**NCERT Solutions Maths Class 10 Exercise 4.3**

**1.
Find the roots of the following quadratic equations, if they exist, by the
method of completing square:**

**(i) 2 x^{2 }**

**− 7**

*x*+ 3 = 0**(ii) 2 x^{2 }+ x **

**− 4 = 0**

**(iii) 4 x^{2 }+ 4√3x + 3 = 0**

**(iv) 2 x^{2 }+ x + 4 = 0**

**Solution: (i) 2 x^{2
}**

**− 7**

*x*+ 3 = 0Following
the procedure of completing square, we
first divide the given equation by 2 to make the coefficient of *x*^{2} equal to 1.

We
divide middle term of the equation by 2*x*, we get

We
add and subtract square of 7/4 from the equation

[Since
(*a* **−** *b*)^{2}
= *a*^{2} **−**
2*ab* + *b*^{2}]

Taking square root on both
sides, we get

Therefore, *x* = 1/2, 3

**(ii) 2 x^{2 }+ x **

**− 4 = 0**

Following
the procedure of completing square, we
first divide the given equation by 2 to make the coefficient of *x*^{2} equal to 1.

We
divide middle term of the equation by 2*x*, we get *x*/2 ÷ 2*x* = 1/4

We add and subtract square of 1/4 from the equation

*a*+

*b*)

^{2}=

*a*

^{2}+ 2

*ab*+

*b*

^{2}]

Taking square root on both
sides, we get

**(iii) 4 x^{2 }+ 4√3x + 3 = 0**

Following
the procedure of completing square, we
first divide the given equation by 4 to make the coefficient of *x*^{2} equal to 1.

We
divide middle term of the equation by 2*x*, we get √3*x* ÷ 2*x*
= √3/2

We
add and subtract square of √3/2 from the equation

[Since (*a*
+ *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}]

**(iv) 2 x^{2 }+ x + 4 = 0**

Following
the procedure of completing square, we
first divide the given equation by 2 to make the coefficient of *x*^{2} equal to 1.

We
divide middle term of the equation by 2*x*, we get *x*/2 ÷ 2*x* = 1/4

We
add and subtract square of 1/4 from the equation

[Since
(*a* + *b*)^{2} = *a*^{2}
+ 2*ab* + *b*^{2}]

Taking square root on both sides, right hand
side does not exist because square root of negative number does not exist.

Therefore, there is no
solution for the quadratic equation 2*x*^{2 }+ *x* + 4 = 0.

**2.
Find the roots of the following quadratic equations by applying quadratic
formula.**

**(i) 2 x^{2 }**

**− 7**

*x*+ 3 = 0**(ii) 2 x^{2 }+ x **

**− 4 = 0**

**(iii) 4 x^{2 }+ 4√3x + 3 = 0**

**(iv) 2 x^{2 }+ x + 4 = 0**

**Solution:
(i) 2 x^{2 }**

**− 7**

*x*+ 3 = 0Comparing
the equation 2*x*^{2
}− 7*x* + 3 = 0 with the general
form of a quadratic equation *ax*^{2
}+ *bx* + *c* = 0, we get

*a* = 2, *b* =
−7 and *c* = 3

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 3, 1/2

**(ii) 2 x^{2 }+ x **

**− 4 = 0**

Comparing
the equation 2*x*^{2
}+ *x* − 4 = 0 with the general
form of a quadratic equation *ax*^{2
}+ *bx* + *c* = 0, we get

*a *= 2, *b *= 1 and *c *= −4

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

**(iii) 4 x^{2 }+ 4√3x + 3 = 0**

Comparing
the equation 4*x*^{2
}+ 4√3*x* + 3 = 0 with the general
form of a quadratic equation *ax*^{2
}+ *bx* + *c* = 0, we get

*a *= 4, *b *= 4√3 and *c *= 3

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

**(iv) 2 x^{2 }+ x + 4 = 0**

Comparing
the equation 2*x*^{2
}+ *x* + 4 = 0 with the general
form of a quadratic equation *ax*^{2
}+ *bx* + *c* = 0, we get

*a *= 2, *b *= 1 and *c *= 4

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

Since the square root of
negative number is not defined.

Therefore, the quadratic equation 2*x*^{2
}+ *x* + 4 = 0** **has
no solution.

**3. Find the roots of the following equations:**

⇒ *x*^{2 }− 1 = 3*x*

⇒ *x*^{2 }− 3*x* − 1 = 0

Comparing
the equation *x*^{2 }− 3*x* − 1 = 0 with the general
form of a quadratic equation *ax*^{2
}+ *bx* + *c* = 0, we get

*a *= 1, *b *= −3 and *c *= −1

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ −30 = *x*^{2 }− 7*x* + 4*x*
− 28

⇒ *x*^{2 }− 3*x* + 2 = 0

Comparing
the equation *x*^{2 }− 3*x* + 2 = 0 with the general
form of a quadratic equation *ax*^{2
}+ *bx* + *c* = 0, we get

*a *= 1, *b *= −3 and *c *= 2

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 2, 1

**4.
The sum of reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from
now is 1/3. Find his present age.**

**Solution:
**Let the present age of Rehman be *x* years.

3
years ago, the age of Rehman = (*x *− 3) years.

