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CBSE Questions for Class 10 Maths Quadratic Equations Quiz 9 - MCQExams.com
CBSE
Class 10 Maths
Quadratic Equations
Quiz 9
Solve the quadratic equation by using quadratic formula
9
x
2
−
3
(
a
+
b
)
x
+
a
b
=
0
Report Question
0%
x
=
a
3
,
b
3
0%
x
=
−
a
3
,
−
b
3
0%
x
=
−
a
3
,
b
3
0%
x
=
a
3
,
−
b
3
Explanation
⇒
Here,
a
=
9
,
b
=
−
3
(
a
+
b
)
,
c
=
a
b
⇒
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
−
[
−
3
(
a
+
b
)
]
±
√
[
−
3
(
a
+
b
)
]
2
−
4
(
9
)
(
a
b
)
2
(
9
)
=
3
a
+
3
b
±
√
9
(
a
2
+
b
2
+
2
a
b
)
−
36
a
b
18
=
3
a
+
3
b
±
√
9
a
2
+
9
b
2
+
18
a
b
−
36
a
b
18
=
3
a
+
3
b
±
√
9
a
2
+
9
b
2
−
18
a
b
18
3
a
+
3
b
±
√
9
(
a
2
+
b
2
−
2
a
b
)
18
=
3
a
+
3
b
±
√
9
(
a
−
b
)
2
18
=
3
a
+
3
b
±
3
(
a
−
b
)
18
=
3
a
+
3
b
±
3
a
−
3
b
18
x
=
3
a
+
3
b
+
3
a
−
3
b
18
and
x
=
3
a
+
3
b
−
3
a
+
3
b
18
x
=
6
a
18
and
x
=
6
b
18
∴
x
=
a
3
,
b
3
The solution for
2
x
2
−
9
x
+
7
=
0
is
Report Question
0%
2
,
3
0%
1
,
9
4
0%
1
4
,
13
4
0%
1
,
13
4
Explanation
Given Quadratic equation is
2
x
2
−
9
x
+
7
=
0
Comparing it with the general form of quadratic equation
a
x
2
+
b
x
+
c
=
0
we write,
a
=
2
,
b
=
−
9
,
c
=
7
Quadratic formula is
x
=
−
b
±
√
b
2
−
4
a
c
2
a
⟹
x
=
−
(
−
9
)
±
√
(
−
9
)
2
−
4
(
2
)
(
7
)
2
(
2
)
⟹
x
=
9
±
√
81
−
56
4
⟹
x
=
9
±
√
25
4
⟹
x
=
9
±
5
4
⟹
x
=
9
+
5
4
x
=
9
−
5
4
⟹
x
=
1
,
7
2
Find the roots of the following equations, if they exist, by the completing the square:
Report Question
0%
2
x
2
−
7
x
+
3
=
0
0%
2
x
2
+
x
−
4
=
0
0%
4
x
2
+
√
4
3
x
+
3
=
0
0%
2
x
2
+
x
+
4
=
0
Explanation
s
o
t
h
e
r
o
o
t
o
f
q
u
a
d
r
a
t
i
c
e
q
u
a
t
i
o
n
a
r
e
X
=
3
a
n
d
X
=
1
2
The root of
x
2
+
k
x
+
k
=
0
are real and equal , find
k
.
Report Question
0%
0
0%
4
0%
0
or
4
0%
2
Explanation
The roots of
x
2
+
k
x
+
k
=
0
are real and equal
i.e.,
b
2
−
4
a
c
=
0
⟹
k
2
−
4
(
1
)
k
=
0
⟹
k
2
−
4
k
=
0
⟹
k
(
k
−
4
)
=
0
∴
k
=
0
or
4
Find the root of the following quadratic equation, if they exist, by the method of completing the square.
2
x
2
+
x
−
4
=
0
.
