EXERCISE 4.1
Q1(i). Check whether (x + 1)^2 = 2(x – 3) is a quadratic equation.
Answer:
(x + 1)^2 = 2(x – 3) => x^2 + 2x + 1 = 2x – 6 => x^2 + 2x + 1 – 2x + 6 = 0 => x^2 + 7 = 0. This is of the form ax^2 + bx + c = 0 (with a=1, b=0, c=7). Therefore, it is a quadratic equation.
Q1(ii). Check whether x^2 – 2x = (–2)(3 – x) is a quadratic equation.
Answer:
x^2 – 2x = (–2)(3 – x) => x^2 – 2x = –6 + 2x => x^2 – 2x – 2x + 6 = 0 => x^2 – 4x + 6 = 0. This is of the form ax^2 + bx + c = 0. Therefore, it is a quadratic equation.
Q1(iii). Check whether (x – 2)(x + 1) = (x – 1)(x + 3) is a quadratic equation.
Answer:
(x – 2)(x + 1) = (x – 1)(x + 3) => x^2 + x – 2x – 2 = x^2 + 3x – x – 3 => x^2 – x – 2 = x^2 + 2x – 3 => – x – 2 – 2x + 3 = 0 => –3x + 1 = 0. This is not of the form ax^2 + bx + c = 0 (the x^2 term is cancelled, so it is linear). Therefore, it is not a quadratic equation.
Q1(iv). Check whether (x – 3)(2x + 1) = x(x + 5) is a quadratic equation.
Answer:
(x – 3)(2x + 1) = x(x + 5) => 2x^2 + x – 6x – 3 = x^2 + 5x => 2x^2 – 5x – 3 = x^2 + 5x => 2x^2 – 5x – 3 – x^2 – 5x = 0 => x^2 – 10x – 3 = 0. This is of the form ax^2 + bx + c = 0. Therefore, it is a quadratic equation.
Q1(v). Check whether (2x – 1)(x – 3) = (x + 5)(x – 1) is a quadratic equation.
Answer:
(2x – 1)(x – 3) = (x + 5)(x – 1) => 2x^2 – 6x – x + 3 = x^2 – x + 5x – 5 => 2x^2 – 7x + 3 = x^2 + 4x – 5 => 2x^2 – 7x + 3 – x^2 – 4x + 5 = 0 => x^2 – 11x + 8 = 0. This is of the form ax^2 + bx + c = 0. Therefore, it is a quadratic equation.
Q1(vi). Check whether x^2 + 3x + 1 = (x – 2)^2 is a quadratic equation.
Answer:
x^2 + 3x + 1 = (x – 2)^2 => x^2 + 3x + 1 = x^2 – 4x + 4 => 3x + 1 + 4x – 4 = 0 => 7x – 3 = 0. This is not of the form ax^2 + bx + c = 0 (the x^2 term is cancelled). Therefore, it is not a quadratic equation.
Q1(vii). Check whether (x + 2)^3 = 2x(x^2 – 1) is a quadratic equation.
Answer:
(x + 2)^3 = 2x(x^2 – 1) => x^3 + 6x^2 + 12x + 8 = 2x^3 – 2x => 0 = 2x^3 – 2x – x^3 – 6x^2 – 12x – 8 => x^3 – 6x^2 – 14x – 8 = 0. The highest degree is 3, not 2. Therefore, it is not a quadratic equation.
Q1(viii). Check whether x^3 – 4x^2 – x + 1 = (x – 2)^3 is a quadratic equation.
Answer:
x^3 – 4x^2 – x + 1 = (x – 2)^3 => x^3 – 4x^2 – x + 1 = x^3 – 6x^2 + 12x – 8 => – 4x^2 – x + 1 + 6x^2 – 12x + 8 = 0 => 2x^2 – 13x + 9 = 0. The x^3 terms cancel, leaving degree 2. This is of the form ax^2 + bx + c = 0. Therefore, it is a quadratic equation.
Q2(i). The area of a rectangular plot is 528 m^2. The length of the plot (in metres) is one more than twice its breadth. Represent this situation in the form of a quadratic equation to find the length and breadth of the plot.
