or x2 – 7x + 10 = 0 Area of rectangular mango grove = 800 m2 ∴ Length of cloth purchase 64 + x2 – 16x = 9 (6 – x) (b) 4 Hence, p = 0, Question 2. ∴ Put y = 6 in (i) we get (By factorization method) Quadratic Equations Class 10 Extra Questions Short Answer Type 1. (b) ≥ 0, Question 15. If one root of quadratic equation x2 + kx + 3 = 0 is 1, then the value of k will be: Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? Answer: Answer: Solution. ⇒ y2 – 4y + 3 = 0 Let the speed of train be x km/ Hr. ⇒ 4 – 4 – p = 0 Given that – 2 is a root of given quadratic equation x 2 + 2x – p = 0. ⇒ x(x – 13) + 2(x – 13) = 0 Solution. Find the number. ∴ x – y = -4 Solution. (a) 3 (c) 3, Question 4. Given equation is: (d) 1, 3 ⇒ x = 20 x + 15 = 0 ⇒ x (x + 11) = 132 × 11 Now, x – 12 = 0 x = 12 Solution. The zero of the polynomial x2 + 2x – 3 are: (b) -1 x2 – (90x – 30x) – 2700 = 0 According to the question, 2x = 3 ⇒ 4y2 + 18y – 10y – 45 = 0 Suppose two digit number is 10x + y. If = \(\frac {x}{6}\) = \(\frac {6}{x}\), then the value of x will be: ⇒ (2x2 – x2) + (60x – 120x) + (900 – 3600) = 0 Solution. A quadratic equation ax2 + bx + c = 0, a ≠ 0 has equal roots if b2 – 4ac is: Find the lengths of the base if its height is 7 cm more then base. ⇒ 200 = 225 – x2 ⇒ x(x – 18) + 10(x – 18) = 0 ∴ y = 3 or y = –\(\frac{5}{2}\) Let the present age of Rehman – x year ∴ x2 – 2bx + b2 – 4a2 = 0, Question 7. where y = 3, x = 2 + 3 = 5 But (∵ breadth never be negative) (b) -1, 3 (c) -4, Question 7. = Length + Breadth = 40 m x(2x +3) = 90 Four years ago, the product of their ages in years was 48. ⇒ 4 – 4 – p = 0. Given quadratic equation Difference of squares of two numbers = 180 Solution. Let Number of vessel manufacture in a day = x Answer: (d) none of these. (a) b > \(\frac {8}{5}\) and the number of birds moving on a hill An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). Hence, the solutions of the given equation are. (d) 2 (a) -4 [∵ (a + b)2 = a2 + 2ab + b2] Now substitute x + \(\frac {1}{x}\) = y 5p = 35 ⇒ p = 7 If so, determine their present ages. ⇒ (x – 18)(x + 10) = 0 Length = (40 – x) m (a) ±2 Find the roots of the quadratic equation: Therefore, the given equation can be written as (c) ≤ 0 ⇒ x2 + 44x – 33x – 1452 = 0 Suppose the width of Verandah is x metre. Find the number. Also, the number of birds in vakula tree = 56 (b) not real, Question 5. 3x2 – 2√6 x + 2 = 0 If- 5 is a root of the quadratic equation 2x 2 + px -15 = 0 and the quadratic equation p(x 2 + … ∴ x = 2 and x = 5. ⇒ x2 + 100x – 2400 = 0 y2 – 18y – 144 = 0 (d) 8 (c) -1, -3 (√3x – √2)2 = 0 (d) \(\frac {4}{3}\) Breadth of verandah = (2x + 12) m. According two question ∴ x2 – (sum of roots) x + product of roots = 0 ⇒ x2 + 12x – 5x – 60 = 0 Where (b) two zeroes Solve the quadratic equation ⇒ x = – 30 which is not possible because side cannot be negative. ⇒ 2 × 30 × 30 = 9(225 – x2) (a) 1 Solution. (∵ The sum of the ages of two friends is 20 yr) ⇒ y2 = [8 (-10)] = -80 ∴ x = 18 or x = – 10 (i) and (ii), we have On a particular day it is observed that the cost of manufacturing each item was 3 more than twice the number of vessels manufactured on that day. If one root of quadratic equation ax2 + bx + c = 0 is 1, then: Solution. (b) ≥ 0 (b) b < \(\frac {-8}{5}\) (c) ±4, Question 10. ⇒ x2 + 300x – 250x – 75000 = 0 speed to stream = 5 km/Hr. ∴ (-2)2 + 2 (-2) – p = 0 ⇒ -7x + 126 √x + 1008 = 0 Find the speed of stream. ∴ y = 24 The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is a. ⇒ x2 + 11x – 1452 = 0 ⇒ x2 – 2x × 20 + (20)2 = 0 (a) real and unequal x (6) = 12 ⇒ (2x)2 + 2.2xb + (b)2-(a)2 = 0 y – 6 = 0 or y + 2 = 0 We have, Had the price of cloth been ₹ 50 per metre higher, then he would have been able to purchase the cloth less by 1.5 metre. Let the breadth of the park = x metre Area of right angle ∆ = \(\frac {1}{2}\) × base × height (d) ∴ 6x + 6 = x2 + 2x – 15 2(x2 + 1) = 5x If the longer side is 30 m more than the shorter side, find the sides of the field. 