Solution:

LHS = (x+1)² = x² + 2x + 1

RHS = 2(x-3) = 2x - 6

Equation: x² + 2x + 1 = 2x - 6

x² + 2x + 1 - 2x + 6 = 0

x² + 7 = 0

This is of the form ax² + bx + c = 0 where a = 1, b = 0, c = 7

✅ Yes, it is a quadratic equation.

Solution:

RHS = (-2)(3-x) = -6 + 2x

Equation: x² - 2x = -6 + 2x

x² - 2x - 2x + 6 = 0

x² - 4x + 6 = 0

This is of the form ax² + bx + c = 0 where a = 1, b = -4, c = 6

✅ Yes, it is a quadratic equation.

Solution:

LHS = (x-2)(x+1) = x² + x - 2x - 2 = x² - x - 2

RHS = (x-1)(x+3) = x² + 3x - x - 3 = x² + 2x - 3

Equation: x² - x - 2 = x² + 2x - 3

x² - x - 2 - x² - 2x + 3 = 0

-3x + 1 = 0

This is a linear equation (degree 1), not quadratic.

❌ No, it is not a quadratic equation.

Solution:

LHS = (x-3)(2x+1) = 2x² + x - 6x - 3 = 2x² - 5x - 3

RHS = x(x+5) = x² + 5x

Equation: 2x² - 5x - 3 = x² + 5x

2x² - 5x - 3 - x² - 5x = 0

x² - 10x - 3 = 0

This is of the form ax² + bx + c = 0 where a = 1, b = -10, c = -3

✅ Yes, it is a quadratic equation.

Solution:

LHS = (2x-1)(x-3) = 2x² - 6x - x + 3 = 2x² - 7x + 3

RHS = (x+5)(x-1) = x² - x + 5x - 5 = x² + 4x - 5

Equation: 2x² - 7x + 3 = x² + 4x - 5

2x² - 7x + 3 - x² - 4x + 5 = 0

x² - 11x + 8 = 0

✅ Yes, it is a quadratic equation.

Solution:

RHS = (x-2)² = x² - 4x + 4

Equation: x² + 3x + 1 = x² - 4x + 4

x² + 3x + 1 - x² + 4x - 4 = 0

7x - 3 = 0

This is a linear equation, not quadratic.

❌ No, it is not a quadratic equation.

Solution:

LHS = (x+2)³ = x³ + 6x² + 12x + 8

RHS = 2x(x²-1) = 2x³ - 2x

Equation: x³ + 6x² + 12x + 8 = 2x³ - 2x

0 = 2x³ - 2x - x³ - 6x² - 12x - 8

x³ - 6x² - 14x - 8 = 0

This is a cubic equation (degree 3), not quadratic.

❌ No, it is not a quadratic equation.

Solution:

RHS = (x-2)³ = x³ - 6x² + 12x - 8

Equation: x³ - 4x² - x + 1 = x³ - 6x² + 12x - 8

Cancel x³ from both sides:

-4x² - x + 1 = -6x² + 12x - 8

-4x² - x + 1 + 6x² - 12x + 8 = 0

2x² - 13x + 9 = 0

✅ Yes, it is a quadratic equation.

Solution:

Let breadth = x metres

Length = 2x + 1 metres

Area = Length × Breadth = x(2x + 1) = 528

2x² + x = 528

📝 Quadratic equation: 2x² + x - 528 = 0

Solution:

Let first integer = x

Next consecutive integer = x + 1

Product = x(x + 1) = 306

x² + x = 306

📝 Quadratic equation: x² + x - 306 = 0

Solution:

Let Rohan's present age = x years

Mother's present age = x + 26 years

After 3 years: Rohan's age = x + 3, Mother's age = x + 29

Product = (x + 3)(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x + 87 - 360 = 0

📝 Quadratic equation: x² + 32x - 273 = 0

Solution:

Let speed of train = x km/h

Time taken = Distance/Speed = 480/x hours

If speed = (x - 8) km/h, time = 480/(x-8) hours

Difference in time = 3 hours:

480/(x-8) - 480/x = 3

Multiply by x(x-8): 480x - 480(x-8) = 3x(x-8)

480x - 480x + 3840 = 3x² - 24x

3840 = 3x² - 24x

3x² - 24x - 3840 = 0

Divide by 3: x² - 8x - 1280 = 0

📝 Quadratic equation: x² - 8x - 1280 = 0

Solution:

x² - 3x - 10 = 0

Find two numbers whose product = -10 and sum = -3

Numbers: -5 and +2 (since -5 × 2 = -10, -5 + 2 = -3)

x² - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x - 5)(x + 2) = 0

x - 5 = 0 → x = 5

x + 2 = 0 → x = -2

✅ Roots: x = 5, x = -2

Solution:

2x² + x - 6 = 0

Multiply a × c = 2 × (-6) = -12

Find two numbers whose product = -12 and sum = 1

Numbers: 4 and -3 (4 × -3 = -12, 4 + (-3) = 1)

2x² + 4x - 3x - 6 = 0

2x(x + 2) - 3(x + 2) = 0

(x + 2)(2x - 3) = 0

x + 2 = 0 → x = -2

2x - 3 = 0 → 2x = 3 → x = 3/2

✅ Roots: x = -2, x = 3/2

Solution:

