Exercise 4.1 (11)
Question: 1
Choose the correct answer from the given four
options in the following questions:
Which of the following is a quadratic equation?
a.
2
2
x 2x 1 4 x 3
b.
2
2
2x 5 x 2x
5
c. , where
2
3
k 1 x x 7
2
k 1
d.
3
3 2
x x x 1
Solution:
(d)
(a) Given:
2
2
x 2x 1 4 x 3
Simplify the equation,
2
2
x 2x 1 4 x 3
2
2 2
x 2x 1 4 x 2 4 x 3
2 2
x 2x 1 16 x 8x 3
2x 1 8x 19
This is not in the form of .
Hence, it is not a quadratic equation.
(b) Given:
2
2x 5 x
2
2x
5
Simplify the equation,
2
2
2x 5 x 2x
5
2 2
2
2x 10x 2 2x x
5
2
10x 2 x 0
5
This is not in the form of .
Hence, it is not a quadratic equation.
(c) Given:
2
3
k 1 x x 7
2
Also,
k 1
Simplify the equation,
2
3
k 1 x x 7
2
2
3
1 1 x x 7
2
3
x 7
2
This is not in the form of .
Hence, it is not a quadratic equation.
(d) Given:
3
3 2
x x x 1
Simplify the equation,
3
3 2
x x x 1
3 3 2 2
3 2
x x x 1 3 x 1 3 x 1
3 2 3 2
x x x 1 3x 3x
2 2
x 1 3x 3x
2 2
x 1 3x 3x 0
2
2x 3x 1 0
This is in the form of .
Hence, it is a quadratic equation.
Question: 2
Which of the following is not a quadratic equation?
a.
2
2
2 x 1 4x 2x 1
b.
2 2
2x x x 5
c.
2
2 2
2x 3 x 3x 5x
d.
2
2 4 3
x 2x x 3 4x
Solution:
(c)
a. Given:
2
2
2 x 1 4x 2x 1
Simplify the equation,
2
2x 3x 1 0
2 2
2
2 x 1 2 x 1 4x 2x 1 0
2 2
2x 2 4x 4x 2x 1 0
2
2x 1 2x 0
This is in the form of .
Hence, it is a quadratic equation.
b. Given:
2 2
2x x x 5
Simplify the equation,
2 2
2x x x 5
2 2
2x x x 5 0
2
2x 2x 5 0
2
2x 2x 5 0
This is in the form of .
Hence, it is a quadratic equation
c. Given:
2
2 2
2x 3 x 3x 5x
Simplify the equation,
2
2 2
2x 3 x 3x 5x
2 2
2 2
2x 3 2 2x 3 x 3x 5x 0
2 2 2
2x 3 2 6x x 3x 5x 0
This is not in the form of .
Hence, it is not a quadratic equation.
d. Given:
2
2 4 3
x 2x x 3 4x
Simplify the equation,
2
2 4 3
x 2x x 3 4x
2
2
2 2 4 3
x 2x 2 x 2x x 3 4x
4 2 3 4 3
x 4x 4x x 3 4x
2
4x 3 0
This is in the form of .
Hence, it is a quadratic equation.
Question: 3
Which of the following equations has as a root?
2
a.
2
x 4x 5 0
b.
2
x 3x 12 0
c.
2
2x 7x 6 0
d.
2
3x 6x 2 0
Solution:
(c)
a. Given:
2
x 4x 5 0
Substitute in the given equation,
x 2
2
x 4x 5 0
2
2 4 2 5 0
4 8 5 0
9 8 0
1 0
Which is false.
Thus, does not satisfy the given equation.
x 2
Hence, is not the root of the given equation.
2
b. Given:
2
x 3x 12 0
Substitute in the given equation,
x 2
2
x 3x 12 0
2
2 3 2 12 0
4 6 12 0
10 12 0
2 0
Which is false.
Thus, does not satisfy the given equation.
x 2
Hence, is not the root of the given equation.
