Hence, the solutions of the given equation are. Product of roots = αß = (b – 2a)(b + 2a) (a) 0 Answer: x = 2 ∴ root of the equation are real. (c) 1, Question 1. ∴ y = 3 or 1. If x2 + 5bx + 16 = 0 has no real roots, then: Solve \(\sqrt{3 x+10}\) + \(\sqrt{6-x}\) = 6. ∴ x = 6 Solve: 2x4 – x3 – 11x2 + x + 2 = 0. According to the question, Find his present age. Answer: If the roots of 5x2 – px + 1 = 0 are real and distinct, then: Let the required numbers be x and y, where x > y (c) 6. Justify your answer. Side of the rectangle PQ = 30 m more than the shorter side = (x + 30) m ∴ 2x2 + 30x – 3x – 45 = 0 (b) –\(\frac {7}{5}\) x = 3/2 3x2 – 2√6 x + 2 = 0 The given equation is: Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? (d) 3. Solution. x = √25 = ± 5 If one root of quadratic equation x3 – 3x + 2 = 0 is 2, then the second root is: ∴ 4x2 + 54x – 90 = 0 Then, according to the question, Perimeter of a rectangular park = 80 m ∴ Number of birds moving about in lotus plants = \(\frac {x}{4}\) We have Now, new price of cloth per metre become = ₹ (x + 50) New speed of the train = (x + 5) km/Hr. Given that – 2 is a root of given quadratic equation x2 + 2x – p = 0 ∴ According to the given information, we have Let the shorter side QR of the rectangle = x m. ⇒ x(x – 7) + 3(x – 7) = 0 Let the average speed of the passenger train = x kmh-1. [∵ (a + b)2 = a2 + 2ab + b2] Let the base of the right triangle = x ∴ roots of quadratic equation are real. (a) 5 ∴ Area of a rectangular park = Length × Breadth = (40 – x)x m2 x ≥ –\(\frac {10}{3}\) and x ≤ 6 Solution. 4 yr ago age of one of the two friends = (x – 4) yr Hence, the numbers are (18 and 12) or (18 and -12). ∴ x – y = 2 …..(i) (b) +2, Question 11. Putting the value of p = 7 in given (ii) quadratic equation. (By fatorization method) takes 1 hour more to go 32 km up stream than to return down stream to the same spot. ⇒ (x + 2)(2x + 1) = 0 x2 – 60x – 2700 = 0 Question 5. (c) ≤ 0 ⇒ (7)2 = 4 × 7 × k ⇒ x2 + 1 = 3x ⇒ x2 – 40x + 400 = 0 ∴ p = 0 ⇒ x(x + 45) – 40(x + 45) = 0 as the roots of this equation are equal (b) -3, Question 23. ⇒ 360 × 5 = x2 + 5x (b) p > – 2√5 Answer: (a) ±2 Question 8. Suppose two digit number is 10x + y. ⇒ (2 + y)2 + y2 = 34 = 400 – 448 = 48 < 0 a = 1, b = -20 and c = 112 Question 5. ∴ Length of a rectangular mango grove = 2x metre(By given condition) ∴ Length of cloth purchased (a) real (c) may be (a) and (b) 9x2 + 49y2 – 42xy The square of a positive integer is greater than 11 times the integer by 26. Area of rectangular mango grove = 800 m2 (c) 3 Answer: ∴ x2 – 2bx + b2 – 4a2 = 0, Question 7. Find the speed of the train. Question 14. \(\frac{2250}{x}\) = metre. One person purchased some cloth for ₹ 2250. 10x + y = 10(2) + 6 = 26. which is not possible i.e., imaginary value. ⇒ 16x – x2 – 64 + 4x = 48 64 + x2 – 16x = 9 (6 – x) or (x − 2)(x – 5) = 0 (d) none of these. (a) real and unequal, Question 17. Question 6. (c) \(\frac {-8}{5}\) < b < \(\frac {8}{5}\) ⇒ x2 + 50x – 75000 = 0 Question 3. Solution. The square of the smaller number is 8 times the larger number. Suppose the width of Verandah is x metre. (c) not real The diagonal of a rectangular field is 60 m more than the shorter side. ⇒ 2x2 + 3x – 90 = 0 …(i) x2 + kx + 64 = 0. ⇒ (x – 2)(2x – 1) = 0 Is the following situation possible ? So, present age of Rehman = 7 yr. Question 16. Answer: 9x2 + 7x – 2 = 0 (not acceptable). \(\sqrt{3 x+10}\) = 6 – \(\sqrt{6-x}\) = (-20)2 – 4 × 1 × 112 (d) p > 2√5 or p < – 2√5, Question 20. Given equation is ⇒ 30 = \(\frac {1}{2}\)x. Let the speed of stream be x km/h. ⇒ x2 + 12x – 5x – 60 = 0 (d) none of these.