(i) (x + 1)² = 2(x − 3)

Solution:

Expanding both sides of the given equation:

x² + 2x + 1 = 2x − 6

Bringing all terms to one side:

x² + 2x − 2x + 1 + 6 = 0

x² + 7 = 0

Since the simplified equation is of the form ax² + bx + c = 0, where a ≠ 0, it is a quadratic equation.

(ii) x² − 2x = (−2)(3 − x)

Solution:

Simplifying the right-hand side of the equation:

x² − 2x = −6 + 2x

Rearranging the terms:

x² − 2x − 2x + 6 = 0

x² − 4x + 6 = 0

Since the equation is of the form ax² + bx + c = 0, where a ≠ 0, it is a quadratic equation.

(iii) (x − 2)(x + 1) = (x − 1)(x + 3)

Solution:

Multiplying the binomials on both sides:

x² + x − 2x − 2 = x² + 3x − x − 3

x² − x − 2 = x² + 2x − 3

Subtracting x² from both sides and rearranging terms:

−x − 2 − 2x + 3 = 0

−3x + 1 = 0

The highest power of the variable x is 1, so it is a linear equation. Therefore, it is not a quadratic equation.

(iv) (x − 3)(2x + 1) = x(x + 5)

Solution:

Expanding both sides of the equation:

2x² + x − 6x − 3 = x² + 5x

2x² − 5x − 3 = x² + 5x

Bringing all terms to the left-hand side:

2x² − x² − 5x − 5x − 3 = 0

x² − 10x − 3 = 0

Since the equation is of the form ax² + bx + c = 0, where a ≠ 0, it is a quadratic equation.

(v) (2x − 1)(x − 3) = (x + 5)(x − 1)

Solution:

Expanding the expressions on both sides:

2x² − 6x − x + 3 = x² − x + 5x − 5

2x² − 7x + 3 = x² + 4x − 5

Collecting all terms on the left-hand side:

2x² − x² − 7x − 4x + 3 + 5 = 0

x² − 11x + 8 = 0

Since the expression is of the form ax² + bx + c = 0, where a ≠ 0, it is a quadratic equation.

(vi) x² + 3x + 1 = (x − 2)²

Solution:

Expanding the right-hand side using the algebraic identity (a − b)² = a² − 2ab + b²:

x² + 3x + 1 = x² − 4x + 4

Subtracting x² from both sides and shifting terms to the left side:

3x + 4x + 1 − 4 = 0

7x − 3 = 0

Since the degree of this equation is 1, it is a linear equation. Therefore, it is not a quadratic equation.

(vii) (x + 2)³ = 2x(x² − 1)

Solution:

Expanding the left side using (a + b)³ = a³ + 3a²b + 3ab² + b³ and multiplying out the right side:

x³ + 3(x²)(2) + 3(x)(4) + 8 = 2x³ − 2x

x³ + 6x² + 12x + 8 = 2x³ − 2x

Rearranging all terms to one side:

2x³ − x³ − 6x² − 2x − 12x − 8 = 0

x³ − 6x² − 14x − 8 = 0

Since the highest power of the variable is 3, this is a cubic equation. Therefore, it is not a quadratic equation.

(viii) x³ − 4x² − x + 1 = (x − 2)³

Solution:

Expanding the right-hand side expression using (a − b)³ = a³ − 3a²b + 3ab² − b³:

x³ − 4x² − x + 1 = x³ − 3(x²)(2) + 3(x)(4) − 8

x³ − 4x² − x + 1 = x³ − 6x² + 12x − 8

Canceling x³ from both sides and shifting terms to the left-hand side:

−4x² + 6x² − x − 12x + 1 + 8 = 0

2x² − 13x + 9 = 0

Since the simplified equation is of the form ax² + bx + c = 0, where a ≠ 0, it is a quadratic equation.

(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(Note- Important Question for board exam)

Solution:

Let the breadth of the rectangular plot be x metres.

According to the given condition, the length of the plot is one more than twice its breadth. Therefore, the length of the plot is (2x + 1) metres.

We know that the area of a rectangle is given by the product of its length and breadth:

Area = Length × Breadth

528 = (2x + 1)x

528 = 2x² + x

Rearranging the equation into the standard form:

2x² + x − 528 = 0

Hence, the required quadratic equation representing the given situation is 2x² + x − 528 = 0.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Solution:

Let the first positive integer be x.

Since the integers are consecutive, the second positive integer will be x + 1.

According to the given condition, the product of these two integers is 306:

x(x + 1) = 306

x² + x = 306

Rearranging the terms into standard form:

x² + x − 306 = 0

Hence, the required quadratic equation representing the given situation is x² + x − 306 = 0.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

Solution:

Let Rohan’s present age be x years.

Since his mother is 26 years older than him, her present age is (x + 26) years.

Three years from now:

Rohan’s age will be (x + 3) years.

Mother’s age will be (x + 26 + 3) = (x + 29) years.

According to the given condition, the product of their ages 3 years from now is 360:

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

x² + 29x + 3x + 87 = 360

x² + 32x + 87 − 360 = 0

x² + 32x − 273 = 0

Hence, the required quadratic equation representing the given situation is x² + 32x − 273 = 0.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

(Note- Important Question for board exam)

Solution:

Let the uniform speed of the train be x km/h.

The total distance to be traveled is 480 km.

We know that Time = Distance / Speed. Therefore, the time taken at normal speed is:

T₁ = 480 / x

When the speed is reduced by 8 km/h, the new speed becomes (x − 8) km/h. The time taken at this reduced speed is:

T₂ = 480 / (x − 8)

According to the given condition, the new time taken is 3 hours more than the original time:

T₂ − T₁ = 3

[480 / (x − 8)] − [480 / x] = 3

Dividing both sides by 3 to simplify:

[160 / (x − 8)] − [160 / x] = 1

Taking the common denominator:

160[x − (x − 8)] / [x(x − 8)] = 1

160(8) / (x² − 8x) = 1

1280 = x² − 8x

Rearranging into standard form:

x² − 8x − 1280 = 0

Hence, the required quadratic equation representing the given situation is x² − 8x − 1280 = 0.