**NCERT Solutions Maths Class 10 Exercise 4.2**

**1. Find the roots of the following quadratic equations
by factorization:**

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

**− 3**

*x*− 10 = 0**(ii) 2 x^{2
}+ x **

**− 6 = 0**

**(iii) √2 x^{2
}+ 7x + 5√2 = 0**

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

**–**

*x*+ 1/8 = 0**(v) 100 x^{2
}**

**− 20**

*x*+ 1 = 0**Solution: (i) x^{2 }**

**− 3**

*x*− 10 = 0⇒
*x*^{2 }− 5*x
*+ 2*x *− 10 = 0

⇒
*x*(*x *− 5) + 2(*x *− 5) = 0

⇒
(*x *− 5) (*x *+ 2) = 0

⇒
(*x *− 5) = 0 or (*x *+ 2) = 0

⇒
*x *= 5 or *x *= −2

Thus,
*x *= 5, −2

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

**− 6 = 0**

⇒ 2*x*^{2 }+ 4*x* − 3*x *−
6 = 0

⇒ 2*x *(*x *+ 2) – 3(*x *+ 2) = 0

⇒
(2*x *− 3) (*x *+ 2) = 0

⇒
(2*x *− 3) = 0 or (*x *+ 2) = 0

⇒
*x *= 3/2 or *x *= −2

Thus,
*x *= 3/2, −2

**(iii) √2 x^{2
}+ 7x + 5√2 = 0**

⇒
√2*x*^{2 }+ 2*x* + 5*x* + 5√2 = 0

⇒ √2*x*(*x* + √2) + 5(*x* + √2) = 0

⇒ (√2*x* + 5)(*x* + √2) = 0

⇒ *x* = −5/√2,
−√2

or *x* = −5√2/2, −√2

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

**–**

*x*+ 1/8 = 0⇒ (16*x*^{2}
– 8*x* + 1)/8 = 0

⇒ 16*x*^{2} − 8*x*
+ 1 = 0

⇒ 16*x*^{2} − 4*x*
− 4*x* + 1 = 0

⇒
4*x*(4*x *− 1) – 1(4*x *− 1) = 0

⇒
(4*x *− 1) (4*x *− 1) = 0

⇒ *x* = 1/4, 1/4

**(v) 100 x^{2
}**

**− 20**

*x*+ 1 = 0⇒ 100*x*^{2
}− 10*x *–
10*x* + 1 = 0

⇒
10*x*(10*x *− 1) – 1(10*x *− 1) = 0

⇒
(10*x *− 1) (10*x *− 1) = 0

⇒*x *= 1/10, 1/10

**2. Solve the following problems
given:**

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

**– 45**

*x*+ 324 = 0**(ii) x^{2}**

**− 55**

*x*+ 750 = 0**Solution: (i) x^{2}**

**– 45**

*x*+ 324 = 0⇒
*x*^{2}* *– 36*x* – 9*x* + 324 = 0

⇒
*x*(*x *− 36) – 9(*x *− 36) = 0

⇒
(*x *− 9) (*x *− 36) = 0

⇒
*x *= 9, 36

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

**− 55**

*x*+ 750 = 0⇒ *x*^{2 }– 25*x* – 30*x* + 750 = 0

⇒
*x*(*x *− 25) – 30(*x *− 25) = 0

⇒
(*x *− 30) (*x *− 25) = 0

⇒
*x *= 30, 25

**3. Find two numbers whose sum
is 27 and product is 182.**

**Solution:
**Let the first
number be *x*.

Then,
the second number is (27 − *x*).

According
to the question, the product of the two numbers is 182.

Therefore,
*x*(27 − *x*) = 182

⇒ 27*x* – *x*^{2}
= 182

⇒ *x*^{2 }– 27*x* + 182 = 0

⇒ *x*^{2 }– 14*x* – 13*x* + 182 = 0

⇒
*x*(*x *− 14) – 13(*x *− 14) = 0

⇒
(*x *− 14) (*x *− 13) = 0

⇒
*x *= 14, 13

Therefore,
the first number is 14 or 13.

