Chapter 4: Quadratic Equations (NCERT Solutions)

Exercise 4.1: Introduction

Q1. Check whether the following are quadratic equations:

(i) (x+1)² = 2(x-3)
x² + 2x + 1 = 2x - 6.
x² + 7 = 0. Degree is 2. Yes.

(ii) x² - 2x = (-2)(3-x)
x² - 2x = -6 + 2x.
x² - 4x + 6 = 0. Degree is 2. Yes.

(iii) (x-2)(x+1) = (x-1)(x+3)
x² - x - 2 = x² + 2x - 3.
-3x + 1 = 0. Degree is 1. No.

(iv) (x-3)(2x+1) = x(x+5)
2x² - 5x - 3 = x² + 5x.
x² - 10x - 3 = 0. Degree is 2. Yes.

(v) (2x-1)(x-3) = (x+5)(x-1)
2x² - 7x + 3 = x² + 4x - 5.
x² - 11x + 8 = 0. Degree is 2. Yes.

(vi) x² + 3x + 1 = (x-2)²
x² + 3x + 1 = x² - 4x + 4.
7x - 3 = 0. Degree is 1. No.

(vii) (x+2)³ = 2x(x²-1)
x³ + 8 + 6x(x+2) = 2x³ - 2x.
x³... Degree is 3. No.

(viii) x³ - 4x² - x + 1 = (x-2)³
x³... = x³ - 8 - 6x(x-2).
Terms of x³ cancel. Degree of remaining is 2. Yes.

Q2. Represent the following situations in the form of quadratic equations:

(i) Area of rectangular plot is 528 m². Length is one more than twice breadth.
Let breadth = x. Length = 2x + 1.
x(2x + 1) = 528 ⇒ 2x² + x - 528 = 0.

(ii) Product of two consecutive positive integers is 306.
Let integers be x, x+1.
x(x+1) = 306 ⇒ x² + x - 306 = 0.

(iii) Rohan's mother is 26 years older than him...
Product (x+3)(x+26+3) = 360.
(x+3)(x+29) = 360 ⇒ x² + 32x + 87 = 360.
x² + 32x - 273 = 0.

(iv) Train travels 480 km... speed 8 km/h less...
Time difference = 3 hours.
480/(x-8) - 480/x = 3.
480x - 480(x-8) = 3x(x-8).
3840 = 3x² - 24x.
x² - 8x - 1280 = 0.

Exercise 4.2: Solution by Factorization

Q1. Find the roots of the following quadratic equations by factorisation:

(i) x² - 3x - 10 = 0
(x - 5)(x + 2) = 0. Roots: 5, -2.

(ii) 2x² + x - 6 = 0
2x² + 4x - 3x - 6 = 0 ⇒ 2x(x+2) - 3(x+2) = 0.
(2x-3)(x+2) = 0. Roots: 3/2, -2.

(iii) √2x² + 7x + 5√2 = 0
√2x² + 5x + 2x + 5√2 = 0.
x(√2x + 5) + √2(√2x + 5) = 0.
(x+√2)(√2x+5) = 0. Roots: -√2, -5/√2.

(iv) 2x² - x + 1/8 = 0
16x² - 8x + 1 = 0 ⇒ (4x-1)² = 0.
Roots: 1/4, 1/4.

(v) 100x² - 20x + 1 = 0
(10x-1)² = 0. Roots: 1/10, 1/10.

Q3. Find two numbers whose sum is 27 and product is 182.

x(27-x) = 182 ⇒ x² - 27x + 182 = 0.
(x-13)(x-14) = 0.
Numbers: 13, 14.

Q4. Find two consecutive positive integers sum of whose squares is 365.

x² + (x+1)² = 365.
2x² + 2x - 364 = 0 ⇒ x² + x - 182 = 0.
(x+14)(x-13) = 0. x must be positive.
Numbers: 13, 14.

