Given: The algebraic equation $(x - 2)(x + 1) = (x - 1)(x + 3)$.

To Find: Determine whether the given equation is a quadratic equation.

Definition: A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

Step 1: Expanding the Left-Hand Side (LHS)

The LHS is $(x - 2)(x + 1)$. We apply the distributive property of multiplication over addition: $(a + b)(c + d) = ac + ad + bc + bd$.

$(x - 2)(x + 1) = x(x) + x(1) - 2(x) - 2(1)$

$= x^2 + x - 2x - 2$

$= x^2 - x - 2$

Step 2: Expanding the Right-Hand Side (RHS)

The RHS is $(x - 1)(x + 3)$. Similarly, applying the distributive property:

$(x - 1)(x + 3) = x(x) + x(3) - 1(x) - 1(3)$

$= x^2 + 3x - x - 3$

$= x^2 + 2x - 3$

Step 3: Equating LHS and RHS and Simplifying

Now, set the expanded LHS equal to the expanded RHS:

$x^2 - x - 2 = x^2 + 2x - 3$

To bring all terms to one side, subtract $(x^2 + 2x - 3)$ from both sides:

$x^2 - x^2 - x - 2x - 2 + 3 = 0$

Combine like terms:

$(1 - 1)x^2 + (-1 - 2)x + (-2 + 3) = 0$

$0x^2 - 3x + 1 = 0$

$-3x + 1 = 0$

Step 4: Conclusion

The resulting equation is $-3x + 1 = 0$. This is a linear equation, not a quadratic equation, because the coefficient of $x^2$ is $0$ (i.e., $a = 0$). Since the definition of a quadratic equation requires $a \neq 0$, this equation does not satisfy the condition.

Final Answer: The given equation $(x - 2)(x + 1) = (x - 1)(x + 3)$ is not a quadratic equation.

Given: The equation $(x^3 - 4x^2 - x + 1) = (x - 2)^3$.

To Find: Determine whether the given equation is a quadratic equation.

Step 1: Understanding the definition of a Quadratic Equation
A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$. The highest power (degree) of the variable $x$ must be exactly 2.

Step 2: Expanding the right-hand side (RHS)
The RHS is $(x - 2)^3$. We use the algebraic identity for the cube of a binomial:
$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
Here, $a = x$ and $b = 2$.
$(x - 2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - (2)^3$
$(x - 2)^3 = x^3 - 6x^2 + 3(x)(4) - 8$
$(x - 2)^3 = x^3 - 6x^2 + 12x - 8$ [Applying the identity $(a-b)^3$]

Step 3: Equating LHS and RHS
Substitute the expanded RHS back into the original equation:
$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$

Step 4: Simplifying the equation
To simplify, we bring all terms to the left-hand side (LHS):
$(x^3 - 4x^2 - x + 1) - (x^3 - 6x^2 + 12x - 8) = 0$
$x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0$ [Distributing the negative sign]

Step 5: Combining like terms
Group the terms by their powers of $x$:
$(x^3 - x^3) + (-4x^2 + 6x^2) + (-x - 12x) + (1 + 8) = 0$
$0x^3 + 2x^2 - 13x + 9 = 0$
$2x^2 - 13x + 9 = 0$

Step 6: Conclusion
The resulting equation is $2x^2 - 13x + 9 = 0$. This equation is in the form $ax^2 + bx + c = 0$, where $a = 2$, $b = -13$, and $c = 9$. Since the highest degree of the variable $x$ is 2 and $a \neq 0$, the equation satisfies the definition of a quadratic equation.

Final Answer: Yes, the given equation is a quadratic equation.

Given:

1. The shape of the plot is rectangular.
2. The area of the rectangular plot ($A$) = $528$ $m^2$.
3. The relationship between length ($l$) and breadth ($b$): The length is one more than twice its breadth.

To Find:

Represent the given situation in the form of a quadratic equation.

Step 1: Defining the Variables

Let the breadth of the rectangular plot be $x$ metres.
According to the problem, the length is one more than twice the breadth.
Therefore, the length of the plot $l = (2x + 1)$ metres.

Step 2: Applying the Formula for Area

The area of a rectangle is given by the formula:
$Area = Length \times Breadth$
$A = l \times b$

Step 3: Formulating the Equation

Substitute the given values and the expressions defined in Step 1 into the area formula:
$528 = (2x + 1) \times x$

Step 4: Simplifying the Expression

Distribute $x$ into the parentheses:
$528 = 2x^2 + x$

Step 5: Rearranging into Standard Quadratic Form

The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Subtract $528$ from both sides of the equation:
$2x^2 + x - 528 = 0$

Justification:

The equation $2x^2 + x - 528 = 0$ is a quadratic equation because it is a polynomial equation of degree 2, where $a = 2$, $b = 1$, and $c = -528$.

Final Answer: The quadratic equation representing the situation is $2x^2 + x - 528 = 0$, where $x$ represents the breadth of the plot in metres.

