CBSE Class 10 Mathematics Chapter 4 - Quadratic Equations - Exercise 4.2

 CBSE Class 10 Mathematics  Chapter 4 - Quadratic Equations - Exercise 4.2

Q1.	Find the roots of the following quadratic equations by factorisation:
(i)	x2 – 3x – 10 = 0
Solution.	x2 – 3x – 10 = 0 Split middle term x2 − 5x + 2x – 10 = 0  (because (−5) × 2 = −10 and (−5) + 2 = −3) (x2 − 5x) + (2x – 10) = 0  (group terms) x(x − 5) + 2(x − 5) = 0 (x − 5)(x + 2) = 0 Now, set each factor equal to zero and solve for x: x – 5 = 0 x = 5 x + 2 = 0 x = −2 So, the roots of the equation are 5 and −2.

(ii)	2x2 + x – 6 = 0
Solution.	2x2 + x – 6 = 0 Split middle term 2x2 + 4x − 3x  – 6 = 0    (because (-3) × 4 = -12 and (-3) + 4 = 1) (2x2 + 4x) − (3x − 6) = 0  (group terms) 2x(x + 2) − 3(x + 2) = 0 (2x − 3)(x + 2) = 0 Now, set each factor equal to zero and solve for x: 2x – 3 = 0 2x = 3 x = 3/2 x + 2 = 0 x = −2 So, the roots of the equation are 3/2 and −2.

(iii)	√2 x2 + 7x + 5√2 = 0
Solution.	√2 x2 + 7x + 5√2 = 0 Split middle term √2 x2 + 2x + 5x + 5√2 = 0 (because 5√2 × √2 = 5 × 2 = 10 and 2 + 5 = 7) √2 x2 + (√2 × √2)x + 5x + 5√2 = 0 √2x(x + √2) + 5(x + √2) = 0 (√2x + 5)(x + √2) = 0 Now, set each factor equal to zero and solve for x: √2x + 5 = 0 √2x = — 5 x = —5/√2 x + √2 = 0 x = — √2 So, the roots of the equation are —5/√2 and — √2.

(iv)

2x2 — x + ⅛ = 0

Solution.

2x2 — x + ⅛ = 0 Multiply the entire equation by 8 to eliminate fractions: (2x2 — x + ⅛)8 = 0×8 16x2 — 8x + 1 = 0 Split middle term 16x2 — 4x — 4x + 1 = 0 (because 1 × 16 = 16 and (—4) + (—4) = —8) (16x2 — 4x) — (4x + 1) = 0 (group terms) 4x(4x — 1) — 1(4x — 1) = 0  (4x — 1)(4x — 1) = 0 Now, set each factor equal to zero and solve for x: 4x — 1 = 0 4x = 1 x = ¼ So, the roots of the equation are ¼ and ¼.

(v)	100 x2 – 20x + 1 = 0
Solution.	100 x2 – 20x + 1 = 0 Split middle term 100 x2 – 10x – 10x + 1 = 0 (because 1 × 100 = 100 and (—10) + (—10) = —20) (100 x2 – 10x) – (10x + 1) = 0 (group terms) 10x(10x – 1) – 1(10x – 1) = 0  (10x – 1)(10x – 1) = 0 Now, set each factor equal to zero and solve for x: 10x – 1 = 0 10x = 1 x = 1/10 So, the roots of the equation are 1/10 and 1/10.

Q2.

Solve the problems given in Example 1.

Solution.

Example 1 Question (i) is: 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. After solve example 1 Question (i) we found the eqauaton: x2 —45x + 324 = 0 For find root of equation, Split middle term – ⇒ x2 —36x—9x + 324 = 0 (because 1 × 324 = 324 and (—36) + (—9) = —45) ⇒ x(x —36)—9(x — 36) = 0 ⇒ (x —9)(x — 36) Now, set each factor equal to zero and solve for x: x – 9 = 0 x = 9 x — 36 = 0 x = 36 So, the roots of the equation are 9 and 36.   Example 1 Question (ii) is: 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. After solve example 1 Question (ii) we found the eqauaton: x2 — 55x + 750 = 0 For find root of equation, Split middle term – ⇒ x2 —25x—30x + 750 = 0  (because 1 × 750 = 750 and (—25) + (—30) = —55) ⇒ x(x —25)—30(x — 25) = 0 ⇒ (x —25)(x — 30) Now, set each factor equal to zero and solve for x: x — 25 = 0 x = 25 x — 30 = 0 x = 30 So, the roots of the equation are 25 and 30.

Q3.

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

Solution.