Report Question
0%
−
1
−
√
3
3
4
0%
−
1
−
√
3
3
44
0%
−
1
+
√
3
3
44
0%
1
−
√
3
3
4
If both roots of quadratic equation
(
α
+
1
)
x
2
−
2
(
1
+
3
α
)
x
+
1
+
8
α
=
0
are real and distinict, the
α
be-
Report Question
0%
-2
0%
1
0%
2
0%
3
Explanation
(
α
+
1
)
x
2
−
2
(
1
+
3
α
)
x
+
1
+
8
α
=
0
Here,
a
=
α
+
1
b
=
−
2
(
1
+
3
α
)
c
=
1
+
8
α
If both roots are real and distinct,
D
>
0
⇒
b
2
−
4
a
c
>
0
⇒
(
−
2
(
1
+
3
α
)
)
2
−
4
(
1
+
α
)
(
1
+
8
α
)
>
0
⇒
4
(
1
+
9
α
2
+
6
α
)
−
4
(
1
+
9
α
+
8
α
2
)
>
0
⇒
4
(
1
+
9
α
2
+
6
α
−
1
−
9
α
−
8
α
2
)
>
0
⇒
α
2
−
3
α
>
0
⇒
α
(
α
−
3
)
>
0
⇒
α
∈
(
−
∞
,
0
)
∪
(
3
,
∞
)
Hence among the given values,
α
will be
−
2
.
Hence the correct answer is
(
A
)
−
2
.
Identify the standard form of the given equation
(
4
x
−
5
)
(
4
x
+
5
)
=
0
.
Report Question
0%
16
x
2
−
25
=
0
0%
16
x
3
−
25
=
0
0%
16
x
2
−
2
=
0
0%
6
x
2
−
25
=
0
Explanation
(
4
x
+
5
)
(
4
x
−
5
)
=
0
4
x
(
4
x
−
5
)
+
5
(
4
x
−
5
)
=
0
16
x
2
−
20
x
+
20
x
−
25
=
0
16
x
2
−
25
=
0
which is in standard form of quadratic equation
If
x
2
−
5
x
+
1
=
0
, then
x
10
+
1
x
5
has the value
Report Question
0%
2524
0%
2525
0%
2424
0%
2010
Explanation
x
2
−
5
x
+
1
=
0
∴
x
2
+
1
=
5
x
∴
x
+
1
x
=
5
∴
x
2
+
1
x
2
=
25
−
2
=
23
∴
x
3
+
1
x
3
=
(
x
+
1
x
)
3
−
3
(
x
+
1
x
)
∴
x
3
+
1
x
3
=
125
−
15
∴
x
3
+
1
x
3
=
110
Now
(
x
3
+
1
x
2
)
(
x
3
+
1
x
3
)
=
56
×
23
∴
x
5
+
x
+
1
x
+
1
x
5
=
110
×
23
∴
x
5
+
1
x
5
=
2530
−
5
∴
x
5
+
1
x
5
=
2525
.
If
a
,
b
,
c
,
x
are real numbers and
(
a
2
+
b
2
)
x
2
−
2
b
(
a
+
c
)
x
+
(
b
2
+
c
2
)
=
0
has real & equal roots, then
a
,
b
,
c
are in
Report Question
0%
A
.
P
0%
G
.
P
0%
H
.
P
0%
N
o
n
e
o
f
t
h
e
s
e
Explanation
Since
(
a
2
+
b
2
)
x
2
−
2
b
(
a
+
c
)
x
+
(
b
2
+
c
2
)
=
0
has real & equal roots.
