CBSE Class 10 Mathematics Quadratic Equations Important Questions Set 01

Answer: D = b² − 4ac = 25 − 20 = 5.

Teacher's Note:
1. Use D = b² − 4ac.
2. Here a = 1, b = 5, c = 5.

Answer: Sum = −b/a = −4.

Teacher's Note:
1. Use the sum of roots formula.
2. The roots are 4 and −8, which add to −4.

Answer: Product = c/a = −4/2 = −2.

Teacher's Note:
1. Use product = c/a.
2. Keep the sign of c.

Answer: D = 4k² − 36 ≥ 0, so k² ≥ 9. Hence k ≤ −3 or k ≥ 3.

Teacher's Note:
1. Real roots need D ≥ 0.
2. Solve k² ≥ 9 for both signs.

Answer: D = 4(k + 1)² − 4k² = 4(2k + 1) = 0, so k = −1/2.

Teacher's Note:
1. Equal roots mean D = 0.
2. Expand carefully before simplifying.

Answer: a² − (a + b)a + k = 0 gives −ab + k = 0, so k = ab.

Teacher's Note:
1. A solution satisfies the equation.
2. Substitute x = a.

Answer: Let his present age be x. (x − 5)(x + 9) = 15 gives x² + 4x − 45 = 15, i.e. x² + 4x − 60 = 0.

Teacher's Note:
1. Write each age in terms of x.
2. Expand and bring everything to one side.

Answer: x² − 5x + 2x − 10 = (x − 5)(x + 2) = 0, so the roots are 5 and −2.

Teacher's Note:
1. Find two numbers with product −10 and sum −3.
2. These are −5 and 2.

Answer: Let them be x and x + 2. x² + 2x − 483 = 0, so (x + 23)(x − 21) = 0 and x = 21 (taking the positive case).
The numbers are 21 and 23 (or −23 and −21).

Teacher's Note:
1. Consecutive odd numbers differ by 2.
2. Both the positive and negative pairs satisfy the condition.

Answer: (b) real and distinct. D = 36 − 24 = 12 > 0.

Teacher's Note:
1. Compute the discriminant.
2. A positive D that is not a perfect square gives irrational distinct roots.

Answer: (c) real and equal. The formula gives x = −b/2a twice.

Teacher's Note:
1. D = 0 removes the square root term.
2. Both roots coincide.

Answer: (d) −8. D = 16 − 24 = −8, so the equation has no real roots.

Teacher's Note:
1. Use b² − 4ac with a = 2, b = −4, c = 3.
2. A negative value means no real roots.

Answer: Sum of roots = 5 = −2k/3, so k = −15/2. Product = 6 = m/3, so m = 18.
k = −15/2, m = 18.

Teacher's Note:
1. Use the sum and product of roots.
2. Check by substituting x = 2: 12 − 30 + 18 = 0.

Answer: (x + 1)(x + 2) + (x − 2)(x − 1) = 3(x − 1)(x + 2)
2x² + 4 = 3x² + 3x − 6, so x² + 3x − 10 = 0, i.e. (x + 5)(x − 2) = 0.
x = 2 or x = −5.

Teacher's Note:
1. Multiply through by the common denominator.
2. Both values are allowed since neither is 1 or −2.

Answer: D = (p² − q²)² + 4p²q² = (p² + q²)².
x = [−(p² − q²) ± (p² + q²)] / 2p², giving x = q²/p² or x = −1.
x = q²/p², −1.

Teacher's Note:
1. The discriminant is a perfect square.
2. Factorising as (p²x − q²)(x + 1) gives the same result.

Answer: D = 576 + 64 = 640 = 64 × 10. x = (24 ± 8√10)/32 = \( \dfrac{3\pm\sqrt{10}}{4} \).

Teacher's Note:
1. Simplify the square root of the discriminant.
2. Cancel the common factor 8.

Answer: D = (−4)² − 4(2)(3) = 16 − 24 = −8 < 0. The equation has no real roots.

Teacher's Note:
1. The sign of D decides the nature of roots.
2. Negative D means no real roots.

Answer: D = 529 + 1320 = 1849 = 43². x = (23 ± 43)/6 gives x = 11 or x = −10/3.
x = 11, −10/3.

Teacher's Note:
1. Compute D first and see if it is a perfect square.
2. Then apply the formula.

Answer: Sum = 4 and product = 4 − 3 = 1.
Equation: x² − 4x + 1 = 0.

Teacher's Note:
1. Use x² − (sum)x + product = 0.
2. The product is a difference of squares.

