NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 4.1
Q1. Check whether the following are quadratic equations
(i) (x + 1)² = 2(x – 3) → x² + 2x + 1 = 2x – 6 → x² + 7 = 0. Yes, degree 2.
(ii) x² – 2x = (–2)(3 – x) → x² – 2x = –6 + 2x → x² – 4x + 6 = 0. Yes.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3) → x² – x – 2 = x² + 2x – 3 → –3x + 1 = 0. No, degree 1.
(iv) (x – 3)(2x + 1) = x(x + 5) → 2x² – 5x – 3 = x² + 5x → x² – 10x – 3 = 0. Yes.
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) → 2x² – 7x + 3 = x² + 4x – 5 → x² – 11x + 8 = 0. Yes.
(vi) x² + 3x + 1 = (x – 2)² → x² + 3x + 1 = x² – 4x + 4 → 7x – 3 = 0. No, degree 1.
(vii) (x + 2)³ = 2x(x² – 1) → x³ + 6x² + 12x + 8 = 2x³ – 2x → x³ – 6x² – 14x – 8 = 0. No, degree 3.
(viii) x³ – 4x² – x + 1 = (x – 2)³ → expanded and simplified → 2x² – 13x + 9 = 0. Yes.
Represent the following situations in the form of quadratic equations:
Answer
(i) Rohan's mother is 26 years older. Product of their ages 3 years from now will be 360.
Let Rohan's age = x. After 3 years: (x+3)(x+29) = 360
⇒ x² + 32x – 273 = 0
(ii) A train travels 480 km at uniform speed. If speed were 8 km/h less, it would take 3 hours more.
Let speed = x. 480/x – 480/(x–8)... wait: 480/(x–8) – 480/x = 3
⇒ x² – 8x – 1280 = 0
(iii) Two water taps together fill a tank in 75/8 hours. Larger tap takes 10 hours less. Find time taken by each.
Let smaller tap = x hours. 1/x + 1/(x–10) = 8/75
⇒ 8x² – 230x + 750 = 0 (simplified: 4x² – 115x + 375 = 0)
Exercise 4.2
Find the roots of the following quadratic equations by factorisation:
(i) x² – 3x – 10 = 0: (x–5)(x+2) = 0 → x = 5 or x = –2
(ii) 2x² + x – 6 = 0: (2x–3)(x+2) = 0 → x = 3/2 or x = –2
(iii) √2x² + 7x + 5√2 = 0: (√2x + 5)(x + √2) = 0 → x = –5/√2 or x = –√2
(iv) 2x² – x + 1/8 = 0: Multiply by 8: 16x² – 8x + 1 = 0 → (4x–1)² = 0 → x = 1/4 (repeated)
(v) 100x² – 20x + 1 = 0: (10x–1)² = 0 → x = 1/10 (repeated)
Answer
Q2. Solve the problems
(i) John and Jivanti had 45 marbles. Each lost 5. Product of remaining = 124. Find marbles each had.
x(45–x) after losing 5 each: (x–5)(40–x) = 124 → x² – 45x + 324 = 0 → (x–36)(x–9) = 0 → x = 36 or 9. One had 36, other had 9.
(ii) A cottage industry produces toys. Cost of each toy (in ₹) = 55 – number of toys produced. On a day, total cost was ₹750. Number of toys produced?
