# Class10 NCRT Quadratic Equations Exercise – 4.2 pdf || UP Board

__Quadratic
Equations__

__Exercise
– 4.2__

__Quadratic Equations__

__Exercise – 4.2__

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

(i) x^{2} – 3x – 10 = 0

Simplifying the quadratic equation,

x^{2} – 5x + 2x – 10 = 0

x(x – 5) + 2(x – 5) = 0

(x – 5)(x + 2)
= 0

x – 5 = 0, x + 2 = 0

x = 5, x = -2

Hence, the roots of the quadratic equation are 5 and -2.

(ii) 2x^{2} + x – 6 = 0

Simplifying the quadratic equation,

2x^{2} – 4x + 3x – 6 = 0

2x(x – 4) + 3(x – 6) = 0

(x – 4) (2x + 3) = 0

x – 4 = 0, 2x + 3 = 0

x = 4, x = -3/2

Hence, the roots of the quadratic equation are 2 and
-3/2.

(iii) √2x^{2} + 7x + 5√2 = 0

Simplifying the quadratic equation,

√2x^{2} + 5x + 2x + 5√2 = 0

x(√2x + 5) + √2(√2x + 5) = 0

(√2x + 5)(x + √2) = 0

√2x + 5 = 0, x + √2 = 0

x = -5/√2, x = -√2

Hence, the roots of the quadratic equation are -5/√2
and -√2.

(iv) 2x^{2} – x + 1/8 = 0

Simplifying the quadratic equation,

16x^{2} – 8x + 1 = 0

16x^{2} – 4x – 4x + 1 = 0

4x(4x – 1) – 1(4x – 1) = 0

(4x – 1) (4x – 1) =0

4x – 1 = 0, 4x – 1 = 0

x = ¼, x = ¼

Hence, the roots of the quadratic equation are ¼ and
¼.

(v) 100x^{2} – 20x + 1 = 0

Simplifying the quadratic equation,

100x^{2} – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1)( 10x – 1) = 0

10x – 1 = 0, 10x – 1 = 0

x = 1/10, x = 1/10

Hence, the roots of the quadratic equation are 1/10
and 1/10.

**2. ****Solve
the problems given in Example 1.**** **

**[****The
problems given in the example**** ****x ^{2} – 45x + 324 = 0**

**and**

**x**

^{2}– 55x + 750 = 0.**]**

x^{2} – 45x + 324 = 0

Simplifying the quadratic equation,

x^{2} – 36x – 9x + 324 = 0

x(x – 36) – 9(x – 36) = 0

(x – 36)(x – 9) = 0

x – 36 = 0, x – 9 = 0

x = 36, x = 9

So, John and Jivanti initially had 36 and 9 marbles.

x^{2} – 55x + 750 = 0

Simplifying the quadratic equation,

x^{2} – 30x – 25x + 750 = 0

x(x – 30) – 25(x – 30) = 0

(x – 30)(x – 25) = 0

x = 30, x = 25

Hence, the
number of toys manufactured on that day is 30
or 25.

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

** **

Let the
first number = x

Therefore,
second number = 27 - x

According
to Question,

Product =
x(27 –
x) = 182

27x – x^{2}
= 182

x^{2}
– 27x – 182 = 0

x^{2}
– 13x – 14x + 182 = 0

x(x – 13)
– 14(x – 13) = 0

(x – 13)(x
– 14) = 0

x - 13 =
0, x – 14 = 0

x = 13, x
= 14

Hence, the
numbers are 13 and 14.

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

Let the
first number = x

Therefore,
second number = x + 1

According
to Question,

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

x^{2}
+ x^{2} + 2x + 1 = 365

2x^{2}
+ 2x – 364 =0

x^{2}
+ x – 182 = 0

x^{2}
– 13x + 14x - 182 = 0

x(x – 13)
+ 14(x – 13) = 0

x – 13 =
0, x + 14 = 0

x = 13, x
= -14 या
14 [positive integer]

Hence, two
consecutive positive integers are 13 and 14.

**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.****
**

** **

Let base =
x cm

Therefore,
height = x – 7
cm

Given,
hypotenuse = 13 cm

By the
Pythagoras theorem,

x^{2}
+ (x – 7)^{2} = 13^{2}

x^{2}
+ x^{2} – 14x + 49 = 169

2x^{2}
– 14x – 120 = 0

x^{2}
– 7x – 60 = 0

x^{2}
– 12x + 5x – 60 = 0

x(x – 12)
+ 5(x – 12) = 0

(x – 12)(x
+ 5) = 0

x – 12 =
0, x + 5 = 0

x = 0, x =
-5

But x ≠ -5
[Since x is a side of the
triangle, the side cannot be negative.]

Therefore,
x = 12
and other side (height) = x – 7 = 12 – 7
= 5

Hence, the other two sides are 12 cm and 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**** ****₹90****,
****find the number of
articles produced and the cost of each article.****
**

Let the
number of pottery articles = x

Hence,
cost of one article = 2x
+ 3

According
to Question,

total
manufacturing cost = x(2x + 3) = 90

2x^{2}
+ 3x = 90

2x^{2}
+ 3x – 90 = 0

2x^{2}
+ 15x – 12x – 90 = 0

x(2x + 15)
– 6(2x + 15) = 0

(2x +
15)(x – 6) = 0

2x + 15 =
0, x – 6 =0

x = -15/2,
x = 6

But x ≠ -15/2
[since x is the number of
articles.]

so x = 6

So, cost
of each article = 2x
+ 3 = 2×6
+ 3 = 15

Hence, the
number of articles manufactured is 6 and the cost of each article is ₹15.

Download Class10 NCRT Quadratic Equations Exercise – 4.2 pdf

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