Answer: D = 25 + 24p = 0, so p = −25/24.
The equal root is x = −b/2a = 5/6.

Teacher's Note:
1. Set D = 0.
2. Equal roots are −b/(2a).

Answer: Let the numbers be x and 15 − x. \( \dfrac{1}{x}+\dfrac{1}{15-x}=\dfrac{3}{10} \) gives 150 = 3x(15 − x), i.e. x² − 15x + 50 = 0, so (x − 5)(x − 10) = 0.
The numbers are 5 and 10.

Teacher's Note:
1. Combine the reciprocals into one fraction.
2. Then clear the denominators.

Answer: Let the parts be x and 29 − x. x² + (29 − x)² = 425 gives x² − 29x + 208 = 0, so (x − 13)(x − 16) = 0.
The parts are 13 and 16.

Teacher's Note:
1. Write the second part as 29 − x.
2. Simplify to a quadratic and factorise.

Answer: Write as (3x − a)² − b² = 0, so (3x − a − b)(3x − a + b) = 0.
x = (a + b)/3 or x = (a − b)/3.

Teacher's Note:
1. Spot the perfect-square pattern (3x − a)².
2. Then use the difference of two squares.

Answer: Let there be x students. \( \dfrac{300}{x}-\dfrac{300}{x+10}=1 \) gives 3000 = x² + 10x, so x² + 10x − 3000 = 0.
D = 100 + 12000 = 12100 = 110², so x = (−10 + 110)/2 = 50 (the negative root is rejected).
There are 50 students.

Teacher's Note:
1. Each student's share is total divided by number.
2. Reject negative values of a count.

Answer: Let Mathematics = x, English = 30 − x. (x + 2)(27 − x) = 210 gives x² − 25x + 156 = 0.
D = 625 − 624 = 1, so x = 13 or 12.
Marks: (13, 17) or (12, 18).

Teacher's Note:
1. Form the equation from the changed marks.
2. Both roots give valid positive marks.

Answer: Let the digits be x (tens) and y (units): xy = 35 and 10x + y + 18 = 10y + x, so y = x + 2.
x(x + 2) = 35 gives x² + 2x − 35 = 0, so x = 5 and y = 7.
The number is 57.

Teacher's Note:
1. Use the interchange condition to relate the digits.
2. Reject the negative root for a digit.

Answer: Let the tour be x days. \( \dfrac{360}{x}-\dfrac{360}{x+4}=3 \) gives 480 = x² + 4x, so x² + 4x − 480 = 0.
D = 16 + 1920 = 1936 = 44², so x = (−4 + 44)/2 = 20.
The tour was 20 days.

Teacher's Note:
1. Daily expense is total divided by days.
2. Reject the negative root.

Answer: Let the price be ₹x. \( \dfrac{1740}{x}-\dfrac{1740}{x+2}=1 \) gives 3480 = x² + 2x, so x² + 2x − 3480 = 0.
D = 13924 = 118², so x = (−2 + 118)/2 = 58.
The original price was ₹58 per litre.

Teacher's Note:
1. Litres bought equal money divided by price.
2. Reject the negative root.

Answer: Let the length be x, so the breadth is 40 − x. x(40 − x) = 400 gives x² − 40x + 400 = 0, i.e. (x − 20)² = 0.
Yes, it is possible: length = breadth = 20 m (the park is a square).

Teacher's Note:
1. Half the perimeter is 40 m.
2. A perfect square discriminant of zero gives equal sides.

Answer: D = (b − c)² − 4(a − b)(c − a) = 0.
Expanding: b² − 2bc + c² − 4(ac − a² − bc + ab) = (b + c)² + 4a² − 4a(b + c) = (b + c − 2a)².
So (b + c − 2a)² = 0, giving 2a = b + c.

Teacher's Note:
1. Set the discriminant equal to zero.
2. The expansion collapses to a perfect square.

Answer: D = 4m²c² − 4(1 + m²)(c² − a²) = 0.
Dividing by 4: m²c² − c² + a² − m²c² + m²a² = 0, so a²(1 + m²) = c².
Hence c² = a²(1 + m²).

Teacher's Note:
1. Equal roots mean D = 0.
2. The m²c² terms cancel.

Answer: D = 81(a + b)² − 36(2a² + 5ab + 2b²) = 9(a − b)².
x = [9(a + b) ± 3(a − b)] / 18, so x = (2a + b)/3 or x = (a + 2b)/3.

Teacher's Note:
1. Expand and simplify the discriminant.
2. It is a perfect square in a and b.