Answer:
Let the breadth of the plot be x metres. Then, the length = (2x + 1) metres. Area = length × breadth = (2x + 1)x = 2x^2 + x m^2. Given area = 528 m^2. So, 2x^2 + x = 528 => 2x^2 + x – 528 = 0. This is the required quadratic equation.
Q2(ii). The product of two consecutive positive integers is 306. Represent this situation in the form of a quadratic equation to find the integers.
Answer:
Let the two consecutive positive integers be x and x+1. Their product = x(x+1) = x^2 + x. Given product = 306. So, x^2 + x = 306 => x^2 + x – 306 = 0. This is the required quadratic equation.
Q2(iii). Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. Represent this situation in the form of a quadratic equation to find Rohan's present age.
Answer:
Let Rohan's present age be x years. Then, his mother's present age = (x + 26) years. After 3 years, Rohan's age = x + 3 years, mother's age = x + 26 + 3 = x + 29 years. Product of their ages after 3 years = (x + 3)(x + 29) = x^2 + 29x + 3x + 87 = x^2 + 32x + 87. Given product = 360. So, x^2 + 32x + 87 = 360 => x^2 + 32x – 273 = 0. This is the required quadratic equation.
Q2(iv). A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Represent this situation in the form of a quadratic equation to find the speed of the train.
Answer:
Let the original speed of the train be x km/h. Time taken at original speed = distance/speed = 480/x hours. New speed = (x – 8) km/h. Time taken at new speed = 480/(x – 8) hours. Given that the new time is 3 hours more than the original time: 480/(x – 8) = 480/x + 3 => 480/(x – 8) – 480/x = 3 => [480x – 480(x – 8)] / [x(x – 8)] = 3 => (480x – 480x + 3840) / (x^2 – 8x) = 3 => 3840 / (x^2 – 8x) = 3 => 3840 = 3(x^2 – 8x) => 1280 = x^2 – 8x => x^2 – 8x – 1280 = 0. This is the required quadratic equation.
Exercises
Q1. 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
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 or x = –2.
(ii) 2x² + x – 6 = 0 ⇒ 2x² + 4x – 3x – 6 = 0 ⇒ 2x(x + 2) – 3(x + 2) = 0 ⇒ (x + 2)(2x – 3) = 0 ⇒ x = –2 or x = 3/2.
(iii) √2 x² + 7x + 5√2 = 0 ⇒ √2 x² + 2x + 5x + 5√2 = 0 ⇒ √2 x² + 2x + 5x + 5√2 = 0. Multiply by √2: 2x² + 7√2 x + 10 = 0 ⇒ 2x² + 4√2 x + 3√2 x + 10 = 0 ⇒ 2x(x + 2√2) + 3√2(x + 2√2) = 0 ⇒ (x + 2√2)(2x + 3√2) = 0 ⇒ x = –2√2 or x = –3√2/2.
(iv) 2x² – x + 1/8 = 0 ⇒ Multiply by 8: 16x² – 8x + 1 = 0 ⇒ 16x² – 4x – 4x + 1 = 0 ⇒ 4x(4x – 1) – 1(4x – 1) = 0 ⇒ (4x – 1)(4x – 1) = 0 ⇒ x = 1/4, 1/4.
(v) 100x² – 20x + 1 = 0 ⇒ 100x² – 10x – 10x + 1 = 0 ⇒ 10x(10x – 1) – 1(10x – 1) = 0 ⇒ (10x – 1)(10x – 1) = 0 ⇒ x = 1/10, 1/10.
Q2. Solve the problems given in Example 1.
Answer:
Example 1 problems: (i) Represent x² – 45x + 324 = 0, roots 36 and 9; (ii) Represent x² – 55x + 750 = 0, roots 30 and 25. Solutions: (i) x² – 45x + 324 = 0 ⇒ x² – 36x – 9x + 324 = 0 ⇒ x(x – 36) – 9(x – 36) = 0 ⇒ (x – 36)(x – 9) = 0 ⇒ x = 36 or 9. (ii) x² – 55x + 750 = 0 ⇒ x² – 30x – 25x + 750 = 0 ⇒ x(x – 30) – 25(x – 30) = 0 ⇒ (x – 30)(x – 25) = 0 ⇒ x = 30 or 25.
Q3. Find two numbers whose sum is 27 and product is 182.