After
5 years, the age of Rehman = (*x *+ 5) years.

According to the question,
we have

⇒ 3(2*x *+ 2) = (*x *− 3) (*x *+ 5)

⇒ 6*x *+ 6 = *x*^{2
}− 3*x* + 5*x* – 15

⇒ 6*x *+ 6 = *x*^{2
}+ 2*x* − 15

⇒ *x*^{2 }+ 2*x* − 6*x*
– 15 − 6 = 0

⇒ *x*^{2 }− 4*x* – 21 = 0

Comparing
the equation *x*^{2 }− 4*x* – 21 = 0 with the general form of a quadratic
equation *ax*^{2 }+ *bx* + *c*
= 0, we get

*a *= 1, *b *= −4 and *c *= −21

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 7, −3

Here,
*x *= 7 [We
discard *x *= −3, because the age cannot be negative.]

Therefore, the present age
of Rehman is 7 years.

**5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30.
Had she got 2 marks more in Mathematics and 3 marks less in English, the
product of their marks would have been 210. Find her marks in the two subjects.**

**Solution:
**Let Shefali’s marks in Mathematics be *x.*

Then,
Shefali’s marks in English is 30 – *x.*

If
she had got 2 marks more in Mathematics, her marks in Mathematics would be
= *x *+ 2

If
she had got 3 marks less in English, her marks in English would be = 30 – *x *− 3 = 27 − *x*

According to the question,
we have

(*x *+ 2) (27 − *x*) = 210

⇒ 27*x* − *x*^{2 }+ 54 − 2*x* = 210

⇒ *x*^{2 }− 25*x* + 156 = 0

Comparing
the equation *x*^{2 }− 25*x* + 156 = 0 with the general form of a quadratic
equation *ax*^{2 }+ *bx* + *c*
= 0, we get

*a *= 1, *b *= −25 and *c *= 156

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 13, 12

Therefore, Shefali’s marks
in Mathematics = 13 or 12

Thus, Shefali’s marks in
English = 30 – *x *= 30 – 13 = 17

Or Shefali’s marks in
English = 30 – *x *= 30 – 12 = 18

Hence, Shefali’s marks in Mathematics and
English are (13, 17) or (12, 18).

**6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If
the longer side is 30 metres more than the shorter side, find the sides of the
field.**

**Solution:
**Let the shorter side of the rectangular field be *x *metres.

Then, the diagonal of the rectangular field = (*x *+ 60) metres

And the longer side of the rectangular field = (*x *+ 30) metres

According to Pythagoras
theorem, we have

(*x* + 60)^{2} = (*x *+
30)^{2} + *x*^{2}

⇒ *x*^{2 }+
3600 + 120*x* = *x*^{2 }+ 900 + 60*x*
+ *x*^{2 }[Since (*a* + *b*)^{2} = *a*^{2} + *b*^{2} + 2*ab*]

⇒ *x*^{2 }– 60*x* – 2700 = 0

Comparing
the equation *x*^{2 }– 60*x* – 2700 = 0 with the general form of a quadratic
equation *ax*^{2 }+ *bx* + *c*
= 0, we get

*a *= 1, *b *= −60 and *c *= −2700

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 90, –30

⇒ *x *= 90 [We discard –30, because the
length cannot be in negative.]

Therefore, the length of the shorter side =
90 metres

And
the length of the longer side = *x *+ 30 = 90 +
30 = 120 metres

Hence, the sides of the rectangular field are
90 metres and 120 metres.

**7.
The difference of squares of two numbers is 180. The square of the smaller
number is 8 times the larger number. Find the two numbers.**

**Solution: **Let the smaller number be *x *and the
larger number be *y.*

According to the given
condition, we have

*y*^{2 }– *x*^{2} = 180 …
(1)

It is also given that the square of smaller
number is 8 times the larger number.

⇒
*x*^{2} = 8*y* *… *(2)

Putting the value of *x*^{2} from equation (2) in
equation (1), we get

*y*^{2 }– 8*y* = 180

⇒ *y*^{2 }– 8*y* – 180 = 0

Comparing
the equation *y*^{2 }– 8*y* – 180 = 0 with the general form of a quadratic
equation *ay*^{2 }+ *by* + *c*
= 0, we get

*a *= 1, *b *= −8 and *c *= −180

Using the quadratic formula,
we have

*a*,

*b*and

*c*, we get

⇒ *y *= 18, −10

Putting the values of *y* in equation (2), we get

*x*^{2} = 8*y*

⇒ *x*^{2} = 8*y* = 8 × 18 =
144

⇒ *x *= ±12

And, *x*^{2}
= 8*y* = 8 × −10 = −80

[No real solution for *x*, because square root of negative number cannot be found.]

Therefore, two numbers are
(12, 18) or (−12, 18).

**8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more,
it would have taken 1 hour less for the same journey. Find the speed of the
train.**

**Solution:
**Let the speed of the train be *x* km/h.

Time
taken to cover 360 km = 360/*x* hour

If the speed had been 5 km/h more, the train
would have taken 1 hour less for the same journey.

