HOTS Questions on Chapter 4 Class 10 Quadratic Equations

HOTS Questions on Class 10 Maths Chapter 4 Quadratic Equations help students understand equations, roots, and problem-solving in a simple way. They improve knowledge of factorization, quadratic formula, completing the square, and the nature of roots. This chapter builds a strong base for solving equation-based problems and improving math skills. A downloadable PDF is also available for easy revision and practice.

HOTS Questions Based on Identifying Quadratic Equations

Q1. Determine which of the following are quadratic equations. Justify each answer.

(a) x² + 3x + 1 = 0 (b) x³ − 2x + 1 = 0 (c) (x + 1)² = 2(x − 3) (d) x(x + 1)(x + 2) = 0 (e) (x − 2)² + 3 = 0 (f) 1/x + 1/(x+1) = 3

Ans (a) x² + 3x + 1 = 0: degree 2, a = 1 ≠ 0, Quadratic

Ans (b) x³ − 2x + 1 = 0: degree 3, Cubic, not quadratic

Ans (c) (x+1)² = 2(x−3): expand: x² + 2x + 1 = 2x − 6

x² + 7 = 0

degree 2, a = 1 ≠ 0, Quadratic

Ans (d) x(x+1)(x+2) = 0: expands to x³ + 3x² + 2x = 0, degree 3, Cubic, not quadratic

Ans (e) (x−2)² + 3 = 0: expands to x² − 4x + 7 = 0, degree 2, a = 1 ≠ 0, Quadratic (but D < 0, so no real roots)

Ans (f) 1/x + 1/(x+1) = 3: multiply through by x(x+1): (x+1) + x = 3x(x+1), 2x+1 = 3x² + 3x, 3x² + x − 1 = 0, Quadratic

Q2. A student claims that (x − 3)² = x² − 9 is a quadratic equation. Is this correct? Find the error.

Answer: Expand left side: (x−3)² = x² − 6x + 9.

So the equation becomes:

x² − 6x + 9 = x² − 9

−6x + 9 = −9

−6x = −18, x = 3.

After simplification, the x² terms cancel. The equation reduces to a linear equation, not a quadratic. The student is incorrect what appears quadratic can simplify to linear if the leading terms cancel.

Q3. For what value of k is kx² + 3x + 2 = 0 a quadratic equation? What happens when k = 0?

Answer: For it to be quadratic: k ≠ 0.

When k = 0: the equation becomes 3x + 2 = 0, x = −2/3. This is a linear equation with exactly one solution. The quadratic "degenerates" to a linear equation when the coefficient of x² vanishes.

Q4. If the equation (a² − 1)x² + (a − 1)x + 2 = 0 is quadratic, what values of a are not allowed? If a = 1, what type of equation results?

Answer: For quadratic: a² − 1 ≠ 0, a ≠ 1 and a ≠ −1. a ∈ ℝ, a ≠ ±1.

When a = 1: (1−1)x² + (1−1)x + 2 = 0

 0 + 0 + 2 = 0, 2 = 0, which is a contradiction no solution exists.

This equation has no valid solution for any x.

Q5. Can a quadratic equation have more than two roots? Justify your answer rigorously.

Answer: No. A quadratic equation ax² + bx + c = 0 (a ≠ 0) cannot have more than two roots.

Proof by contradiction: Suppose α, β, and γ are three distinct roots.

Then: aα² + bα + c = 0, aβ² + bβ + c = 0, aγ² + bγ + c = 0.

Subtracting the first from the second: a(β² − α²) + b(β − α) = 0, (β − α)[a(β + α) + b] = 0.

Since β ≠ α, we get a(α + β) + b = 0, i.e., α + β = −b/a … (i)

Similarly: α + γ = −b/a … (ii)

From (i) and (ii): β = γ contradicting the assumption that they are distinct.

Therefore, a quadratic equation has at most two distinct roots.

HOTS Questions Based on Nature of Roots

Q1. Find the values of k for which the equation 2x² + kx + 3 = 0 has equal roots.