2
c. Given:
2
2x 7x 6 0
Substitute in the given equation,
x 2
2
2x 7x 6 0
2
2 2 7 2 6 0
8 14 6 0
0 0
Which is true.
Thus, satisfies the given equation.
x 2
Hence, is the root of the given equation.
2
d. Given:
2
3x 6x 2 0
Substitute in the given equation,
x 2
2
3x 6x 2 0
2
3 2 6 2 2 0
12 12 2 0
2 0
Which is false.
Thus, does not satisfy the given equation.
x 2
Hence, is not the root of the given equation.
2
Question: 4
If is a root of the equation , then
1
2
2
5
x kx 0
4
the value of is
k
a.
2
b.
2
c.
1
4
d.
1
2
Solution:
(a)
Given: is a root of the equation ,
1
2
2
5
x kx 0
4
So, satisfies the given equation.
1
x
2
2
5
x kx 0
4
2
1 1 5
k 0
2 2 4
 
1 k 5
0
4 2 4
 
1 2k 5 0
2k 4
k 2
Thus, the value of is .
k
2
Question: 5
Which of the following equations has the sum of its
roots as ?
3
a.
2
2x 3x 6 0
b.
2
x 3x 3 0
c.
2
3
2x x 1 0
2
d.
2
3x 3x 3 0
Solution:
(b)
a. Given:
2
2x 3x 6 0
Now, find the sum of roots.
3
3
2
Thus, the sum of roots of the equation is not .
3
b. Given:
2
x 3x 3 0
Now, find the sum of roots.
3
3
1
Thus, the sum of roots of the equation is .
3
c. Given:
2
3
2x x 1 0
2
Now, find the sum of roots.
3
3 3
2
3
2
2 2 2
Thus, the sum of roots of the equation is not .
3
d. Given:
2
3x 3x 3 0
Now, find the sum of roots.
3
1 3
3
Thus, the sum of roots of the equation is not .
3
Question: 6
Values of for which the quadratic equation
k
has equal roots is
a. only
0
b.
4
c. only
8
d.
0,8
Solution:
Given:
Condition for equal roots,
2
b 4ac 0
2
b 4ac 0
2
k 4 2 k 0
2
k 8k 0
k k 8 0
That is,
or
k 0
k 8 0
or
k 0
k 8
Thus, the value of are and .
k
0
8
Question: 7
Which constant must be added and subtracted to
2
3
9x x 2 0
4
the method of completing the square?
a.
1
8
b.
1
64
c.
1
4
d.
9
64
Solution:
(a)
Given:
2
3
9x x 2 0
4
By the completing square method, add and subtract
.
2
1
8
2
2
1 3 1
9x x 2 0
8 4 64
2
2
1 3 1
3x x 2
8 4 64
2
1 1
3x 2
8 64
Thus, the required solution.
Question: 8
2
2x 5x 1 0
a. two distinct real roots
b. two equal real roots
c. no real roots
d. more than real roots
2
Solution:
(c)
Given:
2
2x 5x 1 0
Now, find the roots.
2
D b 4ac
2
5 4 2 1
5 8
3
So, .
D 0
This equation has no real roots.
Question: 9
Which of the following equations has two distinct
real roots?
a.
2
9
2x 3 2x 0
4
b.
2
x x 5 0
c.
2
x 3x 2 2 0
d.
2
5x 3x 1 0
Solution:
(b)
a. Given:
2
9
2x 3 2x 0
4
Now, find the roots.
2
D b 4ac
2
9
3 2 4 2
4
18 18
0
So, .
D 0
This equation has two real and equal roots.
b. Given:
2
x x 5 0
Now, find the roots.
2
D b 4ac
2
1 4 1 5
1 20
21
So, .
D 0
This equation has two real and distinct roots.
c. Given:
2
x 3x 2 2 0
Now, find the roots.
2
D b 4ac
2
3 4 1 2 2
9 8 2
9 8 1 414.
9 11.312
122.3
So, .
D 0
This equation has no real roots.
d. Given:
2
5x 3x 1 0
Now, find the roots.
2
D b 4ac
2
3 4 5 1
9 20
11
So, .
