CBSE - Class 10 - Maths - CH4 - Quadratic Equation by Factorisation (NCERT Ex 4.2)

Quadratic Equation by Factorisation

(NCERT Exercise 4.2)

✏ Any quadratic equation can have atmost two roots.


Q1: Find the roots of the following quadratic equations by factorisation:

(i) x² – 3x – 10 = 0
(ii) 2x² + x – 6 = 0
(iii) √2 x² + 7 x + 5√2 = 0
(iv) 2x² – x + 1/8 = 0
(v) 100 x² – 20x + 1 = 0

 Answer:


(i) x² – 3x – 10 = 0
     x² – 5x + 2x – 10 = 0
⇒ x(x - 5) + 2(x - 5) = 0
⇒ (x - 5)(x + 2) = 0
∴ x - 5 = 0  and x + 2 = 0
∴ x = 5 and x = -2
∴ 5 and -2 are the roots of the quadratic equation.

(ii)  2x² + x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
∴  2x(x + 2) – 3(x + 2) = 0
∴  (x + 2)(2x – 3) = 0
⇒ x + 2 = 0   and  2x – 3 = 0
∴ x = –2  and x = 3/2
∴ –2 and 3/2 are the roots of the quadratic equation.

(iii) √2 x² + 7x + 5√2 = 0
⇒ √2 x² + 5x + 2x + 5√2 = 0
∴ x (√2x + 5) + √2(√2x + 5) = 0
∴ (√2x + 5)(x + √2) = 0
⇒  √2x + 5 = 0 and  x + √2 = 0
∴  x = –5/√2  and x = –√2
∴  –5/√2 and  –√2 are the roots of the quadratic equation.

(iv) 2x² – x + 1/8 = 0
Taking 1/8 common, we have
 (16x² – 8x + 1) × 1/8 = 0
⇒ 16x² – 8x + 1 = 0
∴ 16x² – 4x – 4x + 1 = 0
∴ 4x(4x – 1) – 1( 4x – 1) = 0
∴ (4x – 1)(4x – 1) = 0
⇒ 4x – 1 = 0 and 4x – 1 = 0
∴ x = 1/4
∴ 1/4 and 1/4 are the roots of the quadratic equation.

(v) 100 x² – 20x + 1 = 0
⇒ 100 x² – 10x – 10x + 1 = 0
∴  10x(10x – 1) – 1(10x – 1) = 0
∴ (10x – 1)(10x – 1) = 0
⇒ 10x -1 = 0
∴ x = 1/10
∴ 1/10, 1/10 are the roots of the quadratic equation.

Q2: (a) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(b) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day.

Answer:
(a) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (since total no. of marbles is 45).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
   = 40 – x
Therefore, their product is: (x – 5) (40 – x) = 124   (Given)
⇒ 40x – x² – 200 + 5x = 124
⇒ – x² + 45x – 200 – 124 = 0
⇒ – x² + 45x – 324 = 0
or   x² – 45x + 324 = 0
∴ x² – 36x – 9x + 324 = 0
∴ x(x – 36) – 9(x - 36) = 0
∴  (x – 36)(x – 9) = 0
∴ x – 36 = 0 or x – 9 = 0
∴ x = 36 or x = 9
⇒ If the number of John’s marbles = 36, then, number of Jivanti’s marbles = 45 − 36 = 9
OR If number of John’s marbles = 9, then, number of Jivanti’s marbles = 45 − 9 = 36

(b) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
Therefore, x (55 – x) = 750
⇒ 55x – x² = 750
⇒ – x² + 55x – 750 = 0
⇒ x² – 55x + 750 = 0
∴ x² – 25x – 30x + 750 = 0
∴ x(x – 25) –30( x – 25) = 0
∴ (x – 25)(x – 30) = 0
∴ x – 25 = 0 or x –30 = 0
∴ x = 25 or x = 30
Thus, the number of toys is either 25 or 30.

Q3: Find two numbers whose sum is 27 and product is 182.

Answer: Let the first number  = y
∴ the second number is = 27 – y
Product of these numbers = 182
⇒ y(27 – y) = 182
⇒ 27y –  y² – 182 = 0
⇒ y² – 27y + 182 = 0
∴ y² – 13y – 14y + 182 = 0
∴ y(y  – 13) – 14 (y – 13) = 0
∴ (y  – 13)( y  – 14) = 0
∴ y  – 13 = 0 or y  – 14 = 0
∴ y = 13 or y = 14
Thus the two numbers are 13 and 14

Q4: Find two consecutive positive integers, sum of whose squares is 365.

Answer: Let the two consecutive positive integers are n and n+1
Sum of the squares of these integers = 365
⇒ n²+ (n + 1)² = 365
∴ n²+ n²+ 2n + 1 – 365 = 0
∴ 2n²+ 2n – 364 = 0
∴ n²+ n – 182 = 0
∴ n²+ 14n – 13n – 182 = 0
∴ n(n + 14) – 13(n + 14) = 0
⇒ (n + 14)(n – 13) = 0
∴ n + 14 = 0 or n – 13 = 0
∴ n = –14 or n = 13
Since the number is positive integer. n = 13 and its consecutive integer is 14
Therefore required numbers are 13 and 14.

Q5: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer:  Let the base = b cm
altitude (a) of rt. triangle is = b - 7 cm
Given hypotenuse h = 13cm

Applying Pythagoras theorem,  b²+ a² = h²
⇒ b²+ (b – 7)² = 13²
∴ b²+ b²+ 49 – 14b = 169
⇒ 2b² – 14b + 48 – 169 = 0
⇒ 2b² – 14b  – 120 = 0
Divide by 2 to both sides,
∴ b² – 7b  – 60 = 0
⇒ b² – 12b + 5b – 60 = 0
⇒ b (b – 12) + 5 (b – 12) = 0
⇒ (b – 12)(b + 5) = 0
∴ b – 12 = 0 or b + 5 = 0
∴ b = 12 or b = –5
Since sides are positive,  ∴ b = 12cm
⇒ altitude = 12 – 7 = 5cm


Q6: 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.

Answer:  Let the number of articles produced = n
∴ cost of production of each article = ₹ 2n + 3 
Given total production cost = ₹ 90

⇒ n(2n + 3) = 90
⇒ 2n² + 3n – 90 = 0
⇒ 2n² + 15n – 12n – 90 = 0
⇒ n (2n + 15) – 6(2n + 15) = 0
⇒ (2n + 15)(n – 6) = 0
∴ 2n + 15 = 0 and n – 6 = 0
∴ n = –15/2 and n = 6
Since the number of articles can't be negative, ⇒ n = 6.
Number of articles produced = 6
Cost of each article = 2 × 6 + 3 =  ₹15