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∴ p = 0 ∴ x = -4 ∴ x = -300 or x = 250 = (-20)2 – 4 × 1 × 112 Solution. ⇒ x2 – 2x × 20 + (20)2 = 0 Question 6. x = y – 4 (d) none of these. x + 2 x + 2 = x – 2 which is not true y = 3 = $$\frac{x-2}{x+2}$$ ⇒ (x + a) (x + b) = 0 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. Square of smaller number = 8 × Larger number New speed of the train = (x + 5) km/Hr. ⇒ x2 + 120x + 3600 = x2 + 60x + 900 + x2 (c) -1, -3 Since, a and ß are the roots of the equation ⇒ 2y(y – 3) + 5(y – 3) = 0 Solution. Since, x ≠ -44 because speed can’t be negative. ⇒ y = ± 12 Find the measure of the cloth purchased and price per metre. as speed is never negative:. But this rectangle will be a square of side 20 m. Question 1. (b) two zeroes (a) a = 1 (a) p > √15 (d) p > 2√5 or p < – 2√5. 4x2 + 46x – (a2 – 12) = 0 Let PQRS be the rectangular field. (c) ±4 ⇒ (x – 7)(x + 3) = 0 (ii) 64 – 4k ⇒ 2x2 = 800 Question 10. (c) -2√5 < p < 2√5 If the roots of 5x2 – px + 1 = 0 are real and distinct, then: (d) ±1. (d) 1, 3 Solution. as speed is never negative Solution: (x + 1)(x – 2) + x = 0 ⇒ x 2 – x – 2 + x = 0 ⇒ x 2 – 2 = 0 D = b 2 – 4ac ⇒ (-4(1)(-2) = 8 > 0 ∴ Given equation has two distinct real roots. (a) 0, Question 22. If the equations have real roots, then D ≥ 0 Discriminant of (i) k2 – 256 and equation. ∴ Discriminant, D = b2 – 4ac Answer: [∵ (a + b)2 = a2 + 2ab + b2] and length of the park = (40 – 20) m = 20 m Question 16. (a) one zero The boat completes 30 km distance in going downstream and 30 km up stream in total 4 Hr 30 minute. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks. Solution. If – 5 is a root of quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots. y2 – 18y – 144 = 0 (b) 4 (c) 1 ⇒ y(y – 3)-1 (y – 3) = 0 If x2 + 5bx + 16 = 0 has no real roots, then: (c) 3 Answer: Then, age of other friend = (20 – x) yr (b) ≥ 0 9x2 + 7x – 2 = 0 4 yr ago age of the other friend An expression in α and β is called symmetrical expression if by interchanging α and β, the expression is: ⇒ x – 18√x – 144 = 0 (b) -3, Question 23. Question 12. Now, x = 18 (a), Question 6. ∴ x = 576. (a) equal to 0 ∴ y(y – 6) + 2(y – 6) = 0 In ∆PQR, PR2 = PQ2 + QR2 (c) 2 Solution. (b) The square of a positive integer is greater than 11 times the integer by 26. ⇒ x2 + (12 – 5)x – 60 = 0 Let the length of base of triangle = x cm (c) –$$\frac {7}{3}$$ Let the shorter side QR of the rectangle = x m. Given equation is: Solve the following equation (x + 7) x2 – 8x = 180 (b) –$$\frac {3}{4}$$ Let the positive integer be x then according to the given condition, Question 7. x = 2 ⇒ (x – 13)(x + 2) = 0 Find the equation whose roots are b – 2a and b + 2a. (c) –$$\frac {4}{3}$$ (b) b = 1 …(i) Answer: ∴ x = -2 or x = 13 If aand Bare the roots of the equation ax2 + bx + b = 0, then prove that: Again, x = -10 (c) 1 ⇒ 2x2 – 4x – x + 2 = 0 The degree of the polynomial x3 – x + 7 is: Now, from (ii), we get, (a) 1 Solution. ∴ roots of quadratic equation are real. ⇒ x(a + b + x) = – ab ∴ x2 – (2b)x + (b2 – 4a2) = 0 (b) ≥ 0, Question 15. which satisfy $$\frac{-10}{3}$$ ≤ x ≤ 6. Solution. x + 15 = 0 Now, x – 12 = 0 x = 12 Hence, k = 16. The diagonal of a rectangular field is 60 m more than the shorter side. Solve the quadratic equation Solution. ∴ By Pythagoras theorem, Some vessels are manufactured