D 0
This equation has no real roots.
Question: 10
Which of the following equations has no real roots?
a.
2
x 4x 3 2 0
b.
2
x 4x 3 2 0
c.
2
x 4x 3 2 0
d.
2
3x 4 3x 4 0
Solution:
(a)
a. Given:
2
x 4x 3 2 0
Now, find the roots.
2
D b 4ac
2
4 4 1 3 2
16 12 2
16 12 1 414.
16 16 68.9
680.9
So, .
D 0
This equation has no real roots.
b. Given:
2
x 4x 3 2 0
Now, find the roots.
2
D b 4ac
2
4 4 1 3 2
16 12 2
16 12 1 414.
16 16 68.9
832.96
So, .
D 0
This equation has two real and distinct roots.
c. Given:
2
x 4x 3 2 0
Now, find the roots.
2
D b 4ac
2
4 4 1 3 2
16 12 2
16 12 1 414.
16 16 68.9
832.96
So, .
D 0
This equation has two real and distinct roots.
d. Given:
2
3x 4 3x 4 0
Now, find the roots.
2
D b 4ac
2
4 3 4 3 4
16 3 48
48 48
0
So, .
D 0
This equation has two real and equal roots.
Question: 11
has
2
2 2
x 1 x 0
a. four real roots
b. two real roots
c. no real roots
d. one real root
Solution:
(c)
Given:
2
2 2
x 1 x 0
Simplify the equation,
2
2 2
x 1 x 0
2
2
2 2 2
x 1 2x x 0
2
2 2
x 1 x 0
Let
2
x y
2
y 1 y 0
Now, find the roots.
2
D b 4ac
2
1 4 1 1
1 4
3
So,
D 0
The value of is not real so, the value of will not
y
2
x
be real.
Thus, the given equation has no real roots.
Exercise 4.2 (7)
Question: 1
State whether the following quadratic equations
(i)
2
x 3x 4 0
(ii)
2
2x x 1 0
(iii)
2
9
2x 6x 0
2
 
(iv)
2
3x 4x 1 0
(v)
2
x 4 8x 0
(vi)
2
x 2 2 x 1 0
(vii)
2
3 1
2x x 0
2 2
(viii)
x 1 x 2 0
(ix)
x 1 x 2 2 0
(x)
x 1 x 2 x 0
Solution:
(i) Given:
2
x 3x 4 0
Now, find the roots.
2
D b 4ac
2
3 4 1 4
9 16
7
So, .
D 0
This equation has no real roots.
(ii) Given:
2
2x x 1 0
Now, find the roots.
2
D b 4ac
2
1 4 2 1
1 8
9
So, .
D 0
This equation has two real and distinct roots.
(iii) Given:
2
9
2x 6x 0
2
 
Now, find the roots.
2
D b 4ac
2
9
6 4 2
2
36 36
0
So, .
D 0
This equation has two real and equal roots.
(iv) Given:
2
3x 4x 1 0
Now, find the roots.
2
D b 4ac
2
4 4 3 1
16 12
4
So, .
D 0
This equation has two real and distinct roots.
(v) Given:
2
x 4 8x 0
Simplify the equation.
2
x 4 8x 0
2 2
x 4 2 x 4 8x 0
2
x 16 8x 8x 0
2
x 16 0
Now, find the roots.
2
D b 4ac
 
2
0 4 1 16
0 64
64
So, .
D 0
This equation has no real roots.
(vi) Given:
2
x 2 2 x 1 0
Simplify the equation.
2
x 2 2 x 1 0
2
x 2 2 x 1 0
2
x 2 2 2x 2x 2 0
2
x 2 2 2 x 0
Now, find the roots.
2
D b 4ac
2
2 2 2 4 1 0
2
2 2 2 0
2
2 2 2
So, .
D 0
This equation has real and distinct roots.
(vii) Given:
2
3 1
2x x 0
2 2
Now, find the roots.
2
D b 4ac
2
3 1
4 2
2 2
9 4
2 1
9 8
2
1
2
So, .