√2x² + 7x + 5√2 = 0

Multiply a × c = √2 × 5√2 = 5 × 2 = 10

Find two numbers whose product = 10 and sum = 7

Numbers: 5 and 2

√2x² + 5x + 2x + 5√2 = 0

x(√2x + 5) + √2(√2x + 5) = 0

(√2x + 5)(x + √2) = 0

√2x + 5 = 0 → √2x = -5 → x = -5/√2

x + √2 = 0 → x = -√2

✅ Roots: x = -5/√2, x = -√2

Solution:

2x² - x + 1/8 = 0

Multiply by 8: 16x² - 8x + 1 = 0

16x² - 4x - 4x + 1 = 0

4x(4x - 1) - 1(4x - 1) = 0

(4x - 1)(4x - 1) = 0

(4x - 1)² = 0

4x - 1 = 0 → x = 1/4

✅ Roots: x = 1/4 (repeated)

Solution:

100x² - 20x + 1 = 0

(10x)² - 2(10x)(1) + 1² = 0

(10x - 1)² = 0

10x - 1 = 0 → x = 1/10

✅ Roots: x = 1/10 (repeated)

Solution:

Let the numbers be x and y

x + y = 27, xy = 182

Quadratic: t² - (sum)t + product = 0

t² - 27t + 182 = 0

t² - 13t - 14t + 182 = 0

t(t - 13) - 14(t - 13) = 0

(t - 13)(t - 14) = 0

t = 13 or t = 14

✅ Numbers are 13 and 14

Solution:

Let integers = x and x + 1

x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x + 1 - 365 = 0

2x² + 2x - 364 = 0

Divide by 2: x² + x - 182 = 0

x² + 14x - 13x - 182 = 0

x(x + 14) - 13(x + 14) = 0

(x + 14)(x - 13) = 0

x = -14 (reject, positive) or x = 13

✅ Integers are 13 and 14

Solution:

Let base = x cm, altitude = x - 7 cm

By Pythagoras: x² + (x-7)² = 13²

x² + x² - 14x + 49 = 169

2x² - 14x + 49 - 169 = 0

2x² - 14x - 120 = 0

Divide by 2: x² - 7x - 60 = 0

x² - 12x + 5x - 60 = 0

x(x - 12) + 5(x - 12) = 0

(x - 12)(x + 5) = 0

x = 12 or x = -5 (reject)

Base = 12 cm, Altitude = 5 cm

✅ Sides: 12 cm, 5 cm, 13 cm

Solution:

Let number of articles = x

Cost per article = 2x + 3

Total cost = x(2x + 3) = 90

2x² + 3x - 90 = 0

2x² + 15x - 12x - 90 = 0

x(2x + 15) - 6(2x + 15) = 0

(2x + 15)(x - 6) = 0

x = 6 or x = -15/2 (reject)

Number of articles = 6

Cost per article = 2(6) + 3 = 15

✅ 6 articles, each ₹15

Solution:

a = 2, b = -3, c = 5

Discriminant D = b² - 4ac = (-3)² - 4(2)(5) = 9 - 40 = -31

Since D < 0

✅ No real roots (imaginary roots)

Solution:

a = 3, b = -4√3, c = 4

D = b² - 4ac = (-4√3)² - 4(3)(4) = 16 × 3 - 48 = 48 - 48 = 0

Since D = 0, roots are real and equal

Roots: x = -b/2a = (4√3)/(2×3) = (4√3)/6 = (2√3)/3

✅ Real and equal roots: x = 2√3/3

Solution:

a = 2, b = -6, c = 3

D = (-6)² - 4(2)(3) = 36 - 24 = 12 > 0

Real and distinct roots

x = [6 ± √12] / 4 = [6 ± 2√3] / 4 = [3 ± √3] / 2

✅ Real roots: x = (3+√3)/2, x = (3-√3)/2

Solution:

For equal roots, D = 0

b² - 4ac = 0

k² - 4(2)(3) = 0

k² - 24 = 0

k² = 24

✅ k = ± 2√6

Solution:

kx(x-2) + 6 = 0 → kx² - 2kx + 6 = 0

a = k, b = -2k, c = 6

D = b² - 4ac = (-2k)² - 4(k)(6) = 4k² - 24k = 0

4k(k - 6) = 0

k = 0 or k = 6

k = 0 not possible (equation becomes 6=0)

✅ k = 6

Solution:

Let breadth = x m, length = 2x m

Area = x × 2x = 2x² = 800

x² = 400 → x = 20 (positive)

Breadth = 20 m, Length = 40 m

✅ Yes possible. Length = 40 m, Breadth = 20 m

Solution:

Let one friend's age = x, other = 20 - x

Four years ago: (x-4)(20-x-4) = 48

(x-4)(16-x) = 48

16x - x² - 64 + 4x = 48

-x² + 20x - 64 - 48 = 0

-x² + 20x - 112 = 0

Multiply by -1: x² - 20x + 112 = 0

D = 400 - 448 = -48 < 0

❌ No real roots. Situation not possible.

Solution:

Let length = l, breadth = b

Perimeter: 2(l + b) = 80 → l + b = 40 → b = 40 - l

Area: l × b = l(40 - l) = 400

40l - l² = 400

-l² + 40l - 400 = 0

Multiply by -1: l² - 40l + 400 = 0

(l - 20)² = 0 → l = 20, b = 20

✅ Yes possible. Length = 20 m, Breadth = 20 m (square)