Q5. Altitude of right triangle is 7cm less than base. Hypotenuse is 13cm.

x² + (x-7)² = 13² = 169.
2x² - 14x + 49 = 169.
2x² - 14x - 120 = 0 ⇒ x² - 7x - 60 = 0.
(x-12)(x+5) = 0.
Base = 12 cm, Altitude = 5 cm.

Q6. Cottage industry... cost of production was 3 more than twice articles. Total 90.

x(2x+3) = 90 ⇒ 2x² + 3x - 90 = 0.
2x² - 12x + 15x - 90 = 0.
2x(x-6) + 15(x-6) = 0.
x = 6.
Articles: 6, Cost: 15.

Exercise 4.3: Nature of Roots

Q1. Find nature of roots. If real, find them.

(i) 2x² - 3x + 5 = 0
D = (-3)² - 4(2)(5) = 9 - 40 = -31 < 0.
No real roots.

(ii) 3x² - 4√3x + 4 = 0
D = (-4√3)² - 4(3)(4) = 48 - 48 = 0.
Real and Equal roots. x = -b/2a = 4√3/6 = 2√3/3.

(iii) 2x² - 6x + 3 = 0
D = (-6)² - 4(2)(3) = 36 - 24 = 12 > 0.
Real and distinct roots. x = (6 \u00B1 √12)/4 = (6 \u00B1 2√3)/4 = (3 \u00B1 √3)/2.

Q2. Find values of k for equal roots.

(i) 2x² + kx + 3 = 0
D = k² - 24 = 0 ⇒ k = ±√24 (±2√6).

(ii) kx(x-2) + 6 = 0 ⇒ kx² - 2kx + 6 = 0
D = (-2k)² - 4(k)(6) = 4k² - 24k = 0.
4k(k-6) = 0. k=0 (reject, not quadratic) or k=6.
k = 6.

Q3. Rectangular mango grove, length twice breadth, area 800. Possible?

l = 2b. l × b = 2b² = 800 ⇒ b² = 400.
b = 20. Real value exists. Yes.
Breadth = 20m, Length = 40m.

Q4. Sum of ages of two friends is 20. Four years ago product was 48.

x, 20-x.
(x-4)(16-x) = 48.
16x - x² - 64 + 4x = 48.
-x² + 20x - 112 = 0 ⇒ x² - 20x + 112 = 0.
D = 400 - 448 = -48 < 0.
Not possible.

Q5. Rectangular park. Perim 80, Area 400.

2(l+b) = 80 ⇒ l+b = 40.
lb = 400.
x(40-x) = 400 ⇒ x² - 40x + 400 = 0.
(x-20)² = 0.
Yes. l = 20m, b = 20m (Square).

Linear Equations - RD Sharma Important Questions

Q1. Find the value of k for which the system kx + 2y = 5, 3x + y = 1 has a unique solution.

For unique solution: a1/a2 ≠ b1/b2.
k/3 ≠ 2/1.
k ≠ 6.

Q2. Solve for x and y: 99x + 101y = 499; 101x + 99y = 501.

Add eqs: 200x + 200y = 1000 ⇒ x + y = 5 ... (i)
Subtract eqs: -2x + 2y = -2 ⇒ -x + y = -1 ⇒ y - x = -1 ... (ii)
Adding (i) and (ii): 2y = 4 ⇒ y = 2.
x = 3.

Q3. Determine the value of k for which the system x + (k+1)y = 5, (k+1)x + 9y = 8k - 1 has infinitely many solutions.

1/(k+1) = (k+1)/9 = 5/(8k-1).
(k+1)² = 9 ⇒ k+1 = ±3 ⇒ k = 2 or -4.
Check 5/(8k-1):
If k=2: 5/15 = 1/3. And 1/(2+1) = 1/3. Fits.
If k=-4: 5/(-33). And 1/-3. Not equal.
So, k = 2.

Q4. Solve: 2/x + 3/y = 13; 5/x - 4/y = -2.