Answer:
Let numbers be x and 27 – x. Then x(27 – x) = 182 ⇒ 27x – x² = 182 ⇒ x² – 27x + 182 = 0 ⇒ x² – 13x – 14x + 182 = 0 ⇒ x(x – 13) – 14(x – 13) = 0 ⇒ (x – 13)(x – 14) = 0 ⇒ x = 13 or 14. Numbers are 13 and 14.
Q4. Find two consecutive positive integers, sum of whose squares is 365.
Answer:
Let integers be n and n+1. Then n² + (n+1)² = 365 ⇒ n² + n² + 2n + 1 = 365 ⇒ 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 = 13 or n = –14 (reject). 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.
Answer:
Let base = x cm, altitude = (x – 7) cm. By Pythagoras: x² + (x – 7)² = 13² ⇒ x² + x² – 14x + 49 = 169 ⇒ 2x² – 14x – 120 = 0 ⇒ x² – 7x – 60 = 0 ⇒ x² – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x – 12)(x + 5) = 0 ⇒ x = 12 or x = –5 (reject). Base = 12 cm, altitude = 5 cm.
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 ₹90, find the number of articles produced and the cost of each article.
Answer:
Let number of articles = n. Cost per article = (2n + 3). Total cost = n(2n + 3) = 90 ⇒ 2n² + 3n – 90 = 0 ⇒ 2n² – 12n + 15n – 90 = 0 ⇒ 2n(n – 6) + 15(n – 6) = 0 ⇒ (n – 6)(2n + 15) = 0 ⇒ n = 6 or n = –15/2 (reject). Number of articles = 6, cost per article = 2(6) + 3 = ₹15.
Q1 (Exercise 4.3). Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x² – 3x + 5 = 0 (ii) 3x² – 4√3 x + 4 = 0 (iii) 2x² – 6x + 3 = 0
Answer:
(i) a = 2, b = –3, c = 5. Discriminant = (–3)² – 4(2)(5) = 9 – 40 = –31 < 0. No real roots.
(ii) a = 3, b = –4√3, c = 4. Discriminant = (–4√3)² – 4(3)(4) = 48 – 48 = 0. Two equal real roots: x = –b/(2a) = (4√3)/(6) = 2√3/3.
(iii) a = 2, b = –6, c = 3. Discriminant = (–6)² – 4(2)(3) = 36 – 24 = 12 > 0. Two distinct real roots: x = [6 ± √12]/(4) = [6 ± 2√3]/4 = (3 ± √3)/2. Roots: (3 + √3)/2 and (3 – √3)/2.
Q2 (Exercise 4.3). Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x² + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
Answer:
(i) For equal roots, discriminant = 0: k² – 4(2)(3) = 0 ⇒ k² – 24 = 0 ⇒ k = ±√24 = ±2√6.
(ii) Rewrite as kx² – 2kx + 6 = 0. For equal roots, discriminant = 0: (–2k)² – 4(k)(6) = 0 ⇒ 4k² – 24k = 0 ⇒ 4k(k – 6) = 0 ⇒ k = 0 or k = 6. But if k = 0, equation becomes 6 = 0, not quadratic. So k = 6.
Q3 (Exercise 4.3). Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Answer:
Let breadth = b m, length = 2b m. Area = 2b × b = 2b² = 800 ⇒ b² = 400 ⇒ b = ±20. Since breadth positive, b = 20 m, length = 40 m. Yes, possible.
Q4 (Exercise 4.3). Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Answer:
Let present ages be x and 20 – x. Four years ago: (x – 4)(20 – x – 4) = (x – 4)(16 – x) = 48 ⇒ 16x – x² – 64 + 4x = 48 ⇒ –x² + 20x – 64 – 48 = 0 ⇒ –x² + 20x – 112 = 0 ⇒ x² – 20x + 112 = 0. Discriminant = 400 – 448 = –48 < 0. No real roots. So situation is not possible.
Q5 (Exercise 4.3). Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Answer:
Let length = l, breadth = b. Perimeter: 2(l + b) = 80 ⇒ l + b = 40; Area: l × b = 400. So l and b are roots of t² – 40t + 400 = 0. Discriminant = 1600 – 1600 = 0. Roots: t = 20, 20. So l = 20 m, b = 20 m (square). Possible.