If the first
number = 14, then, the second number = 27 – *x *= 27 – 14 =
13

If the first
number = 13, then, the second number = 27 – *x *= 27 – 13 =
14

Therefore,
the two numbers are 13 and 14.

**4. Find two consecutive
positive integers, sum of whose squares is 365.**

**Solution:
**Let the two
consecutive positive integers be *x *and (*x *+ 1).

According
to the question, we have

*x*^{2} + (*x*
+ 1)^{2 }= 365

⇒
*x*^{2 }+ *x*^{2 }+ 1 + 2*x* =
365 [Since (*a* + *b*)^{2} = *a*^{2} + 2*ab* + *b*^{2}]

⇒ 2*x*^{2 }+ 2*x* – 364 = 0

Dividing
the above equation by 2, we get

⇒ *x*^{2 }+ *x* – 182 = 0

⇒ *x*^{2 }+ 14*x* – 13*x*
– 182 = 0

⇒
*x*(*x *+ 14) – 13(*x *+ 14) = 0

⇒ (*x *+ 14) (*x *− 13) = 0

⇒
*x *= 13, −14

Therefore, the
first number is 13. [We discard −14
because it is a negative number.]

Thus, the second number = *x *+ 1 = 13 + 1 = 14

Hence, the two
consecutive positive integers are 13 and 14, sum of whose squares is 365.

**5. The altitude of a right triangle is 7 cm
less than its base. If the hypotenuse is 13 cm, find the other two sides.**

**Solution: **Let the base of the triangle be *x* cm.

Then,
the altitude of the triangle is (*x *− 7) cm.

It
is given that the hypotenuse of the triangle is 13 cm.

According
to Pythagoras Theorem, we have

(Hypotenuse)^{2}
= (Base)^{2} + (Altitude)^{2}

13^{2}
= *x*^{2} + (*x *– 7)^{2}

⇒ 169 = *x*^{2}
+ *x*^{2} + 49 – 14*x* [Since
(*a* + *b*)^{2} = *a*^{2}
+ *b*^{2} + 2*ab*]

⇒ 169 = 2*x*^{2}–
14x + 49

⇒ 2*x*^{2
}– 14*x* – 120 = 0

Dividing
the above equation by 2, we get

⇒ *x*^{2 }– 7*x*
– 60 = 0

⇒ *x*^{2 }– 12*x* + 5*x* – 60 = 0

⇒ *x*(*x *− 12) + 5(*x *− 12) = 0

⇒
(*x *− 12) (*x *+ 5) = 0

⇒
*x *= −5, 12

Therefore,
the base of the triangle = 12 cm

[We discard *x *= −5, because
the length of the side of a triangle cannot be negative.]

Thus, altitude of the triangle = (*x *− 7) = 12 – 7 = 5 cm

Hence, the base
is 12 cm and the altitude is 5 cm.

**6. A cottage industry produces a certain number
of pottery articles in a day. It was observed on a particular day that the cost
of production of each article (in rupees) was 3 more than twice the number of
articles produced on that day. If the total cost of production on that day was
Rs 90, find the number of articles produced and the cost of each article.**

**Solution:
**Let the cost of
production of one article be Rs *x*.

It
is given that the total cost of production on that particular day = Rs 90

Therefore,
the total number of articles produced on that day = 90/x

According
to the question, we have

*x* = 2(90/*x*)
+ 3

⇒ *x* = 180/*x* + 3

⇒ *x* = (180 + 3*x*)/*x*

⇒ *x*^{2 }= 180
+ 3*x*

⇒ *x*^{2 }− 3*x*
− 180 = 0

⇒ *x*^{2 }− 15*x* + 12*x* − 180 = 0

⇒ *x*(*x *− 15) + 12(*x *− 15) = 0

⇒ (*x *− 15) (*x *+ 12) = 0

⇒ *x *= 15, −12

Therefore, *x *= Rs* *15, which is the
cost of production of one article.

[We discard *x *= −12,
because the cost cannot be negative.]

The number of articles produced on that particular day = 90/15 = 6

Hence, the number of articles produced
is 6 and the cost of each article is Rs 12.