Let 1/x = u, 1/y = v.
2u + 3v = 13 ... (i)
5u - 4v = -2 ... (ii)
Multiply (i) by 4, (ii) by 3: 8u + 12v = 52; 15u - 12v = -6.
Add: 23u = 46 ⇒ u = 2.
2(2) + 3v = 13 ⇒ 3v = 9 ⇒ v = 3.
x = 1/2, y = 1/3.

Q5. The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits differ by 2, find the number.

Number = 10x + y. Reversed = 10y + x.
Sum = 11(x+y) = 66 ⇒ x + y = 6.
Diff = x - y = 2 or y - x = 2.
Case 1: x=4, y=2 ⇒ 42.
Case 2: x=2, y=4 ⇒ 24.

Q6. Find value of k for no solution: 3x + y = 1; (2k-1)x + (k-1)y = 2k+1.

a1/a2 = b1/b2 ≠ c1/c2.
3/(2k-1) = 1/(k-1).
3k - 3 = 2k - 1 ⇒ k = 2.

Q7. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine speed of stream and boat.

Let 1/(x-y) = u, 1/(x+y) = v.
30u + 44v = 10.
40u + 55v = 13.
Solve to get u = 1/5, v = 1/11.
x - y = 5, x + y = 11.
x = 8 km/h (boat), y = 3 km/h (stream).

Q8. Solve: ax + by = a - b; bx - ay = a + b.

Multiply (i) by b, (ii) by a: abx + b²y = b(a-b); abx - a²y = a(a+b).
Subtract: (b² + a²)y = ab - b² - a² - ab = -(a² + b²).
y = -1.
ax - b = a - b ⇒ ax = a ⇒ x = 1.
Solution: x = 1, y = -1.

Q9. Points A and B are 90 km apart. One car starts from A and another from B at the same time. If they go in same direction they meet in 9 hrs, if opposite direction they meet in 9/7 hrs. Find speeds.

Same direction: 9(x - y) = 90 ⇒ x - y = 10.
Opposite: 9/7(x + y) = 90 ⇒ x + y = 70.
x = 40 km/h, y = 30 km/h.

Q10. 2 men and 7 boys can do a work in 4 days. 4 men and 4 boys can do it in 3 days. Find time taken by 1 man alone.

Let man take x days, boy take y days.
2/x + 7/y = 1/4.
4/x + 4/y = 1/3.
Let 1/x = u, 1/y = v.
Solve to find u = 1/15.
So 1 man takes 15 days.

Q11. Solve for x and y: x/a + y/b = 2; ax - by = a² - b².

x/a + y/b = 2 ⇒ bx + ay = 2ab.
ax - by = a² - b².
Solving gives x = a, y = b.

Q12. The area of a rectangle gets reduced by 9 sq units if length is reduced by 5 and breadth increased by 3. If length increased by 3 and breadth by 2, area increases by 67. Find dimensions.

(x-5)(y+3) = xy - 9 ⇒ 3x - 5y = 6.
(x+3)(y+2) = xy + 67 ⇒ 2x + 3y = 61.
Solving: x = 17, y = 9.

Q13. Solve graphically: 2x + y = 6; 2x - y + 2 = 0. Find area of triangle formed by these lines and x-axis.

Lines intersect at (1, 4).
x-intercepts: (3, 0) and (-1, 0).
Base = 4 units. Height = 4 units.
Area = 1/2 × 4 × 4 = 8 sq units.

Q14. The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. Ages of Cathy and Dharam differ by 30 years. Find ages of Ani and Biju.

Two cases: A - B = 3 or B - A = 3.
D = 2A, B = 2C ⇒ C = B/2.
D - C = 30 ⇒ 2A - B/2 = 30 ⇒ 4A - B = 60.
Solving: A = 19, B = 16 OR A = 21, B = 24.

Q15. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h, it would have taken 3 hours more. Find distance.

Let speed = x, time = y. Dist = xy.
(x+10)(y-2) = xy ⇒ -2x + 10y = 20 ⇒ -x + 5y = 10.
(x-10)(y+3) = xy ⇒ 3x - 10y = 30.
Solving: x = 50 km/h, y = 12 hours.
Distance = 600 km.