∴
4
b
2
(
a
2
+
c
2
+
2
a
c
)
−
4
(
a
2
+
b
2
)
(
b
2
+
c
2
)
=
0
4
b
2
a
2
+
4
b
2
c
2
+
8
a
c
b
2
−
4
a
2
b
2
−
4
a
2
c
2
−
4
b
2
−
4
b
2
c
2
=
0
8
a
c
b
2
−
4
a
2
c
2
+
4
b
4
=
0
(
2
a
c
−
2
b
2
)
2
=
0
a
c
=
b
2
The discriminant of the quadratic equation
5
x
2
−
6
x
+
1
=
0
is
Report Question
0%
16
0%
√
56
0%
4
0%
56
Explanation
The given quadratic equation is
5
x
2
−
6
x
+
1
=
0
is in the form of
a
x
2
+
b
x
+
c
by comparing
a
=
5
,
b
=
−
6
,
c
=
1
Now, Discriminant
D
=
b
2
−
4
a
c
D
=
36
−
4
×
5
×
1
D
=
36
−
20
=
16
Let
λ
1
and
λ
2
be two values of
λ
for which the expression
x
2
+
(
2
−
λ
)
x
+
λ
−
3
becomes a perfect square. The value of
(
λ
2
1
+
λ
2
2
)
equals
Report Question
0%
32
0%
25
0%
50
0%
100
If
α
,
β
are the roots of
a
x
2
+
b
x
+
c
and
α
+
h
,
β
+
h
are the roots of
p
x
2
+
q
x
+
r
=
0
and
D
1
,
D
2
are the respective
discriminants of these equations,
then
D
1
:
D
2
=
Report Question
0%
a
2
p
2
0%
b
2
q
2
0%
c
2
r
2
0%
1
Explanation
Let
A
=
α
+
h
and
B
=
β
+
h
Then,
A
−
B
=
(
α
+
h
)
−
(
β
+
h
)
⇒
A
−
B
=
α
−
β
⇒
(
A
−
B
)
2
=
(
α
−
β
)
2
⇒
(
A
+
B
)
2
−
4
A
B
=
(
α
+
β
)
2
−
4
α
β
⇒
b
2
a
2
−
4
c
a
=
q
2
p
2
−
4
r
p
⇒
b
2
−
4
a
c
a
2
=
q
2
−
4
r
p
p
2
⇒
D
1
a
2
=
D
2
p
2
⇒
D
1
D
2
=
a
2
p
2
Hence
D
1
:
D
2
=
a
2
:
p
2
=
a
2
p
2
Find the root of the quadratic equation
x
2
+
2
√
2
x
+
6
=
0
by using the quadratic formula
Report Question
0%
x
=
√
2
±
2
i
0%
x
=
−
√
2
±
2
i
0%
x
=
−
√
4
±
2
i
0%
x
=
−
√
2
±
4
i
Explanation
According to the given quadratic equation,
a
=
1
,
b
=
2
√
2
and
c
=
6
So, by quadratic formula,
x
=
−
b
±
√
b
2
−
4
a
c
2
a
=
−
2
√
2
±
√
(
2
√
2
)
2
−
4
×
1
×
6
2
×
1
=
−
2
√
2
±
√
8
−
24
2
=
−
2
√
2
±
√
−
16
2
=
−
2
√
2
±
4
i
2
Thus,
x
=
−
√
2
±
2
i
Let f (x) =
x
2
+
λ
x
+
μ
c
o
s
x
,
λ
is +ve integer
μ
is a real number. The number of ordered pairs (
λ
,
μ
) for which f (x) = 0 and f (f(c)) = 0 have same set of real roots.
Report Question
0%
0
0%
1
0%
2
0%
3
If
√
1296
x
=
x
2.25
then value
x
is:
Report Question
0%
0.9
0%
0.09
0%
9
0%
1.9
Explanation
we know that,
√
1296
=
36
36
x
=
x
2.25
x
2
=
36
×
2.25
x
2
=
81
x
=
±
√
81
x
=
±
9
Find the value of
x
:
x
2
−
4
x
+
4
=
0
Report Question
0%
2
0%
1
0%
0
0%
None of the above
Explanation
Given equation is:
x
2
−
4
x
+
4
=
0
It can also be written as:
x
2
−
2
(
2
)
(
x
)
+
(
2
)
2
=
0
…
(
1
)
Above equation is of the form of following identity:
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
Hence, the equation
(
1
)
can be written as follows:
(
x
−
2
)
2
=
0
or
(
x
−
2
)
(
x
−
2
)
=
0
x
=
2
,
2
If
k
>
0
and the product of roots of the equations
x
2
−
3
k
x
+
2
e
log
k
−
1
=
0
is
7
, then find the sum of the roots-
Report Question
0%
6
0%
4
0%
3
0%
−
12
Explanation
Given:
x
2
−
3
k
x
+
2.