Assume that the first number is x. According to question sum of both number is 27 So second number is 27—x As given, product of both number is 182, so we can write as x(27—x) = 182 27x—x2 = 182 27x—182 = x2 0 = x2—27x + 182  x2—27x + 182 = 0 For find root of equation, Split middle term – x2—13x —14x + 182 = 0  (because 1 × 182 = 182 and (—13) + (—14) = —27) (x2—13x) —(14x + 182) = 0 (Group terms) x(x—13)—14(x—13) = 0 (x—14)(x—13) = 0 Now, set each factor equal to zero and solve for x: x — 14 = 0 x = 14 x — 13 = 0 x = 13 So, the roots of the equation are 14 and 13. If x = 14 then first number x = 14 and second number 27—x = 27—14 = 13 If x = 13 then first number x = 13 and second number 27—x = 27—13 = 14 So the two numbers are 13 and 14.

Q4.

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

Solution.

We know that difference between two consecutive positive integers is 1. Let First integer = x Then second integer = x + 1 As per question, sum of square = 365 (First integer)2 + (second integer)2 = 365 x2 + (x + 1)2 = 365 By using (a+b)2 = a2 + b2 + 2ab x2 + x2 + (1)2 + 2 (x)(1) = 365 2x2 + 1 + 2x = 365 2x2 + 2x + 1 = 365 2x2 + 2x + 1 — 365 = 0 2x2 + 2x — 364 = 0 2(x2 + x — 182) = 0 x2 + x — 182 = 0 For find root of equation, Split middle term – x2 + 14x—13x — 182 = 0 (because —182 × 1 = —182 and (—13) + (14) = 1) (x2 + 14x)—(13x — 182) = 0 (Group terms) x(x + 14)—13(x + 14) = 0 (x —13) (x + 14) = 0 Now, set each factor equal to zero and solve for x: x — 13 = 0 x = 13 x + 14 = 0 x = —14 So, the roots of the equation are 13 and —14. Since, we have to find positive integer so —14 is not possible. Therefore, If x = 13 then first integer x = 13 and second integer x + 1 = 13 + 1 = 14 So the two consecutive positive integers 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.

Solution.

Assume that ABC is a right angle triangle with altitude AB, Base BC and hypotenuse AC. According to question, Hypotenuse (AC) = 13 cm And altitude (AB) 7 cm less than base Let Base (BC) = x cm Then, Altitude = Base — 7 = x — 7 Because ABC is a right angle triangle, we use Pythagoras theorem (AC)2 = (AB)2 + (BC)2 Put the value – (13)2 = (x—7)2 + (x)2 By using (a – b)2 = a2 + b2 – 2ab 169 = x2 + 72 – 2(x)(7) + x2 169 = x2 + 49 – 14x + x2 169 = 2x2 + 49 – 14x 169 = 2x2 – 14x + 49 2x2 – 14x + 49 = 169 2x2 – 14x + 49 — 169 = 0 2x2 – 14x — 120 = 0 2(x2 – 7x — 60) = 0 x2 – 7x — 60 = 0 For find root of equation, Split middle term – x2 + 5x—12x — 60 = 0 (because —60 × 1 = —60 and (5) + (—12) = —7) (x2 + 5x)—(12x — 60) = 0 (Group Terms) x(x + 5)—12(x + 5) = 0 (x — 12)(x + 5) = 0 Now, set each factor equal to zero and solve for x: x — 12 = 0 x = 12 x + 5 = 0 x = —5 So, the roots of the equation are 12 and —5. Since, Length cannot be negative so —5 is not possible. Therefore, If x = 12 then base (BC) x = 12 and altitude (AB) x — 7 = 12 — 7 = 5

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.

Solution.

Assume Number of a article = x As per question, the cost of production of each article is 3 more than twice the number of articles produced Hence, cost of article = 2x + 3 Given, total cost of production = Rs. 90 We know that – Total cost of production = Number of article produce × cost of a article Put the value – 90 = x (2x + 3) x (2x + 3) = 90 2x2 + 3x = 90 2x2 + 3x — 90 = 0 For find root of equation, Split middle term – 2x2 + 15x—12x — 90 = 0 (because —90 × 2 = —180 and (15) + (—12) = 3) x(2x + 15)—6(2x + 15) = 0 (x — 6)(2x + 15) = 0 Now, set each factor equal to zero and solve for x: x — 6 = 0 x = 6 2x + 15 = 0 2x = —15 X = —15/2 So, the roots of the equation are 6 and —15/2. Since, number of article cannot be negative so —15/2 is not possible. Therefore, If x = 6 then number of article (x) = 6 and cost of article  2x + 3 = 2(6) + 3 = 15 Rs.