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1. Find the roots of the following quadratic equations by factorisation: (i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0 (iii) √2 x2 + 7x + 5√2 = 0 (iv) 2x2 – x +1/8 = 0 (v) 100x2 – 20x + 1 = 0

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

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

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 8740 Points
4 months ago
Solutions: (i) Given, x2 – 3x – 10 =0 Taking LHS, =>x2 – 5x + 2x – 10 =>x(x – 5) + 2(x – 5) =>(x – 5)(x + 2) The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, x – 5 = 0 or x + 2 = 0 => x = 5 or x = -2 (ii) Given, 2x2 + x – 6 = 0 Taking LHS, => 2x2 + 4x – 3x – 6 => 2x(x + 2) – 3(x + 2) => (x + 2)(2x – 3) The roots of this equation, 2x2 + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, x + 2 = 0 or 2x – 3 = 0 => x = -2 or x = 3/2 (iii) √2 x2 + 7x + 5√2=0 Taking LHS, => √2 x2 + 5x + 2x + 5√2 => x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2) The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, √2x + 5 = 0 or x + √2 = 0 => x = -5/√2 or x = -√2 (iv) 2x2 – x +1/8 = 0 Taking LHS, =1/8 (16x2 – 8x + 1) = 1/8 (16x2 – 4x -4x + 1) = 1/8 (4x(4x – 1) -1(4x – 1)) = 1/8 (4x – 1)2 The roots of this equation, 2x2 – x + 1/8 = 0, are the values of x for which (4x – 1)2= 0 Therefore, (4x – 1) = 0 or (4x – 1) = 0 ⇒ x = 1/4 or x = 1/4 (v) Given, 100x2 – 20x + 1=0 Taking LHS, = 100x2 – 10x – 10x + 1 = 10x(10x – 1) -1(10x – 1) = (10x – 1)2 The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0 ∴ (10x – 1) = 0 or (10x – 1) = 0 ⇒x = 1/10 or x = 1/10 2. Solve the problems given in Example 1. Represent the following situations mathematically: (i) 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. (ii) 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 ` 750. We would like to find out the number of toys produced on that day. Solutions: (i) Let us say, the number of marbles John have = x. Therefore, number of marble Jivanti have = 45 – x After losing 5 marbles each, Number of marbles John have = x – 5 Number of marble Jivanti have = 45 – x – 5 = 40 – x Given that the product of their marbles is 124. ∴ (x – 5)(40 – x) = 124 ⇒ x2 – 45x + 324 = 0 ⇒ x2 – 36x – 9x + 324 = 0 ⇒ x(x – 36) -9(x – 36) = 0 ⇒ (x – 36)(x – 9) = 0 Thus, we can say, x – 36 = 0 or x – 9 = 0 ⇒ x = 36 or x = 9 Therefore, If, John’s marbles = 36, Then, Jivanti’s marbles = 45 – 36 = 9 And if John’s marbles = 9, Then, Jivanti’s marbles = 45 – 9 = 36 (ii) Let us say, number of toys produced in a day be x. Therefore, cost of production of each toy = Rs(55 – x) Given, total cost of production of the toys = Rs 750 ∴ x(55 – x) = 750 ⇒ x2 – 55x + 750 = 0 ⇒ x2 – 25x – 30x + 750 = 0 ⇒ x(x – 25) -30(x – 25) = 0 ⇒ (x – 25)(x – 30) = 0 Thus, either x -25 = 0 or x – 30 = 0 ⇒ x = 25 or x = 30 Hence, the number of toys produced in a day, will be either 25 or 30.

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