D 0
This equation has two real and distinct roots.
(viii) Given:
x 1 x 2 0
Simplify the equation.
x 1 x 2 0
2
x x 2 0
2
x x 2 0
Now, find the roots.
2
D b 4ac
2
1 4 1 2
1 8
7
So, .
D 0
This equation has no real roots.
(ix) Given:
x 1 x 2 2 0
Simplify the equation.
x 1 x 2 2 0
2
x 2x x 2 2 0
2
x x 0
Now, find the roots.
2
D b 4ac
2
1 4 1 0
1 0
1
So, .
D 0
This equation has two real and distinct roots.
(x) Given:
x 1 x 2 x 0
Simplify the equation.
x 1 x 2 x 0
2
x 2x x 2 x 0
2
x 2 0
2
x 0x 2 0
Now, find the roots.
2
x 0x 2 0
2
0 4 1 2
0 8
8
So, .
D 0
This equation has two real and distinct roots.
Question: 2
Write whether the following statements are true or
(i) Every quadratic equation has exactly one root.
(ii) Every quadratic equation has at least one real
root.
(iii) Every quadratic equation has at least two
roots.
(iv) Every quadratic equation has at the most two
roots.
(v) If the coefficient of and the constant term of
2
x
a quadratic equation have opposite signs, then
the quadratic equation has real roots.
(vi) If the coefficient of and the constant term
2
x
have the same sign and if the coefficient of
x
term is zero, then the quadratic equation has
no real roots.
Solution:
(i) False: Let the quadratic equation
2
x 4 0
which has two distinct roots and .
2
2
Thus, the given statement is false.
(ii) False: Let the quadratic equation
2
x 1 0
which has no real root.
Thus, the given statement is false.
(iii) False: Let the quadratic equation
2
x 4x 4 0
which has only as root.
2
Thus, the given statement is false.
(iv) True: Let the quadratic equation
2
x 5x 6 0
which has and only two roots. So, any
2
3
quadratic equation can have at the most two
roots.
(v) True: In quadratic equation , if
and have opposite signs, then .
a
c
ac 0
2
b 4ac 0
equation has real roots.
(vi) True: In quadratic equation , if
and have same signs and , then
a
c
b 0
2
2
b 4ac 0 4ac 4ac 0
So, the quadratic equation has no real roots.
Question: 3
A quadratic equation with integral coefficient has
Solution:
No, in a quadratic equation with integral coefficient
can have roots in non-integral,
0, 1, 2, 3,
i.e., fraction.
For example, has integral coefficient.
2
5x 3x 8 0
Now, simplify the equation.
2
5x 3x 8 0
2
5x 8x 5x 8 0
x 5x 8 1 5x 8 0
5x 8 x 1 0
That is,
or
5x 8 0
x 1 0
or
8
x
5
x 1
Roots are .
8
,1
5
Thus, the given statement is false.
Question: 4
Does there exist a quadratic equation whose
coefficients are rational but both its roots are
Solution:
Yes, a quadratic equation, whose coefficients are
rational, has irrational roots.
For example, has rational
2
2x 3x 15 0
coefficients.
2
D b 4ac
2
3 4 2 15
9 120
129
Roots
b D
x
2a
3 129
2 2
3 129
4
The roots are irrational as is irrational.
129
Question: 5
Does there exist a quadratic equation whose
coefficients are all distinct irrationals but both the
roots are rationals? Why?
Solution:
Yes,there may be a quadratic equation whose
coefficients are all distinct irrationals, but both the
roots are rational.
For example: Let us consider a quadratic equation
with distinct irrational coefficient,
2
3 5 2
3 x x 2 0
2 3
6
Now, .
2
D b 4ac
2
D b 4ac
2
5 3 3 2 2
4
6 2 3
25 24
6 1
25 144
6
169
6
13
D
6
Now, find the roots.
b D
x
2a
5 13
6 6
3 3
2
2
1
5 13 2
6
6 3
5 13 2
6 3 2 3
5 13
18
or
8
x
18
18
18
or
4
x
9
1
Thus, the roots are rational while coefficients
a,b,c
are irrational.