in a day in a small industry. Solution. We have 5p = 35 ⇒ p = 7 ∴ x = 6 (a) real and unequal, Question 17. ⇒ (x + 45) (x – 40) = 0 ⇒ (7)2 = 4 × 7 × k β = b + 2a y = 3 = $$\frac{x^{2}+1}{x}$$ (c) 1, Question 1. Solution. (b) Where Also, solve some extra questions at BYJU'S for better practice. If so, find its length and breadth. Altitude of the triangle = 7 cm ⇒ (y – 3)(2y + 5) = 0 (b) not changed, Question 18. If so, determine their present ages. The roots of ax2 + bx + c = 0, a 50 are real and unequal, if b2 – 4ac is: x2 + (44 – 33)x – 1452 = 0 Short Answer Type Questions I [2 Marks] Question 1. 4 yr ago age of one of the two friends = (x – 4) yr Find the number. If α and β are the roots of the equation x2 – x – 2 = 0, find that quadratic equation whose roots are (2α + 1) and (2β + 1). (d) 1, 2. If the total manufacturing cost on that day was 90, then find the number of manufactured vessels and manufacturing cost of each item. = 25 ⇒ Here a =1, b = -4, and c = -8. x2 – (18x – 10x) – 180 = 0 Length of verandah = (2x + 15) m. ∴ speed of train = 40 km/hour. Question 3. ⇒ ax + bx + x2 + ab = 0 Answer: Solution. Question 10. But y = -6, since √x = y is positive ⇒ (2y + 9) (2y – 5) = 0 Less than its base = (x – 7) cm ⇒ x2 + 72x – 8x – 576 = 0 and 10x + y + 36 = 10y + x. ∴ x – y = -4 Given that the speed of boat in still water is 24 km/h by factorization method or x2 – 7x + 10 = 0 (a) 6 The solution of quadratic equation x2 – x – 2 = 0 are: ∴ length of its height = (x + 7) cm Solve the equation Solution. (d) none of these. The zero of the polynomial p(x) = x2 + 1 are: 2(x2 + 1) = 5x or 8 – x = 3$$\sqrt{6-x}$$ Then, according to the question, Perimeter of a rectangular park = 80 m ⇒ √x = 24 (d) ⇒ 2x2 + 5x + 2 = 0 Let the required numbers be x and y, where x > y ⇒ 3x2 + 16x + 16 = 4(x2 + 3x + 2) (a) real Justify your answer. Hence, the solutions are x = 2 and x = 5. Question 4. ⇒ 2y2 + 4y – 30 = 0 Find the lengths of the base if its height is 7 cm more then base. ⇒ (3x + 4)(x + 4) = 4(x + 1)(x + 2) ⇒ x2 + 12x – 5x – 60 = 0 Amount spend = ₹ 2250 Let the price per metre of cloth be ₹ x (√3x – √2)2 = 0 ⇒ x(x – 90) + 30(x – 90) = 0 (d) 3. ⇒ (x + 300)(x – 250) = 0 Let breadth of a rectangular mango grove = x metre ∴ According to the question, or 64 + x2 – 16x – 54 + 9x = 0 We must look for solutions which satisfy (b) -1 (d) none of these. ∴ roots are. = b2 – 4a2 where y = 3, x = 2 + 3 = 5 ∴ x – 7 = 0 ⇒ x = 7 or x + 3 = 0 ⇒ x = -3 (b) not real Question 14. If one root of quadratic equation x2 + kx + 3 = 0 is 1, then the value of k will be: ∴ y = 3 or – 5 Also, the number of birds in vakula tree = 56 ∴ y = 3 or y = –$$\frac{5}{2}$$ ∴ 2x2 + 27x – 45 = 0 x2 + kx + 64 = 0. Find the speed of the train. (d) none of these. ∴ Length of a rectangular mango grove = 2x metre(By given condition) 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: (d) none of these. ⇒ x2 + 100x – 2400 = 0 x2 – 90x + 30x – 2700 = 0 Question 3. from (i) x = 2 + y (a) 5 Let the average speed of the passenger train = x kmh-1. Solution. ∴ length of base = 5 cm. 