Answer: For equal roots: D = 0.

D = k² − 4(2)(3) = k² − 24 = 0, k² = 24, k = ±2√6

Q2. For what values of m does the equation (m + 1)x² − 2(m − 1)x + 1 = 0 have real and equal roots? (Assume m ≠ −1.)

Answer: D = [−2(m−1)]² − 4(m+1)(1) = 4(m−1)² − 4(m+1) = 0

4[(m−1)² − (m+1)] = 0

(m² − 2m + 1) − (m + 1) = 0

m² − 3m = 0 → m(m − 3) = 0 → m = 0 or m = 3

Verify m = −1 is excluded (given). Both m = 0 and m = 3 are valid.

Q3. Show that the equation x² + 4x + 5 = 0 has no real roots. What does this tell you about the graph of y = x² + 4x + 5?

Answer: D = 4² − 4(1)(5) = 16 − 20 = −4 < 0

Since D < 0, the equation has no real roots.

Graphically: The parabola y = x² + 4x + 5 opens upward (a = 1 > 0) and never crosses the x-axis. The entire parabola lies above the x-axis. The minimum value of y occurs at x = −b/2a = −2, where y = (−2)² + 4(−2) + 5 = 4 − 8 + 5 = 1 > 0, confirming the parabola is always positive.

Q4. If one root of the equation x² − 5x + k = 0 is double the other root, find k.

Answer: Let roots be α and 2α.

Sum of roots: α + 2α = 5

3α = 5

α = 5/3

Product of roots: α × 2α = k

2α² = k

2(25/9) = k

k = 50/9

Q5. The equation px² − 14x + 8 = 0 has two equal roots. Find p and verify the roots.

Answer: For equal roots: D = (−14)² − 4(p)(8) = 0, 196 = 32p, p = 196/32 = 49/8

Equation: (49/8)x² − 14x + 8 = 0

multiply by 8: 49x² − 112x + 64 = 0

Equal roots: x = 112/(2×49) = 112/98 = 8/7

Verify: 49(64/49) − 112(8/7) + 64 = 64 − 128 + 64 = 0

Q6. Without solving, determine the nature of roots of: (i) 3x² − 5x + 2 = 0 (ii) x² + x + 1 = 0 (iii) 4x² − 4x + 1 = 0

Answer:

(i) D = 25 − 4(3)(2) = 25 − 24 = 1 > 0, Two distinct real roots

(ii) D = 1 − 4(1)(1) = 1 − 4 = −3 < 0, No real roots

(iii) D = 16 − 4(4)(1) = 16 − 16 = 0, Two equal real roots

HOTS Questions Based on Relationships Between Roots and Coefficients

Q1. If α and β are roots of 2x² − 5x + 3 = 0, find the value of (i) α + β (ii) αβ (iii) α² + β² (iv) 1/α + 1/β

Answer: From Vieta's formulas:

α + β = 5/2 and αβ = 3/2

(i) α + β = 5/2

(ii) αβ = 3/2

(iii) α² + β² = (α+β)² − 2αβ = (5/2)² − 2(3/2) = 25/4 − 3 = 13/4

(iv) 1/α + 1/β = (α+β)/αβ = (5/2)/(3/2) = 5/3

Q2. If α and β are roots of x² − 3x + 2 = 0, form a new quadratic equation whose roots are (α + 1) and (β + 1).

Answer: α + β = 3, αβ = 2.

New roots: (α+1) and (β+1).

New sum = (α+1) + (β+1) = α+β+2 = 3+2 = 5

New product = (α+1)(β+1) = αβ + α + β + 1 = 2+3+1 = 6

New equation: x² − (sum)x + (product) = 0

 x² − 5x + 6 = 0

Q3. The roots of ax² + bx + c = 0 are in the ratio 3:4. Prove that 12b² = 49ac.

Ans. Let roots be 3k and 4k.