Question: 6
Is a root of the equation ? Justify.
0.2
2
x 00.4
Solution:
If is a root of equation , then
0.2
x 4 00.
0.2
must satisfy the given equation.
2
x 0 0.4
2
0 0.2 .4 0
0.04 0.4 0
0.36 0
Thus, is not a root of the given equation.
0.2
Question: 7
If , is it true that the roots of
b 0,c 0
are numerically equal and opposite in
2
x bx c 0
sign? Justify.
Solution:
Given:
2
x bx c 0
[Given]
b 0
2
x c 0
2
x c
x c
As is negative, becomes positive or is
c
c
c
real.
So, the roots of the given equation.
x c
or and
x c
c
Thus, the roots of the given equation are real, equal
and opposite in sign.
Exercise 4.3 (2)
Question: 1
Find the roots of the quadratic equations by using
the quadratic formula in each of the following:
(i)
2
2x 3x 5 0
(ii)
2
5x 13x 8 0
(iii)
2
3x 5x 12 0
(iv)
2
x 7x 10 0
(v)
2
x 2 2x 6 0
(vi)
2
x 3 5x 10 0
(vii)
2
1
x 11x 1 0
2
Solution:
(i) Given:
2
2x 3x 5 0
Now, find the discriminant.
2
D b 4ac
2
3 4 2 5
9 40
49
Now, find the roots.
b D
x
2a
3 49
2 2
3 7
4
That is,
3 7
,
4
3 7
4
10
,
4
4
4
5
,
2
1
Thus, the roots of the given equations are
5
2
and .
1
(ii) Given:
2
5x 13x 8 0
Now, find the discriminant.
2
D b 4ac
 
2
13 4 5 8
169 160
9
Now, find the roots.
b D
x
2a
 
13 9
2 5
13 3
10
That is,
13 3
,
10
13 3
10
10
,
10
16
10
1,
8
5
Thus, the roots of the given equations are
1
and .
8
5
(iii) Given:
2
3x 5x 12 0
Now, find the discriminant.
2
D b 4ac
2
5 4 3 12
25 144
169
Now, find the roots.
b D
x
2a
5 169
2 3
5 13
6
That is,
5 13
,
6
5 13
6
8
,
6
18
6
4
,
3
3
Thus, the roots of the given equations are
4
3
and .
3
(iv) Given:
2
x 7x 10 0
Now, find the discriminant.
2
D b 4ac
2
7 4 1 10
49 40
9
Now, find the roots.
b D
x
2a
7 9
2 1
7 3
2
That is,
7 3
,
2
7 3
2
4
,
2
10
2
2,
5
Thus, the roots of the given equations are
2
and .
5
(v) Given:
2
x 2 2x 6 0
Now, find the discriminant.
2
D b 4ac
2
2 2 4 1 6
8 24
32
Now, find the roots.
b D
x
2a
2 2 32
2 1
2 2 4 2
2 1
That is,
2 2 4 2
,
2
2 2 4 2
2
2 2
,
2
6 2
2
2,
3 2
Thus, the roots of the given equations are
2
and .
3 2
(vi) Given:
2
x 3 5x 10 0
Now, find the discriminant.
2
D b 4ac
2
3 5 4 1 10
45 40
5
Now, find the roots.
b D
x
2a
3 5 5
2 1
3 5 5
2
That is,
3 5 5
,
2
3 5 5
2
4 5
,
2
2 5
2
2 5,
5
Thus, the roots of the given equations are
2 5
and .
5
(vii) Given:
2
1
x 11x 1 0
2
Now, find the discriminant.
2
D b 4ac
2
1
11 4 1
2
11 2
9
Now, find the roots.
b D
x
2a
11 9
1
2
2
11 3
That is,
11 3,
11 3
Thus, the roots of the given equations are
and .
