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Byju's Answer
Standard IX
Mathematics
By Grouping
Given that ...
Question
Given that
a
≠
±
1
2
, simplify the expression
2
a
2
+
5
a
−
3
4
a
2
−
1
?
A
2
a
−
3
2
a
+
1
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B
a
−
3
2
a
−
1
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C
2
a
+
3
2
a
−
1
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D
a
+
3
2
a
+
1
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Solution
The correct option is
D
a
+
3
2
a
+
1
We need to simplify
2
a
2
+
5
a
−
3
4
a
2
−
1
On factoring, we get
=
2
a
2
−
1
a
+
6
a
−
3
=
a
(
2
a
−
1
)
+
3
(
2
a
−
1
)
The factors are
(
2
a
−
1
)
(
a
+
3
)
⇒
4
a
2
−
1
=
(
2
a
−
1
)
(
2
a
+
1
)
⇒
2
a
2
+
5
a
−
3
4
a
2
−
1
=
(
2
a
−
1
)
(
a
+
3
)
(
2
a
−
1
)
(
2
a
+
1
)
=
a
+
3
2
a
+
1
Suggest Corrections
0
Similar questions
Q.
Show that
(
a
+
1
2
a
+
1
2
a
+
⋯
)
2
+
(
a
−
1
2
a
−
1
2
a
−
⋯
)
2
=
2
a
2
,
and
(
a
+
1
2
a
+
1
2
a
+
⋯
)
(
a
−
1
2
a
−
1
2
a
−
⋯
)
=
a
2
−
1
2
a
2
−
1
2
a
2
−
⋯
.
Q.
If
x
=
3
√
2
a
+
1
+
3
√
2
a
−
1
3
√
2
a
+
1
−
3
√
2
a
−
1
, show that
x
3
−
6
x
2
+
3
x
−
2
a
=
0
Q.
Using properties of determinants, prove that
∣
∣ ∣
∣
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
∣
∣ ∣
∣
=
(
a
−
1
)
3
Q.
Prove that:
∣
∣ ∣
∣
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
∣
∣ ∣
∣
=
(
a
−
1
)
C
then
C
=
?
Q.
∣
∣ ∣
∣
a
2
+
2
a
2
a
+
1
1
2
a
+
1
a
+
2
1
3
3
1
∣
∣ ∣
∣
=
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