4y2 + 8y – 45 = 0 ⇒ (x – 12) (x + 5) = 0 (d) 1 Question 12. Answer: ∴ x = 8 Answer: ⇒ x = -44 or x – 33 = 0 But y = -6, since √x = y is positive Out of a certain number of Saras birds, one-fourth of the number are moving about in lotus plants, \(\frac {1}{9}\)th coupled with \(\frac {1}{4}\)th as well as 7 times the square root of the number move on a hill, 56 birds remain in vakula tree. Solution. ⇒ 2y2 + 4y – 30 = 0 ⇒ y(y – 3)-1 (y – 3) = 0 Find the speed of stream. Question 13. ⇒ 64x = 576 – x2 Given that – 2 is a root of given quadratic equation x2 + 2x – p = 0 as number of vessels are not negative NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Given, quadratic equation = (3.12 – 7.5)2 Here a = 9, b = 7 and c= -2 (a) Question 1. Where ⇒ 2x2 – 14x – 120 = 0 and speed of boat in up stream = (15 – x) km/H Answer: ∴ 4x2 + 54x – 90 = 0 (d) none of these. Quadratic Equations Class 10 Maths Chapter 4 Important questions with solutions for board exam 2019-2020 are available here. and length of the park = (40 – 20) m = 20 m ∴ x = 5 or – 12 Let PQRS be the rectangular field. Time taken by passenger train to cover (c) 6 Answer: (a) 0, Question 22. ⇒ \(\frac {x}{9}\) + \(\frac {x}{4}\) + 7√x Hence, the numbers are (18 and 12) or (18 and -12). ∴ x = 90 ∴ x = 6 or – \(\frac {15}{2}\) Now, we first transform one of the radicals co the R.H.S. ⇒ (13)2 = x2 + (x – 7)2 Answer: Let the positive integer be x then according to the given condition, Question 4. β = b + 2a Answer: x = -300 Question 14. ⇒ Here a =1, b = -4, and c = -8. = 25 Find the equation whose roots are b – 2a and b + 2a. ⇒ (x + 2)(2x + 1) = 0 The speed of boat in still water is 15km/H. (b) -2, Question 8. ⇒ (x + 300)(x – 250) = 0 ∴ (y – 6) (y + 2) = 0 If the average speed of the express train is 11 kmh-1 more than that of the passenger train, find the average speed of the two trains. If 2x2 + 1 = 33, then the value of x will be: = 400 – 448 = 48 < 0 Now, new price of cloth per metre become = ₹ (x + 50) One root of quadratic equation x2 + 3x – 10 = 0: Amount spend = ₹ 2250 ∴ y = 3 or 1. ⇒ (y – 3)(2y + 5) = 0 (b) -1, -2 Solution. Less than its base = (x – 7) cm (d) -5. ⇒ (x – 7)(x + 3) = 0 Question 3. = b2 – 4a2 ⇒ y2 + 3 – 4y = 0 ∴ Equation reduced to Question 1. By Pythagoras theorem, we have (y – 4)y = 12 Solution. (b) two zeroes, (ii) From the equation 4√5 x2 + 7x – 3√5 = 0, the value of x will be: Let the total number of birds be x. x = 2 x2 – (12x – 5x) – 60 = 0 We have: When So, present age of Rehman = 7 yr. Now, substituting y = x + \(\frac {1}{x}\), we get (x – 4)(16 – x) = 48 Solution. Question 5. We must look for solutions which satisfy Answer: ⇒ (2x + b)2-(a)2 = 0 Find the positive integer. Putting the value of p = 7 in given (ii) quadratic equation. ⇒ x(a + b + x) = – ab Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers, Question 1. Question 2. a = 1, b = -20 and c = 112 We have ∴ 2(-5)2 + p(-5) – 15 = 0 Now, from (ii), we get, ∴ 2x(x + 15) – 3(x + 15) = 0 Let the speed of stream = x km/ Hr. ∴ Discriminant, D = b2 – 4ac Answer: i.e., –\(\frac {10}{3}\) ≤ x ≤ 6. Answer: …(i) 9x + 2x + 126 √x – 18x + 1008 = 0 (b) (a) 5 If the sum of the roots of the equation 3x2 + (2k + 1)x – (k + 5) = 0 be equal to their product, the value of k is: By solving the equation (2011OD) Solution: x2 – 3x – m(m + 3) = 0 […] ⇒ (2x + b + a) (2x + b – a) = 0, Question 5. x2 – (7x – 3x) – 21 = 0 Solution. In ∆ABC AC2 = BC2 + Ab2 But according to the question, ⇒ 2y (2y + 9) – 5(2y +9) = 0 y = 3 = \(\frac{x^{2}+1}{x}\) ⇒ Length × Breadth = 2x (x) = 800 ⇒ x2 = 400 ⇒ x = ± 20 (c) three zeroes If so, find its length and breadth. Rehman’s