11 3
11 3
Question: 2
Find the roots of the following quadratic equations
by the factorisation method:
(i)
2
5
2x x 2 0
3
(ii)
2
2 3
x x 0
5 5
(iii)
2
3 2x 5x 2 0
(iv)
2
3x 5 5x 10 0
(v)
2
1
21x 2x 0
21
Solution:
(i) Given:
2
5
2x x 2 0
3
We can rewrite it as,
2
6x 5x 6 0
Split the middle term to find out factors of the
equation.
2
6x 5x 6 0
2
6x 9x 4x 6 0
Take the common terms outside the
parenthesis.
3x 2x 3 2 2x 3 0
2x 3 3x 2 0
That is,
or
2x 3 0
3x 2 0
or
2x 3
3x 2
or
3
x
2
2
x
3
Thus, and are the roots of the equation.
3
2
2
3
(ii) Given:
2
2 3
x x 0
5 5
We can rewrite it as,
2
2x 5x 3 0
Split the middle term to find out factors of the
equation.
2
2x 5x 3 0
2
2x 6x 1x 3 0
Take the common terms outside the
parenthesis.
2x x 3 1 x 3 0
2x 1 x 3 0
That is,
or
2x 1 0
x 3 0
or
2x 1
x 3
or
1
x
2
x 3
Thus, are the roots of the equation.
1
and3
2
(iii) Given:
2
3 2x 5x 2 0
Split the middle term to find out factors of the
equation.
2
3 2x 5x 2 0
2
3 2x 6x 1x 2 0
Take the common terms outside the
parenthesis.
3 2x x 2 1 x 2 0
3 2x 1 x 2 0
That is,
or
3 2x 1 0
x 2 0
or
3 2x 1
x 2
or
1 2
x
3 2 2
x 2
or
2
x
6
x 2
Thus, and are the roots of the
2
6
2
equation.
(iv) Given:
2
3x 5 5x 10 0
Split the middle term to find out factors of the
equation.
2
3x 5 5x 10 0
2
3x 6 5x 5x 10 0
Take the common terms outside the
parenthesis.
3x x 2 5 5 x 2 5 0
x 2 5 3x 5 0
That is,
or
x 2 5 0
3x 5 0
or
x 2 5
3x 5
or
x 2 5
5
x
3
Thus, and are the roots of the
5
3
2 5
equation.
(v) Given:
2
1
21x 2x 0
21
We can rewrite it as,
2
441x 42x 1 0
Split the middle term to find out factors of the
equation.
2
441x 42x 1 0
2
441x 21x 21x 1 0
Take the common terms outside the
parenthesis.
21x 21x 1 1 21x 1 0
21x 1 21x 1 0
That is,
or
21x 1 0
21x 1 0
or
21x 1
21x 1
or
1
x
21
1
x
21
Thus, and are the roots of the equation.
1
21
1
21
Exercise 4.4 (8)
Question: 1
Find whether the following equations have real
roots. If real roots exist, find them.
(i)
2
8x 2x 3 0
(ii)
2
2x 3x 2 0
(iii)
2
5x 2x 10 0
(iv)
1 1 3
1, x ,5
2x 3 x 5 2
(v)
2
x 5 5x 70 0
Solution:
(i) Given:
2
8x 2x 3 0
Now, find the discriminant.
2
D b 4ac
2
2 4 8 3
4 96
100
So, .
D 0
Thus, the roots are real.
Now, find the roots.
b D
x
2a
2 100
2 8
2 10
16
That is,
2 10
,
16
2 10
16
8
,
16
12
16
1
,
2
3
4
Thus, the roots of the given equations are
1
2
and .
3
4
(ii) Given:
2
2x 3x 2 0
Now, find the discriminant.
2
D b 4ac
2
3 4 2 2
9 16
25
So, .
D 0
Thus, the equation has real and distinct roots.
Now, find the roots.
b D
x
2a
3 25
2 2
3 5
4
That is,
3 5
,
4
3 5
4
2
,
4
8
4
1
,
2
2
Thus, the roots of the given equations are
1
2
and .
2
(iii) Given:
2
5x 2x 10 0
Now, find the discriminant.
