
Complex Numbers and Quadratic Equations
Report Error
A
(
2
,
∞
)
B
(
3
,
∞
)
C
(
1
,
∞
)
D
(
−
∞
,
1
)
Solution:
x
2
+
(
a
−
b
)
x
+
(
1
−
a
−
b
)
=
0
∵
D
>
0
⇒
(
a
−
b
)
2
−
4
×
1
×
(
1
−
a
−
b
)
>
0
a
2
+
b
2
−
2
ab
−
4
+
4
a
+
4
b
>
0
∵
b
2
+
2
b
(
2
−
a
)
+
(
a
2
+
4
a
−
4
)
>
0
∵
4
(
2
−
a
)
2
−
4
×
1
×
(
a
2
+
4
a
−
4
)
<
0
4
+
a
2
−
4
a
−
a
2
−
4
a
+
4
<
0
8
a
−
8
>
0
a
>
1