Exercise 4.2 answers: Quadratic Equations

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(i) x² − 3x – 10 = 0

1. We need two numbers with sum −3 and product −10: they are −5 and 2.
2. Split the middle term: x² − 5x + 2x − 10 = 0.
3. Group: x(x − 5) + 2(x − 5) = 0, so (x − 5)(x + 2) = 0.
4. Set each factor to zero.

Answerx = 5 or x = −2.

(ii) 2x² + x – 6 = 0

1. a×c = −12 and b = 1, so we need two numbers with product −12 and sum 1: 4 and −3.
2. Split the middle term: 2x² + 4x − 3x − 6 = 0.
3. Group: 2x(x + 2) − 3(x + 2) = 0, so (2x − 3)(x + 2) = 0.
4. Set each factor to zero.

Answerx = 3/2 or x = −2.

(iii) √2 x² + 7x + 5√2 = 0

1. a×c = √2×5√2 = 10 and b = 7, so we need two numbers with product 10 and sum 7: 2 and 5.
2. Split the middle term: √2x² + 2x + 5x + 5√2 = 0.
3. Group: √2x(x + √2) + 5(x + √2) = 0, so (√2x + 5)(x + √2) = 0.
4. Set each factor to zero.

Answerx = −√2, or x = −5/√2 = −5√2/2.

(iv) 2x² − x + 1/8 = 0

1. Multiply throughout by 8 to clear the fraction: 16x² − 8x + 1 = 0.
2. We need two numbers with product 16 and sum −8: −4 and −4.
3. Split the middle term: 16x² − 4x − 4x + 1 = 0.
4. Group: 4x(4x − 1) − 1(4x − 1) = 0, so (4x − 1)² = 0.

Answerx = 1/4 (a repeated root).

(v) 100x² − 20x + 1 = 0

1. We need two numbers with product 100 and sum −20: −10 and −10.
2. Split the middle term: 100x² − 10x − 10x + 1 = 0.
3. Group: 10x(10x − 1) − 1(10x − 1) = 0, so (10x − 1)² = 0.

Answerx = 1/10 (a repeated root).

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(i) John and Jivanti together have 45 marbles.

1. Example 1 already turns this into x² − 45x + 324 = 0, where x is John's starting number of marbles.
2. We need two numbers with sum −45 and product 324: −9 and −36.
3. Split the middle term: x² − 9x − 36x + 324 = 0.
4. Group: x(x − 9) − 36(x − 9) = 0, so (x − 9)(x − 36) = 0, giving x = 9 or x = 36.
5. Taking John's marbles as x = 36, Jivanti's marbles are 45 − 36 = 9. (The equation is equally happy the other way round.)

AnswerJohn had 36 marbles and Jivanti had 9 (or the reverse — the equation cannot tell the two apart).

(ii) the total cost of production was ₹ 750.

1. Example 1 turns this into x² − 55x + 750 = 0, where x is the number of toys made that day.
2. We need two numbers with sum −55 and product 750: −25 and −30.
3. Split the middle term: x² − 25x − 30x + 750 = 0.
4. Group: x(x − 25) − 30(x − 25) = 0, so (x − 25)(x − 30) = 0, giving x = 25 or x = 30.
5. Both are sensible whole numbers of toys, since the cost per toy (55 − x) stays positive either way.

Answer25 or 30 toys were produced that day — both values fit the given information equally well.

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1. Let one number be x; since the two numbers add to 27, the other is 27 − x.
2. Their product is 182: x(27 − x) = 182.
3. Rearrange: x² − 27x + 182 = 0.
4. We need two numbers with sum −27 and product 182: −13 and −14.
5. Split the middle term: x² − 13x − 14x + 182 = 0, so x(x − 13) − 14(x − 13) = 0, giving (x − 13)(x − 14) = 0.

AnswerThe two numbers are 13 and 14.

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1. Let the smaller positive integer be x; the next consecutive integer is x + 1.
2. Sum of squares: x² + (x + 1)² = 365.
3. Expand: 2x² + 2x + 1 = 365, so x² + x − 182 = 0.
4. We need two numbers with sum 1 and product −182: 14 and −13.
5. Split the middle term: x² + 14x − 13x − 182 = 0, so (x + 14)(x − 13) = 0, giving x = −14 or x = 13.
6. Since the integer must be positive, reject x = −14.

AnswerThe two consecutive positive integers are 13 and 14.

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1. Let the base be x cm; the altitude is 7 cm less, so it is (x − 7) cm.
2. By Pythagoras' theorem: x² + (x − 7)² = 13².
3. Expand: 2x² − 14x + 49 = 169, so x² − 7x − 60 = 0.
4. We need two numbers with sum −7 and product −60: −12 and 5.
5. Split the middle term: x² − 12x + 5x − 60 = 0, so (x − 12)(x + 5) = 0, giving x = 12 or x = −5.
6. Since a side length cannot be negative, reject x = −5.

AnswerThe base is 12 cm and the altitude is 12 − 7 = 5 cm.

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1. Let the number of articles produced be x.
2. The cost of each article is 3 more than twice the number of articles: (2x + 3) rupees.
3. Total cost = x(2x + 3) = 90, so 2x² + 3x − 90 = 0.
4. a×c = −180 and b = 3, so we need two numbers with product −180 and sum 3: 15 and −12.
5. Split the middle term: 2x² + 15x − 12x − 90 = 0, so x(2x + 15) − 6(2x + 15) = 0, giving (2x + 15)(x − 6) = 0.
6. So x = −15/2 or x = 6. Since the number of articles cannot be negative, reject x = −15/2.

Answer6 articles were produced, at a cost of 2(6) + 3 = ₹15 each.

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