2
D b 4ac
2
2 4 5 10
4 200
204
So, .
D 0
Thus, the equation are reals and distinct roots.
Now, find the roots.
b D
x
2a
2 204
2 5
2 2 51
10
1 51
5
That is,
1 51
,
5
1 51
5
Thus, the roots of the given equations are
and .
1 51
5
1 51
5
(iv) Given:
1 1 3
1, x ,5
2x 3 x 5 2
We can rewrite it as,
1 1
1
2x 3 x 5
1 1
1
2x 3 x 5
x 5 2x 3
1
2x 3 x 5
Cross-multiply both the sides,
x 5 2x 3 2x 3 x 5
2
x 5 2x 3 2x 10x 3x 15
2
3x 8 2x 13x 15
2
2x 13x 15 3x 8 0
2
2x 16x 23 0
Now, find the discriminant.
2
D b 4ac
2
16 4 2 23
256 184
72
So, .
D 0
Thus, the equation has real and distinct roots.
Now, find the roots.
b D
x
2a
16 72
2 2
16 6 2
4
That is,
16 6 2
,
4
16 6 2
4
3
4 2,
2
3
4 2
2
Thus, the roots of the given equations are
and .
3
4 2
2
3
4 2
2
(v) Given:
2
x 5 5x 70 0
Now, find the discriminant.
2
D b 4ac
2
5 5 4 1 70
125 280
405
So, .
D 0
Thus, the equation has real and distinct roots.
Now, find the roots.
b D
x
2a
5 5 405
2 1
5 5 9 5
2 1
That is,
5 9 5
,
2
5 9 5
2
4 5
,
2
14 5
2
2 5,
7 5
Thus, the roots of the given equations are
2 5
and .
7 5
Question: 2
Find a natural number whose square diminished by
is equal to thrice of more than the given
84
8
number.
Solution:
Let be the required number.
x
According to the question,
2
x 84 3 x 8
2
x 84 3x 24
2
x 3x 84 24 0
2
x 3x 108 0
Split the middle term to find out the factors of the
equation.
2
x 3x 108 0
2
x 12x 9x 108 0
Take the common terms outside the parenthesis.
2
x 12x 9x 108 0
x x 12 9 x 12 0
x 12 x 9 0
That is,
or
x 12 0
x 9 0
or
x 12 0
x 9 0
or
x 12
x 9
Here, is rejected because it is not a natural
9
number.
Thus, the required number is .
12
Question: 3
A natural number, when increased by , equals
12
times its reciprocal. Find the number.
160
Solution:
Let be the required number.
x
According to the question,
2
x 12x 160 0
Split the middle term to find out the factors of the
equation.
2
x 12x 160 0
2
x 20x 8x 160 0
Take the common terms outside the parenthesis.
2
x 20x 8x 160 0
x x 20 8 x 20 0
x 20 x 8 0
That is,
or
x 20 0
x 8 0
or
x 20
x 8
Here, is rejected because it is not a natural
20
number.
Thus, the required number is .
8
Question: 4
A train, travelling at a uniform speed for km,
360
would have taken 48 minutes less to travel the same
distance if its speed was 5 km/h more. Find the
original speed of the train.
Solution:
Let the original speed of train km/hr
x
The new increase speed of train km/hr
x 5
Time taken by train with original speed
Distance 360
hr
Speed x
Time taken by train with new speed
Distance 360
hr
Speed x 5
According to the question,
360 360 48
x x 5 60
1 1 4
360
x x 5 5
x 5 x 4
360
x x 5 5
5 4
360
x x 5 5
2
90 5
1
x 5x 5
2
x 5x 90 25
2
x 5x 2250 0
Split the middle term to find out the factors of the
equation.
2
x 5x 2250 0
2
x 50x 45x 2250 0
Take the common terms outside the parenthesis.
x x 50 45 x 50 0
x 50 x 45 0
That is,
or
x 50 0
x 45 0
or
x 50 0
x 45 0
or
x 50
x 45
Here, is rejected because it is not a natural
50
number.
Thus, the required number is .