Sum: 3k + 4k = 7k = −b/a, k = −b/7a

Product: 3k × 4k = 12k² = c/a

Substitute k = −b/7a into 12k² = c/a:

12(b²/49a²) = c/a

12b²/49a = c

12b² = 49ac

Q4. If one root of x² + px + 12 = 0 is 4 and the equation x² + px + q = 0 has equal roots, find q.

Ans. Since 4 is a root of x² + px + 12 = 0:

16 + 4p + 12 = 0

4p = −28

p = −7

The equation x² − 7x + q = 0 has equal roots, so D = 0:

49 − 4q = 0

q = 49/4

Q5. If the sum of the roots of kx² + 2x + 3k = 0 equals their product, find k.

Ans. Sum of roots = −2/k. Product of roots = 3k/k = 3.

Set equal: −2/k = 3

k = −2/3

Q6. α and β are roots of x² − 6x + 8 = 0. Find a quadratic equation whose roots are α/β and β/α.

Ans. α + β = 6, αβ = 8.

New roots: α/β and β/α.

Sum = α/β + β/α = (α² + β²)/αβ = [(α+β)² − 2αβ]/αβ = [36−16]/8 = 20/8 = 5/2

Product = (α/β)(β/α) = 1

New equation: x² − (5/2)x + 1 = 0

multiply by 2: 2x² − 5x + 2 = 0

HOTS Questions Based on Real-Life Applications of Quadratic Equations

Q1. A rectangular park has its length 3 metres more than twice its breadth. If the area is 90 m², find the dimensions.

Answer: Let breadth = x. Length = 2x + 3.

Area: x(2x + 3) = 90

2x² + 3x − 90 = 0

D = 9 + 720 = 729. 

√D = 27.

x = (−3 ± 27)/4 

x = 24/4 = 6 (taking positive root)

Breadth = 6 m, Length = 2(6) + 3 = 15 m

Verify: 6 × 15 = 90

Q2. The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field.

Answer: Let shorter side = x. Longer side = x + 30. Diagonal = x + 60.

By Pythagoras: x² + (x+30)² = (x+60)²

x² + x² + 60x + 900 = x² + 120x + 3600

x² − 60x − 2700 = 0

D = 3600 + 10800 = 14400. √D = 120.

x = (60 + 120)/2 = 90 (taking positive root)

Shorter side = 90 m, Longer side = 120 m, Diagonal = 150 m

Verify: 90² + 120²

= 8100 + 14400

= 22500 = 150²

Q3. A shopkeeper buys a number of books for ₹1200. If he had bought 10 more books for the same amount, each book would have cost ₹20 less. How many books did he buy?

Answer: Let number of books = x. Price per book = 1200/x.

New price = 1200/(x + 10). Difference = 20:

1200/x − 1200/(x+10) = 20

1200[(x+10−x)/x(x+10)] = 20

1200 × 10 = 20x(x+10)

12000 = 20x² + 200x

x² + 10x − 600 = 0

(x + 30)(x − 20) = 0

x = 20 (rejecting negative)

The shopkeeper bought 20 books.

Q4. A ball is thrown upward and its height (in metres) after t seconds is given by h = 20t − 5t². When does the ball reach a height of 15 m? When does it hit the ground?

Answer: For h = 15: 20t − 5t² = 15

5t² − 20t + 15 = 0

t² − 4t + 3 = 0

(t−1)(t−3) = 0

t = 1 s (going up) and t = 3 s (coming down)

For ground (h = 0): 20t − 5t² = 0

5t(4 − t) = 0

t = 0 (start) or t = 4 s (lands)

Q5. Two trains leave the same station. Train A travels 300 km at a constant speed. Train B travels 360 km but is 20 km/h faster than A and takes the same time. Find the speed of Train A.

Answer: Let speed of A = x km/h. Speed of B = (x + 20) km/h.

Time for A = 300/x. Time for B = 360/(x+20).

Set equal: 300/x = 360/(x+20)

300(x+20) = 360x

300x + 6000 = 360x

60x = 6000

x = 100 km/h

Speed of Train A = 100 km/h, Speed of Train B = 120 km/h

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