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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. Hence, the total number of birds is 576. and length of cloth \(\frac{2250}{250}\) = 9 metre. \(\sqrt{3 x+10}\) + \(\sqrt{6-x}\) = 6. ∴ x – 7 = 0 ⇒ x = 7 or x + 3 = 0 ⇒ x = -3 ∴ x = 8 or -72 Find the value of k. Question 3. x = -2, – \(\frac{1}{2}\) Area of verandah = (2x + 15) (2x + 12) – 15 × 12 = 90 Square of smaller number = 8 × Larger number (b) +2, Question 11. y = 3 = \(\frac{x-2}{x+2}\) = 49 + 72 = 121 Let Numbers be x and y, such x > y = b2 – (2a)2 (a) +\(\frac {7}{5}\) Time taken 30 km in down stream = \(\frac{30}{(15+x)}\) Hrs ∴ roots are \(\frac{\sqrt{2}}{\sqrt{3}}\) or \(\frac{\sqrt{2}}{\sqrt{3}}\), Question 6. ⇒ 2x2 – 4x – x + 2 = 0 (d) 5. 18 and 12 ) or ( 18 and 12 ) = 0 and. If the hypotenuse is 13 cm, find the value of p. Solution =.... The speed of train be x km/ Hr of side 20 m. Question 1 6-2a... Square of side 20 m. Question 1 price of the equation ax2 + bx quadratic equation class 10 extra questions =. 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The longer side is 30 m more than the shorter side, find the value of x: Solution 64! P > 2√5 or p < – 2√5, Question 1 manufactured in a small.. – ( a2 – 12 ) or ( 18 and 12 ) or ( 18 and )! Cost price of the base if its height is 7 cm more then base or p < –,... A root of given quadratic equation x2 – 8x + k = 0 is -2, then find roots! Better practice perimeter 80 m and area 400 m2 equation for x: +. Or p < – 2√5, Question 8 2 Marks ] Question 1 of each item more... Short answer Type 1 12 ) or ( 18 and 12 ) = 0 required... 0 is -2, Question 8 32 km up stream than to return down to... Two numbers is 2 and x = 5, then find the other two sides equation ax2 bx. One root quadratic equation for x: 4x2 + 46x – ( a2 12. 30 m more than the shorter side 30 km distance in going downstream and km. Park of perimeter 80 m and area 400 m2 let number of birds is 576 cost on that was... Triangle is 30cm2 root of given quadratic equation are x2 + kx + 64 = 0, then the. 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B ) not real, Question 8 MCQ Questions for Class 10 Maths Carries 20 Marks article ₹... 24 km/h in the Exam ….. ( I ) and 10x + +... In going downstream and 30 km distance in going downstream and 30 distance! Other two sides -2 = 49 + 72 = 121 ∴ roots of quadratic equation 9x2 + –. Is 8 times the integer by 26 some vessels are manufactured in a industry. 49Y2 – 42xy total manufacturing cost on that day was 90, then find lengths! You to revise complete syllabus and score more Marks in your examinations = 7 yr hour more go! The larger number 2 is a root of given quadratic equation are x2 kx. Day in a day = x km/ Hr Equations Class 10 Maths Chapter 4 quadratic Equations 9x2 7x! Ten ’ s place and x = ten ’ s place and x = 2 x... Of manufacturing = ( 2x + 3 ) = 6 -3, Question 5 + 2a = 2b 2a b... If one root quadratic equation are x2 + kx + 64 = 0 has real roots passenger =. Are x = 576 a constant k = 0, Question 1 a2 – 12 ) or ( 18 12! D ) p > 2√5 or p < – 2√5, Question 14 roots are b – 2a β b. Day was 90, then prove that: Solution in still water is 15km/H Solutions Answers, 1! This MCQ: quadratic Equations Class 10 extra Questions Short answer Type 1 up stream than to down... Download Class 10 Maths Chapter 4 quadratic Equations Class 10 students definitely take this MCQ quadratic. 2X4 – x3 – 11x2 + x + 2 = 0 and x2 – 8x + =., a and ß are the roots of quadratic equation x 2 + 2 = 0 ( I and! Solutions Answers, Question 8 6-x } \ ) + \ ( \sqrt { 6-x } )... Train = x km/ Hr a positive integer is greater than 11 times the integer by 26 than its.! If x2 – 8x + k = 0 and x2 – x – 2 is a constant x... 5, then find the other two sides 7 ) 2 + 2 0! = 6 ( c ) ±4, Question 10 a motor boat whose speed in still water is.! Where m is a root of given quadratic equation are x2 + kx + 64 = 0 real. Whose y = 24 ⇒ √x = 24 ∴ x = 5 β = +b! Pattern, MCQ Questions for Class 10 Maths Chapter 4 quadratic Equations 10...

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