45
Question: 5
If Zeba was younger by 5 years than what she
really is, then the square of her age (in years) would
have been 11 more than five times her actual age.
What is her age now?
Solution:
Let Zeba’a real age now years
x
Zeba’s age when she was 5 years younger than now
years
x 5
According to question,
2
x 5 5x 11
2
2
x 5 2 5 x 5x 11 0
2
x 25 10x 5x 11 0
2
x 15x 14 0
Split the middle term to find out the factors of the
equation.
2
x 15x 14 0
2
x 14x x 14 0
Take the common terms outside the parenthesis.
x x 14 1 x 14 0
x 14 x 1 0
That is,
or
x 14 0
x 1 0
or
x 14 0
x 1 0
or
x 14
x 1
When is subtracted from , then a negative age is
5
1
obtained.
So, is rejected.
1
Thus, the age of Zeba is years.
14
Question: 6
At present Asha’s age (in years) is 2 more than the
square of her daughter, Nisha’s age. When Nisha
grows to her mother’s present age, Asha’s age
would be one year less than 10 times the present
age of Nisha. Find the present ages of both Asha
and Nisha.
Solution:
Let the present age of Nisha years
x
The present age of her mother Asha years
2
x 2
Let the differences of ages years
y
According to problem,
2
x 2 y 10x 1
2
x 2 y 10x 1 0
But,
y differenceof ages
2
x 2 x
2 2
x 2 x 2 x 10x 1 0
2
2x 11x 5 0
Split the middle term to find out the factors of the
equation.
2
2x 11x 5 0
2
2x 10x x 5 0
Take the common terms outside the parenthesis.
2
2x 10x x 5 0
2x x 5 1 x 5 0
x 5 2x 1 0
That is,
or
x 5 0
2x 1 0
or
x 5 0
2x 1 0
or
x 5
2x 1
or
x 5
1
x
2
is rejected.
1
2
Thus,
Nisha’s present age years
5
Asha’s present age years
2
5 2 25 2 27
Question: 7
In the centre of a rectangular lawn of dimensions
, a rectangular pond has to be
50m 40m
constructed so that the area of the grass
surrounding the pond would be [see Fig.
2
1184m
4.1]. Find the length and breadth of the pond.
Solution:
Let the length of pond m
50 2x
Let the breadth of pond m
40 2x
Area of grass around the pond
2
1184m
Area of Lawn Area of Pond 1184
50 40 50 2x 40 2x 1184
2
2000 2000 100x 80x 4x 1184 0
2
2000 2000 180x 4x 1184 0
2
180x 4x 1184 0
2
4x 180x 1184 0
2
x 45x 296 0
Split the middle term to find out the factors of the
equation.
2
x 45x 296 0
2
x 37x 8x 296 0
Take the common terms outside the parenthesis.
x x 37 8 x 37 0
x 37 x 8 0
That is,
or
x 37 0
x 8 0
or
x 37 0
x 8 0
or
x 37
x 8
When , then
x 37
The length of the pond
50 2 37
50 74
24m
The length cannot be negative.
So, is rejected.
x 37
When , then
x 8
The length of the pond
50 2x
50 2 8
50 16
34m
And the breadth of the pond
40 2x
40 2 8
40 16
24m
Hence, the length and the breadth of the pond is
and , respectively.
34m
24m
Question: 8
At minutes past pm, the time needed by the
t
2
minute hand of a clock to show pm was found to
3
be minutes less than minutes. Find .
3
2
t
4
t
Solution:
Total time taken by the minute hand from 2 pm to 3
pm
60min
According to statement,
2
t
t 3 60
4
2
4t t 12 240
2
t 4t 252 0
Split the middle term to find out the factors of the
equation.
2
t 4t 252 0
2
t 18t 14t 252 0
Take the common terms outside the parenthesis.
t t 18 14 t 18 0
t 18 t 14 0
That is,
or
t 18 0
t 14 0
or
t 18 0
t 14 0
or
t 18
t 14
is rejected.
18
Thus, the value of is min.
t
14