RowQ
The Vault
RowQ
The Vault
CBSE Class 10 Maths · 11 questions · 26 marks
A quadratic equation is any equation that can be tidied into the form ax² + bx + c = 0 with a ≠ 0, and it shows up whenever an area, a speed or a product of two unknowns is involved. You will solve them by factorisation and by the quadratic formula, and learn to read the discriminant to know in advance how many real roots to expect. The word problems are where the marks hide, so always define your variable and reject any root that makes no physical sense.
The nature of the roots of the equation 2x² - 5x + 4 = 0 is:
Answer
The equation has no real roots. Here a = 2, b = -5 and c = 4. D = b² - 4ac = (-5)² - 4(2)(4) = 25 - 32 = -7. Since D < 0, the square root in the quadratic formula is not defined over the real numbers, so there are no real roots.
If x = 3 is a root of the equation x² - kx - 6 = 0, then k equals:
Answer
k = 1. If 3 is a root, substituting x = 3 must satisfy the equation: 3² - k(3) - 6 = 0 9 - 3k - 6 = 0, so 3 - 3k = 0 and k = 1. The equation is then x² - x - 6 = (x - 3)(x + 2) = 0, whose roots are 3 and -2, confirming that 3 is a root.
The value of k for which the equation kx² - 8x + 4 = 0 has two equal roots is:
Answer
k = 4. Equal roots occur exactly when D = 0. D = (-8)² - 4(k)(4) = 64 - 16k. Setting 64 - 16k = 0 gives k = 4. The equation becomes 4x² - 8x + 4 = 4(x - 1)² = 0, so the repeated root is x = 1.
The product of the roots of the equation 3x² + 7x - 6 = 0 is:
Answer
The product is -2. For ax² + bx + c = 0 the product of the roots is c/a. Here c/a = -6/3 = -2. Check by factorising: 3x² + 9x - 2x - 6 = 3x(x + 3) - 2(x + 3) = (x + 3)(3x - 2), so the roots are -3 and 2/3, and (-3)(2/3) = -2.
Assertion (A): The equation x² - 4x + 4 = 0 has two equal roots, each equal to 2. Reason (R): A quadratic equation ax² + bx + c = 0 has two equal roots when b² - 4ac = 0.
Answer
Both A and R are true and R is the correct explanation of A. Reason R states the equal-roots condition correctly. For x² - 4x + 4 = 0, D = (-4)² - 4(1)(4) = 16 - 16 = 0, so the roots are equal. The repeated root is x = -b/(2a) = 4/2 = 2, which matches Assertion A. Since A is exactly the consequence of the condition in R, R explains A.
Solve the quadratic equation 6x² - 7x - 3 = 0 by factorisation.
Answer
Here ac = 6 × (-3) = -18, and we need two numbers with product -18 and sum -7: they are -9 and +2. 6x² - 9x + 2x - 3 = 0 3x(2x - 3) + 1(2x - 3) = 0 (2x - 3)(3x + 1) = 0 So 2x - 3 = 0 or 3x + 1 = 0, giving x = 3/2 or x = -1/3.
Solve 2x² - 7x + 3 = 0 using the quadratic formula.
Answer
Here a = 2, b = -7 and c = 3. D = b² - 4ac = 49 - 4(2)(3) = 49 - 24 = 25, and √25 = 5. Since D > 0 there are two distinct real roots. x = [-b ± √D]/(2a) = [7 ± 5]/4. x = (7 + 5)/4 = 12/4 = 3, or x = (7 - 5)/4 = 2/4 = 1/2. The roots are 3 and 1/2. Check: sum = 3.5 = -b/a = 7/2 ✔ and product = 1.5 = c/a = 3/2 ✔
Find the values of k for which the equation x² + kx + 9 = 0 has real and equal roots.
Answer
Real and equal roots require D = 0. D = k² - 4(1)(9) = k² - 36. Setting k² - 36 = 0 gives k² = 36, so k = 6 or k = -6. For k = 6 the equation is x² + 6x + 9 = (x + 3)², with repeated root -3; for k = -6 it is (x - 3)², with repeated root 3.
A train travels 360 km at a uniform speed. If its speed had been 5 km/h more, it would have taken 1 hour less to cover the same distance. Find the original speed of the train.
Answer
Let the original speed be x km/h, where x > 0. Time taken at the original speed = 360/x hours. Time taken at the increased speed = 360/(x + 5) hours. The faster trip takes 1 hour less: 360/x - 360/(x + 5) = 1 Multiply throughout by x(x + 5): 360(x + 5) - 360x = x(x + 5) 360x + 1800 - 360x = x² + 5x 1800 = x² + 5x, so x² + 5x - 1800 = 0. D = 5² - 4(1)(-1800) = 25 + 7200 = 7225, and √7225 = 85. x = [-5 ± 85]/2, giving x = 40 or x = -45. A speed cannot be negative, so we reject x = -45. The original speed of the train is 40 km/h. Check: 360/40 = 9 hours and 360/45 = 8 hours, a difference of 1 hour. ✔
(a) The area of a rectangular garden is 84 m² and its length is 2 m more than twice its breadth. Find the dimensions of the garden. (b) Find the nature of the roots of 4x² - 12x + 9 = 0 and, if the roots are real, find them.
Answer
(a) Let the breadth be x metres, so the length is (2x + 2) metres. Area: x(2x + 2) = 84 2x² + 2x - 84 = 0, and dividing by 2, x² + x - 42 = 0. Factorise: x² + 7x - 6x - 42 = x(x + 7) - 6(x + 7) = (x + 7)(x - 6) = 0. So x = 6 or x = -7. A breadth cannot be negative, so x = 6. Breadth = 6 m and length = 2(6) + 2 = 14 m. Check: 6 × 14 = 84 m² ✔ (b) For 4x² - 12x + 9 = 0, D = (-12)² - 4(4)(9) = 144 - 144 = 0. Since D = 0, the equation has two equal real roots. The repeated root is x = -b/(2a) = 12/8 = 3/2. Indeed 4x² - 12x + 9 = (2x - 3)².
Read the following and answer the questions that follow: A class plans a heritage walk and hires a bus for a total cost of ₹ 3600, to be shared equally by all the students who go. On the day of the trip, 4 students fell ill and could not come, so each of the remaining students had to pay ₹ 30 more than originally planned. (a) Taking the original number of students as n, form a quadratic equation for the situation. (b) Solve the equation to find how many students had originally planned to go. (c) What was the amount each student was originally supposed to pay? (d) What did each student actually pay on the day of the trip?
Answer
(a) Originally each of the n students would pay 3600/n rupees. With 4 fewer students, each of the (n - 4) students paid 3600/(n - 4) rupees, which is ₹ 30 more: 3600/(n - 4) - 3600/n = 30 Multiplying by n(n - 4): 3600n - 3600(n - 4) = 30n(n - 4) 14400 = 30n² - 120n Dividing by 30: n² - 4n - 480 = 0. (b) D = (-4)² - 4(1)(-480) = 16 + 1920 = 1936, and √1936 = 44. n = [4 ± 44]/2, so n = 24 or n = -20. The number of students cannot be negative, so n = 24. Originally 24 students planned to go. (c) Original share = 3600/24 = ₹ 150 per student. (d) Number who actually went = 24 - 4 = 20, so each paid 3600/20 = ₹ 180. This is ₹ 30 more than ₹ 150, as stated. ✔
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