e
2
log
k
−
1
=
0
⇒
x
2
−
3
k
x
+
2
e
log
k
2
−
1
=
0
⇒
x
2
−
3
k
x
+
2
e
log
k
2
−
1
=
0
⇒
x
2
−
3
k
x
+
(
2
k
2
−
1
)
=
0
Let
α
and
β
be the roots
Hence
α
+
β
=
3
k
and
α
β
=
2
k
2
−
1
=
7
R
i
g
h
t
a
r
r
o
w
2
k
2
=
8
=
4
⇒
k
=
±
2
⇒
k
=
2
So,
k
>
0
Now,
α
+
β
=
3
k
=
3
(
2
)
=
6
If
x
2
−
5
x
+
1
=
0
then
x
10
+
1
x
5
is equal to:-
Report Question
0%
2424
0%
3232
0%
2525
0%
None of these
(
x
+
1
)
2
+
(
y
−
1
)
2
+
(
x
−
5
)
2
+
(
y
−
1
)
2
has the value equal to
Report Question
0%
16
0%
25
0%
36
0%
49
The ratio of the roots of the equation
a
x
2
+
b
x
+
c
=
0
is same as the ratio of the roots of equation
p
x
2
+
q
x
+
r
=
0
. If
D
1
and
D
2
are the discriminants of
a
x
2
+
b
x
+
c
=
0
and
p
x
2
+
q
x
+
r
=
0
respectively, then
D
1
:
D
2
=
Report Question
0%
b
2
q
2
0%
b
q
0%
b
q
2
0%
none of these
Explanation
Step 1: Using quadratic formula,
x
=
−
b
±
√
D
2
a
a
x
2
+
b
x
+
c
=
0
x
=
−
b
±
√
D
1
2
a
[
Using Quadratic Formula
]
p
x
2
+
q
x
+
r
=
0
x
=
−
q
±
√
D
2
2
a
[
Using Quadratic Formula
]
As per given question, We have
−
b
+
√
D
1
−
b
−
√
D
1
=
−
q
+
√
D
2
−
q
−
√
D
2
Step 2: Using Componendo and Dividendo, we get
−
√
D
1
b
=
−
√
D
2
q
On squaring both side
D
1
b
2
=
D
2
q
2
⇒
D
1
D
2
=
b
2
q
2
Which of the following is a quadratic equation ?
Report Question
0%
6
x
2
=
20
−
x
3
0%
x
2
−
(
1
x
2
)
=
7
2
0%
3
x
=
4
x
2
0%
5
x
2
+
7
=
3
x
Explanation
For
(
A
)
6
x
2
=
20
−
x
3
⇒
x
3
+
6
x
2
−
20
=
0
In this, the maximum power of
x
is
3
.
So, it is not a quadratic equation.
For
(
B
)
x
2
−
(
1
x
2
)
=
7
2
⇒
2
x
4
−
2
=
7
x
2
⇒
2
x
4
−
7
x
2
−
2
=
0
In this, the maximum power of
x
is
4
.
So, it is not a quadratic equation.
For
(
C
)
3
x
=
4
x
2
⇒
3
=
4
x
3
In this, the maximum power of
x
is
3
.
So, it is not a quadratic equation.
For
(
D
)
5
x
2
+
7
=
3
x
⇒
5
x
2
−
3
x
+
7
=
0
In this, the maximum power of
x
is
2
.
So, it is a quadratic equation.
Hence,
O
p
−
D
is correct.
Solve the quadratic equation by completing the square method:
x
2
+
8
x
−
9
=
0
Report Question
0%
−
9
,
−
1
0%
2
,
1
0%
−
9
,
1
0%
9
,
1
Explanation
x
2
+
8
x
−
9
=
0
⇒
x
2
+
2
×
(
4
)
×
x
+
16
−
25
=
0
(
x
+
4
)
2
=
25
(
x
+
4
)
2
=
5
2
x
+
4
=
±
5
⇒
x
=
1
and
−
9
Hence, the answer is
1
,
−
9.