According to the given
condition, we have

⇒ 360 × 5 = *x*^{2}
+ 5*x*

⇒ *x*^{2 }+ 5*x* − 1800 = 0

Comparing
the equation *x*^{2 }+ 5*x* − 1800 = 0 with the general form of a quadratic
equation *ax*^{2 }+ *bx* + *c*
= 0, we get

*a *= 1, *b *= 5 and *c *= −1800

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 40, −45

[We
discard *x *= −45, because the speed of the train cannot be
negative.]

Therefore, the speed of the
train is 40 km/h.

**9.
Two water taps together can fill a tank in ****hours. The tap of larger
diameter takes 10 hours less than the smaller one to fill the tank separately.
Find the time in which each tap can separately fill the tank.**

**Solution: **Let the time taken by tap of smaller diameter to fill the tank be *x* hours.

Then, the time taken by tap
of larger diameter to fill the tank = (*x *– 10) hours.

It means that the tap of
smaller diameter fills (1/*x*)^{th} part of tank in 1 hour … (1)

And, the tap of larger diameter fills part of tank in 1 hour … (2)

When both the taps are opened
together, they fill the tank in 75/8 hours.

In 1 hour, they fill (8/75)^{th} part
of the tank … (3)

From equations (1), (2) and
(3), we have

⇒
75(2*x *− 10) = 8(*x*^{2}− 10*x*)

⇒ 150*x *– 750 = 8*x*^{2 }− 80*x*

⇒ 8*x*^{2 }− 80*x* − 150*x* + 750 = 0

⇒ 8*x*^{2 }− 230*x* + 750 = 0

Dividing the above equation
by 2, we get

⇒ 4*x*^{2 }− 115*x* + 375 = 0

Comparing
the equation 4*x*^{2 }− 115*x* +
375 = 0 with the general form of a
quadratic equation *ax*^{2 }+ *bx* + *c*
= 0, we get

*a *= 4, *b *= −115 and *c *= 375

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 25, 3.75

The time taken by tap of larger
diameter to fill the tank = *x *– 10 = 3.75 –
10 = −6.25 hours

Time cannot be in negative.
Therefore, we ignore this value.

The time taken by tap of larger
diameter to fill the tank = *x *– 10 = 25 –
10 = 15 hours

Hence, the time taken by tap of larger diameter
to fill the tank is 15 hours and by tap of smaller diameter is 25 hours.

**10.
An express train takes 1 hour less than a passenger train to travel 132 km
between Mysore and Bangalore (without taking into consideration the time they
stop at intermediate stations). If the average speed of the express train is 11
km/h more than that of the passenger train, find the average speed of the two
trains.**

**Solution:
**Let the average speed of the passenger train be *x* km/h.

Then, the average speed of
the express train = (*x *+ 11) km/h

Time taken by the passenger
train to travel 132 km = 132/*x* hours

Time taken by the express
train to travel 132 km = 132/(*x*
+ 11) hours

According to the question,
we have

⇒ 132(11)
= *x*(*x *+ 11)

⇒ 1452 = *x*^{2 }+
11*x*

⇒ *x*^{2 }+ 11*x* – 1452 = 0

Comparing
the equation *x*^{2 }+ 11*x* – 1452 = 0 with the general form of a quadratic equation
*ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 1, *b *= 11
and *c *= −1452

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 33, −44

Therefore, the speed of the passenger train =
33 km/h

[We discard −44, because the
speed cannot be in negative.]

And,
the speed of the express train = *x *+ 11 = 33 +
11 = 44 km/h

**11.
Sum of the areas of two squares is 468 m ^{2}. If the difference of
their perimeters is 24 m, find the sides of the two squares.**

**Solution: **Let the perimeter of the first square be *x* metres.

Then,
the perimeter of the second square = (*x *+ 24) metres

The length of side of the first square = *x*/4 metres

[Since the perimeter of a square = 4 × length of a side]

The length of side of the
second square = (*x *+ 24)/4 metres

Area of the first square =
side × side = *x*/4 × *x*/4 = *x*^{2}/16 sq. m

According to the question,
we have

⇒ 2*x*^{2 }+ 576 +
48*x* = 468 × 16

⇒ 2*x*^{2 }+ 48*x* + 576 = 7488

⇒ 2*x*^{2 }+ 48*x* – 6912 = 0

⇒ *x*^{2 }+ 24*x* – 3456 = 0

Comparing
the equation *x*^{2 }+ 24*x* – 3456 = 0 with the general form of a quadratic equation
*ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 1, *b *=24 and *c *= −3456

Using the quadratic formula,
we have

Putting the values of *a*, *b*
and *c*, we get

⇒ *x *= 48, −72

Therefore, the perimeter of the
first square = 48 metres

[We
discard *x *= −72, because the perimeter of a square cannot be negative.]

And, the perimeter of the second
square = *x *+ 24 = 48 + 24 = 72 metres

Therefore,
the side of the first square =

And, the side of the second
square =

Hence, the
sides of the two squares are 12 m and 18 m.