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Byju's Answer
Standard XII
Mathematics
Existence of Limit
1x-2+2x-1=6 x...
Question
1
(
x
-
2
)
+
2
(
x
-
1
)
=
6
x
,
(
x
≠
2
,
1
)
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Solution
Given
:
1
(
x
−
2
)
+
2
(
x
−
1
)
=
6
x
⇒
(
x
−
1
)
+
2
(
x
−
2
)
(
x
−
1
)
(
x
−
2
)
=
6
x
⇒
3
x
−
5
x
2
−
3
x
+
2
=
6
x
⇒
3
x
2
−
5
x
=
6
x
2
−
18
x
+
12
[
O
n cross multiplying
]
⇒
3
x
2
−
13
x
+
12
=
0
⇒
3
x
2
−
(
9
+
4
)
x
+
12
=
0
⇒
3
x
2
−
9
x
−
4
x
+
12
=
0
⇒
3
x
(
x
−
3
)
−
4
(
x
−
3
)
=
0
⇒
(
3
x
−
4
)
(
x
−
3
)
=
0
⇒
3
x
−
4
=
0
or
x
−
3
=
0
⇒
x
=
4
3
or
x
=
3
H
ence, the roots of the equation are
4
3
and 3.
Suggest Corrections
0
Similar questions
Q.
Using the properties of determinants, solve the following for x:
⎡
⎢
⎣
X
+
2
X
+
6
X
−
1
X
+
6
X
−
1
X
+
2
X
−
1
X
+
2
X
+
6
⎤
⎥
⎦
=
0
Q.
Using the properties of determinants, solve the following for
x
:
∣
∣ ∣
∣
x
+
2
x
+
6
x
−
1
x
+
6
x
−
1
x
+
2
x
−
1
x
+
2
x
+
6
∣
∣ ∣
∣
=
0
Q.
Solve the following equation by using properties of determinants
∣
∣ ∣
∣
x
+
2
x
+
6
x
−
1
x
+
6
x
−
1
x
+
2
x
−
1
x
+
2
x
+
6
∣
∣ ∣
∣
=
0
Q.
Which of the following are quadratic equations?
(i) x
2
+ 6x − 4 = 0
(ii)
3
x
2
-
2
x
+
1
2
=
0
(iii)
x
2
+
1
x
2
=
5
(iv)
x
-
3
x
=
x
2
(v)
2
x
2
-
3
x
+
9
=
0
(vi)
x
2
-
2
x
-
x
-
5
=
0
(vii) 3x
2
− 5x + 9 = x
2
− 7x + 3
(viii)
x
+
1
x
=
1
(ix) x
2
− 3x = 0
(x)
x
+
1
x
2
=
3
x
+
1
x
+
4
(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)
x
+
1
x
=
x
2
,
x
≠
0
(xiii) 16x
2
− 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)
3
= x
3
− 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)