x(55–x) = 750 → x² – 55x + 750 = 0 → (x–25)(x–30) = 0 → x = 25 or 30
Exercise 4.3
Q1. Find the roots by completing the square
(i) 2x² – 7x + 3 = 0: x² – 7x/2 + 3/2 = 0 → (x – 7/4)² = 49/16 – 24/16 = 25/16 → x – 7/4 = ±5/4 → x = 3 or 1/2
(ii) 2x² + x – 4 = 0: (x + 1/4)² = 33/16 → x = (–1 ± √33)/4 → x = (–1+√33)/4 or (–1–√33)/4
(iii) 4x² + 4√3x + 3 = 0: (2x + √3)² = 0 → x = –√3/2 (repeated)
(iv) 2x² + x + 4 = 0: (x + 1/4)² = –31/16 < 0 → No real roots
Q2. Find the roots using the quadratic formula
Quadratic formula: x = [–b ± √(b²–4ac)] / 2a
(i) 2x² – 7x + 3 = 0: D = 49 – 24 = 25. x = (7±5)/4 → x = 3 or 1/2
(ii) 2x² + x – 4 = 0: D = 1 + 32 = 33. x = (–1±√33)/4 → x = (–1±√33)/4
(iii) 4x² + 4√3x + 3 = 0: D = 48 – 48 = 0. x = –4√3/8 = –√3/2 (repeated)
(iv) 2x² + x + 4 = 0: D = 1 – 32 = –31 < 0. No real roots
Find the roots: x – 1/x = 3, x ≠ 0
Answer
x² – 1 = 3x
⇒ x² – 3x – 1 = 0. D = 9 + 4 = 13. x = (3 ± √13)/2
A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream. Find the speed of the stream.
Answer
Let stream speed = x. 24/(18–x) – 24/(18+x) = 1
⇒ 48x = 324 – x²
⇒ x² + 48x – 324 = 0
⇒ (x+54)(x–6) = 0
⇒ x = 6 km/h
Exercise 4.4
Find the nature of the roots of the following quadratic equations. If real roots exist, find them:
(i) 2x² – 3x + 5 = 0: D = 9 – 40 = –31 < 0. No real roots.
(ii) 3x² – 4√3x + 4 = 0: D = 48 – 48 = 0. Two equal real roots. x = 4√3/6 = 2√3/3 = 2/√3
(iii) 2x² – 6x + 3 = 0: D = 36 – 24 = 12 > 0. Two distinct real roots. x = (6±2√3)/4 = (3±√3)/2
Answer
Find the values of k for each of the following quadratic equations, so that they have two equal roots:
(i) 2x² + kx + 3 = 0: D = 0 → k² – 24 = 0 → k = ±2√6
(ii) kx(x – 2) + 6 = 0 → kx² – 2kx + 6 = 0: D = 0 → 4k² – 24k = 0 → 4k(k–6) = 0 → k = 6 (k ≠ 0)
Answer
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Answer
Let breadth = x, length = 2x. Area = 2x² = 800
⇒ x² = 400
⇒ x = 20. Breadth = 20 m, Length = 40 m. Yes, it is possible.
Is the following situation possible? The sum of the ages of two friends is 20 years. Four years ago, the product of their ages (in years) was 48.
Answer
Let ages be x and (20–x). Four years ago: (x–4)(16–x) = 48
⇒ x² – 20x + 112 = 0. D = 400 – 448 = –48 < 0. Not possible (no real roots).
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Answer
Perimeter = 2(l+b) = 80
⇒ l + b = 40, l × b = 400. x² – 40x + 400 = 0
⇒ (x–20)² = 0
⇒ x = 20. Length = Breadth = 20 m (it's a square). Yes, it is possible.
Important Extra Questions
What is the discriminant? What does it tell us about the roots?
Answer
For ax² + bx + c = 0, the discriminant D = b² – 4ac.
• D > 0: Two distinct real roots (if D is a perfect square, roots are rational)
• D = 0: Two equal real roots (one repeated root) x = –b/2a
• D < 0: No real roots (roots are complex/imaginary)
The sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1/3. Find his present age.
Answer
Let age = x. 1/(x–3) + 1/(x+5) = 1/3
⇒ 3(2x+2) = (x–3)(x+5)
⇒ 6x + 6 = x² + 2x – 15
⇒ x² – 4x – 21 = 0
⇒ (x–7)(x+3) = 0
⇒ x = 7 years
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
⇒ (x–90)(x+30) = 0
⇒ x = 90.