Q16. Solve: 152x - 378y = -74; -378x + 152y = -604.

Add: -226(x+y) = -678 ⇒ x + y = 3.
Subtract: 530(x-y) = 530 ⇒ x - y = 1.
Solving: x = 2, y = 1.

Q17. For what value of k will kx + 3y = k-3, 12x + ky = k have no solution?

k/12 = 3/k ≠ (k-3)/k.
k² = 36 ⇒ k = ±6.
If k=6: 6/6 ≠ 3/6 (1 ≠ 1/2), Holds.
If k=-6: -6/-6 = 1. (-9)/-6 = 3/2. Holds.
Usually positive k is asked, check options if MCQ. Both valid for 'no solution' condition relative to c1/c2.

Q18. Solve x/2 + 2y/3 = -1 and x - y/3 = 3.

3x + 4y = -6.
3x - y = 9.
Subtract: 5y = -15 ⇒ y = -3.
3x - (-3) = 9 ⇒ 3x = 6 ⇒ x = 2.

Q19. Find a, b if 2x + 3y = 7 and (a-b)x + (a+b)y = 3a+b-2 have infinite solutions.

2/(a-b) = 3/(a+b) = 7/(3a+b-2).
2a+2b = 3a-3b ⇒ a = 5b.
3(3a+b-2) = 7(a+b) ⇒ 9a+3b-6 = 7a+7b ⇒ 2a-4b=6.
10b - 4b = 6 ⇒ 6b = 6 ⇒ b = 1.
a = 5.

Q20. A part of monthly hostel charges is fixed and the remaining depends on number of days one has taken food. Student A takes food for 20 days pays 1000. Student B takes food for 26 days pays 1180. Find fixed charges and cost of food per day.

x + 20y = 1000.
x + 26y = 1180.
Subtract: 6y = 180 ⇒ y = 30.
x + 600 = 1000 ⇒ x = 400.
Fixed = 400, Food per day = 30.

Linear Equations - Formulas & PYQs

Key Formulas & Concepts

1. Standard Form

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

2. Nature of Solutions

Intersecting Lines (Unique Solution):

a₁/a₂ ≠ b₁/b₂

Coincident Lines (Infinitely Many Solutions):

a₁/a₂ = b₁/b₂ = c₁/c₂

Parallel Lines (No Solution):

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

3. Algebraic Methods

1. Substitution Method

2. Elimination Method

3. Cross-Multiplication Method (Deleted in New NCERT but useful for competitive exams)

Previous Year Questions (CBSE/JKBOSE)

Q1. Find the value of k for which the system of equations x + 2y = 3 and 5x + ky = 7 has no solution. (CBSE 2020)

a₁/a₂ = b₁/b₂ ≠ c₁/c₂
1/5 = 2/k
k = 10.

Q2. Solve for x and y: 2x + y = 6, 2x - y = 2. (CBSE 2012)

Adding: 4x = 8 ⇒ x = 2.
Subtracting: 2y = 4 ⇒ y = 2.
Solution: x = 2, y = 2.

Q3. The sum of digits of a two-digit number is 9. If 27 is added to the number, the digits get reversed. Find the number. (CBSE 2013)

x + y = 9.
(10x + y) + 27 = 10y + x.
9x - 9y = -27 ⇒ x - y = -3.
Adding: 2x = 6 ⇒ x = 3.
y = 6.
Number is 36.

Q4. For what value of k does the pair of linear equations x + y = 3 and 2x + 2y = k have infinitely many solutions? (CBSE 2023)

1/2 = 1/2 = 3/k.
k = 6.

Q5. Solve the pair of linear equations: x/10 + y/5 - 1 = 0 and x/8 + y/6 = 15. (Standard)

Multiply 1st by 10: x + 2y = 10.
Multiply 2nd by 24: 3x + 4y = 360.
Multiply 1st by 2: 2x + 4y = 20.
Subtract: x = 340.
340 + 2y = 10 ⇒ 2y = -330 ⇒ y = -165. (Values are large but consistent with equations provided).

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