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. = (7)2 – 4 × 9 × -2 ⇒ 169 = x2 + x2 – 14x + 49 ∴ Number of birds moving about in lotus plants = $$\frac {x}{4}$$ (b) b < $$\frac {-8}{5}$$ as speed is never negative ⇒ x2 – 20x + 112=0 ⇒ $$\frac {x}{9}$$ + $$\frac {x}{4}$$ + 7√x If the longer side is 30 m more than the shorter side, find the sides of the field. Question 1. When On comparing the above equation with ax2 + bx + c = 0, we get (2011OD) Solution: x2 – 3x – m(m + 3) = 0 […] ⇒ k ≥ 16 and 16 ≥ k. (c) $$\frac {-8}{5}$$ < b < $$\frac {8}{5}$$, Question 19. (a) -4 ⇒ x(x – 18) + 10(x – 18) = 0 ⇒ 2y (2y + 9) – 5(2y +9) = 0 ∴ b2 = 4ac Question 6. (∵ The sum of the ages of two friends is 20 yr) Given that – 2 is a root of given quadratic equation x2 + 2x – p = 0 Let the speed of train be x km/ Hr. ⇒ y2 = [8 (-10)] = -80 Question 5. Find the nature of roots. ⇒ (13)2 = x2 + (x – 7)2 Neglecting the negative value of x, we have x = 13. Solution. Solution. D = b2 – 4ac . Let the breadth of the park = x metre Question 2. The width of verandah = 3/2 m. Question 11. Question 9. Find the positive value of k. Using formula, solve the quadratic equation. The required quadratic equation is Given equation is ⇒ x( 2x + 15)-6 (2x + 15) = 0 y + $$\frac {3}{y}$$ – 4 = 0 Answer: Diagonal of the rectangle PR = 60 m more than the shorter side = (x + 60) m 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). (c) ±4, Question 10. x ≥ –$$\frac {10}{3}$$ and x ≤ 6 Hence, cost of cloth per metre = ₹ 250. (x – 4)(16 – x) = 48 According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks. (d) 5. But x +-3 age cannot be negative. Answer: (x + 15) (2x – 3) = 0 ⇒ x(x + 72) – 8(x + 72) = 0 (x – 8)(x + 72) = 0 x2 – (7x – 3x) – 21 = 0 The product of the roots of quadratic equation 3x2 – 4x = 0) is: Hence, base of the triangle = 12 cm Answer: (d) –$$\frac {4}{3}$$ ⇒ 2x2 + 15x – 12x – 90 = 0 ⇒ x2 – 3x + 1 = 0 (d) 1 Now, we first transform one of the radicals co the R.H.S. Answer: ∴ x2 – (sum of roots) x + product of roots = 0 ⇒ x2 + 45x – 40x – 1800 = 0 By solving the equation ⇒ x(x + 300) – 250(x + 300) = 0 Let the speed of stream be x km/h. Again, squaring both sides, we get ⇒ x = 20 Answer: Hence, the total number of birds is 576. Product of roots = αß = (b – 2a)(b + 2a) Where (c) –$$\frac {4}{3}$$ (b) 4 (d) 2 If one root of quadratic equation 2x2 + px – 4 = 0 is 2, then the value of p will be: ⇒ y2 – 3y – y + 3 = 0 Given that – 2 is a root of given quadratic equation x 2 + 2x – p = 0. ∴ 4x2 + 54x – 90 = 0 ⇒ -7x + 126 √x + 1008 = 0 (d) none of these. Hence, the average speed of the passenger train = 33 km/h ⇒ x2 + 5x – 1800 = 0 ⇒ x2 = 400 Question 8. ∴ New time taken to cover 360 km (c) $$\frac {-8}{5}$$ < b < $$\frac {8}{5}$$ (c) (a) and (b) both ⇒ y2 = 8 × 18 = 144 Solution. Now, substituting y = x + $$\frac {1}{x}$$, we get A motor boat whose speed in still water is 24 km/h. Let the speed of stream = x km/ Hr. ⇒ (2 + y)2 + y2 = 34 Solution. Question 11. Is the following situation possible ? Quadratic Equations Class 10 Extra Questions Very Short Answer Type. Quadratic Equations Class 10 Extra Questions Short Answer Type 1. Answer: From Eqs. ⇒ 30 = $$\frac {1}{2}$$x. Here, a = 1, b = -3 and c = 1 = (3.12 – 7.5)2 ∴ x = 40 or -45 Solution. Answer: (c) 6. ⇒ y2 – 4y + 3 = 0 Solution. A two-digit number is such that the product of the digits is 12, when 36 is added to the number, the digit interchange their places. 10X + y + 36 = 10y + x the Equations x2 + 2x p. Is 60 m more than the shorter side 3 ) and x 5. 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