Discriminant is ................. for the equation
5
x
−
6
=
−
1
x
Report Question
0%
-56
0%
16
0%
-16
0%
0
If
(
1
+
m
2
)
x
2
−
2
(
1
+
3
m
)
x
+
(
1
+
8
m
)
=
0
. Then numbers of value(s) of m for which this equation has no solution.
Report Question
0%
3
0%
infinitely many
0%
2
0%
1
Explanation
We know that
D
=
b
2
−
4
a
c
D
=
4
(
1
+
3
m
)
2
−
4
(
1
+
m
2
)
(
1
+
8
m
)
=
4
(
1
+
9
m
2
+
6
m
−
(
1
+
8
m
+
m
2
+
8
m
3
)
)
=
4
(
8
m
2
−
2
m
−
8
m
3
)
=
−
8
(
4
m
3
−
4
m
2
+
m
)
=
−
8
m
(
4
m
2
−
4
m
+
1
)
=
−
8
m
(
2
m
−
1
)
2
<
0
.........
∵
(
2
m
−
1
)
2
i
s
a
l
w
a
y
s
+
v
e
=
m
>
0
Hence infinitely many values.
Quadratic equation among the following is:
Report Question
0%
x
2
+
1
=
x
2
+
2
x
0%
x
(
x
+
1
)
=
x
(
x
−
3
)
0%
x
2
−
2
x
=
3
(
5
x
+
2
)
0%
x
2
+
2
x
+
3
=
x
2
+
5
x
For what value of
c
, the root of
(
c
−
2
)
x
2
+
2
(
c
−
2
)
x
+
2
=
0
are not real
Report Question
0%
(
1
,
2
)
0%
(
2
,
3
)
0%
(
3
,
4
)
0%
(
2
,
4
)
Explanation
Given the quadratic equation is
(
c
−
2
)
x
2
+
2
(
c
−
2
)
x
+
2
=
0
.
Now the roots of the equation will be not real if the discriminant of the equation is less than
0
.
Then,
{
2
(
c
−
2
)
}
2
−
4.2.
(
c
−
2
)
<
0
or,
4
{
(
c
−
2
)
(
c
−
4
)
}
<
0
or,
(
c
−
2
)
(
c
−
4
)
<
0
or,
2
<
c
<
4
.
So
c
∈
(
2
,
4
)
.
Which constant must be added and subtracted to solve the quadratic equation
9
x
2
+
3
4
x
−
√
2
=
0
by the method of completing the square ?
Report Question
0%
1
8
0%
1
64
0%
1
4
0%
9
64
Explanation
Given equation is
9
x
2
+
3
4
x
−
√
2
=
0
⇒
(
3
x
)
2
+
2
(
1
8
)
(
3
x
)
−
√
2
=
0
Hence, to make it a perfect square we must add and subtract
(
1
8
)
2
=
1
64
If
α
and
β
are roots of the equation
x
2
+
x
sin
θ
−
2
sin
θ
=
0
then
α
12
+
β
12
(
α
−
12
+
β
−
12
)
(
α
−
β
)
24
is equal to
Report Question
0%
2
24
(
8
+
sin
θ
)
12
0%
2
12
(
8
+
sin
θ
)
12
0%
2
12
(
8
−
sin
θ
)
12
0%
1
2
24
Explanation
x
2
+
x
sin
θ
−
2
sin
θ
=
0
, has roots
α
and
β
α
12
+
β
12
(
α
−
12
+
β
−
12
)
(
α
−
β
)
24
=
α
12
.
β
12
(
α
−
β
)
24
=
(
α
β
)
12
(
α
−
β
)
24
=
(
−
2
sin
θ
)
12
(
√
sin
2
θ
+
8
sin
θ
)
24
=
2
12
(
8
+
sin
θ
)
12
If the discriminant of
3
x
2
−
2
x
+
K
=
0
is zero then
K
=
_____
Report Question
0%
3
0%
1
3
0%
−
3
0%
−
1
3
Which constant must be added and subtracted to solve the Quadratic equation
9
x
2
+
3
4
x
−
√
2
by the method of
completion the square.
Report Question
0%
1
8
0%
1
4
0%
1
64
0%
9
64
0:0:1
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Practice Class 10 Maths Quiz Questions and Answers
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