age, after 5 yr = (x + 5) yr ⇒ x2 + 50x – 75000 = 0 You can find other MCQ : Quadratic Equations - 1 extra questions, long questions & short questions for Class 10 on EduRev as well by searching above. An expression in α and β is called symmetrical expression if by interchanging α and β, the expression is: Nature of root, D = b2 – 4ac = (4)2 – 4 × 1 × 0 = 16 > 0 from (i) x = 2 + y (b) -3, Question 23. If one root of x2 + kx + 3 = 0 is 1, then the value of k will be: (∵ In rectangle every adjacent side makes an angle 90° to each other) Now, x = 18 ⇒ (x + 60)2 = (x + 30)2 + x2 (d) none of these. (a) equal to 0 (d) none of these. Answer: Answer: ⇒ x2 + 7x – 60 = 0 ∴ x2 – (2b)x + (b2 – 4a2) = 0 ⇒ 169 = x2 + x2 – 14x + 49 (a) changed x2 – 4x − 21 = 0 Let = \(\frac{x-2}{x+2}\) = y Find the nature of roots. = (√3x)2 – 2√3x × √2 +(√2)2 = 0 The roots of ax2 + bx + c = 0, a 50 are real and unequal, if b2 – 4ac is: Solution. If the speed of a train is increased by 5 km/H then train takes 1 hour less time to cover a distance of 360 km. Question 7. ⇒ (x + 12) (x – 5) = 0 x ≠ 0. Squaring both sides, we get Question 2. ⇒ y = 12 or -12 [From EQuestion (ii)] y + \(\frac {3}{y}\) – 4 = 0 ⇒ x = 20 (d) none of these. Again, x = -10 ⇒ x(x + 300) – 250(x + 300) = 0 4x2 + 4bx – a2 + b2 = 0 Let age of one of the two friends = x yr ∴ k = \(\frac {7}{4}\). Diagonal of the rectangle PR = 60 m more than the shorter side = (x + 60) m Determine whether the quadratic equation 9x2 + 7x – 2 = 0 has real roots. ∴ According to the given information, we have takes 1 hour more to go 32 km up stream than to return down stream to the same spot. ⇒ 3x2 + 16x + 16 = 4(x2 + 3x + 2) Solution. Side of the rectangle PQ = 30 m more than the shorter side = (x + 30) m ∴ According to the question, ⇒ k2 ≥ 256 and 64 ≥ 4k ⇒ y = 24 or y = -6. and the average speed of the express train = (33 + 11) km/h = 44 km/h. (a) 6 Is the following situation possible ? (d) a + b + c = 0 If x2 – 5 = 0, then find the value of x: Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations. and x2 – 8x + k = 0…(ii) ∴ 2x2 + 30x – 3x – 45 = 0 ∴ 2x2 + 27x – 45 = 0 (c) –\(\frac {4}{3}\), Question 21. Solution. In ∆PQR, PR2 = PQ2 + QR2 ⇒ (x + 44)(x – 33) = 0 = \(\frac{2250}{x + 50}\) metre. (b) The altitude of a right triangle is 7 cm less than its base. (c) 1 If one root quadratic equation x 2 + 2x – p = 0 is -2, then find the value of p. Solution. (c) ≤ 0 Putting √x = y, we get as speed is never negative:. ⇒ x(x +12) – 5(x + 12) = 0 ⇒ x(x + 72) – 8(x + 72) = 0 (x – 8)(x + 72) = 0 which implies that the real roots are not possible because this condition represents imaginary roots. ⇒ y2 = 8 × 18 = 144 x = 3/2 Answer: If aand Bare the roots of the equation ax2 + bx + b = 0, then prove that: ∴ x + 44 = 0 (c) 1 = (36 – 35)2 = 1, Question 4. Question 8. If one root of quadratic equation x3 – 3x + 2 = 0 is 2, then the second root is: Short Answer Type Questions I [2 Marks] Question 1. ∴ Length of cloth purchased 3x + 10 = 36 + 6 – x – 12\(\sqrt{6-x}\) (d) none of these. Question 9. The zero of the polynomial p(x) = x2 + 1 are: (a) 1, -3, Question 3. (c) –\(\frac {4}{3}\) which satisfy \(\frac{-10}{3}\) ≤ x ≤ 6. ∴ Area of a rectangular park = Length × Breadth = (40 – x)x m2 x2 = 11x + 26 ∴ roots of quadratic equation are real. as speed is never negative Length of verandah = (2x + 15) m. (c) ±4 ⇒ x( 2x + 15)-6 (2x + 15) = 0 ⇒ x(x – 7) + 3(x – 7) = 0 xy = 12 …..(i) ∴ Number of birds moving about in lotus plants = \(\frac {x}{4}\) ∴ length of its height = (x + 7) cm (b) ±3 Question 11. (c) not real (a) real x2 – (18x – 10x) – 180 = 0 If the total manufacturing cost on that day was 90, then find the number of manufactured vessels and manufacturing cost of each item. Question 2. 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