Given:

• Total distance to be covered by the train, $d = 480$ km.
• The train travels at a uniform speed.
• If the speed were $8$ km/h less, the time taken to cover the same distance would increase by $3$ hours.

To Find:

Represent the given situation in the form of a quadratic equation in terms of the speed of the train.

Step 1: Defining Variables

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

We know the fundamental relationship between distance, speed, and time is given by the formula:

$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$

Step 2: Formulating Time Expressions

Case 1: When the train travels at the original uniform speed $x$ km/h.

Time taken ($t_1$) = $\frac{480}{x}$ hours.

Case 2: When the speed is reduced by $8$ km/h.

New speed = $(x - 8)$ km/h.

Time taken ($t_2$) = $\frac{480}{x - 8}$ hours.

Step 3: Establishing the Relationship

According to the problem, the train takes $3$ hours more when the speed is reduced. Therefore, the difference between the new time ($t_2$) and the original time ($t_1$) is $3$ hours:

$t_2 - t_1 = 3$

Substituting the expressions from Step 2:

$\frac{480}{x - 8} - \frac{480}{x} = 3$

Step 4: Simplifying the Equation

Divide the entire equation by $3$ to simplify the coefficients:

$\frac{160}{x - 8} - \frac{160}{x} = 1$

Find a common denominator for the left-hand side, which is $x(x - 8)$:

$\frac{160x - 160(x - 8)}{x(x - 8)} = 1$

Expand the numerator:

$\frac{160x - 160x + 1280}{x^2 - 8x} = 1$

Simplify the numerator ($160x - 160x = 0$):

$\frac{1280}{x^2 - 8x} = 1$

Step 5: Final Algebraic Form

Multiply both sides by $(x^2 - 8x)$:

$1280 = x^2 - 8x$

Rearrange the terms to the standard form of a quadratic equation $ax^2 + bx + c = 0$:

$x^2 - 8x - 1280 = 0$

Final Answer: The quadratic equation representing the situation is $x^2 - 8x - 1280 = 0$, where $x$ is the speed of the train in km/h.

Given:

1. Rohan’s mother is $26$ years older than Rohan.

2. The product of their ages $3$ years from now will be $360$.

To Find:

Represent the given situation in the form of a quadratic equation in terms of Rohan's present age.

Step 1: Defining the Variables

Let the present age of Rohan be $x$ years.

Since Rohan’s mother is $26$ years older than him, the present age of Rohan’s mother is $(x + 26)$ years.

Step 2: Determining Ages 3 Years from Now

After $3$ years, the age of each person will increase by $3$ years.

Rohan’s age after $3$ years = $(x + 3)$ years.

Rohan’s mother’s age after $3$ years = $(x + 26) + 3 = (x + 29)$ years.

Step 3: Formulating the Equation

According to the problem, the product of their ages $3$ years from now is $360$.

Therefore, we set up the equation:

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

Step 4: Expanding and Simplifying the Equation

Using the distributive property (FOIL method) to expand the left side:

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

$x^2 + 29x + 3x + 87 = 360$

[Combining like terms $29x$ and $3x$]:

$x^2 + 32x + 87 = 360$

Step 5: Bringing the Equation to Standard Form

The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Subtract $360$ from both sides:

$x^2 + 32x + 87 - 360 = 0$

$x^2 + 32x - 273 = 0$

Justification:

The resulting equation $x^2 + 32x - 273 = 0$ is a polynomial of degree $2$, which satisfies the definition of a quadratic equation where $a=1$, $b=32$, and $c=-273$.

Final Answer: The quadratic equation representing the situation is $x^2 + 32x - 273 = 0$, where $x$ is Rohan's present age.

Given: The algebraic equation $(x - 3)(2x + 1) = x(x + 5)$.

To Find: Determine whether the given equation is a quadratic equation.

Definition: A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$. The highest power (degree) of the variable in a quadratic equation must be 2.

Step 1: Expanding the Left-Hand Side (LHS)
The LHS is $(x - 3)(2x + 1)$. We apply the distributive property (FOIL method):
$(x - 3)(2x + 1) = x(2x) + x(1) - 3(2x) - 3(1)$
$= 2x^2 + x - 6x - 3$
$= 2x^2 - 5x - 3$ [Combining like terms $x$ and $-6x$]

Step 2: Expanding the Right-Hand Side (RHS)
The RHS is $x(x + 5)$. We apply the distributive property:
$x(x + 5) = x(x) + x(5)$
$= x^2 + 5x$

Step 3: Equating LHS and RHS and Simplifying
Now, set the expanded LHS equal to the expanded RHS:
$2x^2 - 5x - 3 = x^2 + 5x$

To bring the equation into the standard form $ax^2 + bx + c = 0$, we subtract $(x^2 + 5x)$ from both sides:
$(2x^2 - x^2) + (-5x - 5x) - 3 = 0$
$x^2 - 10x - 3 = 0$

Step 4: Verification against the Standard Form
Comparing $x^2 - 10x - 3 = 0$ with the standard form $ax^2 + bx + c = 0$:
Here, $a = 1$, $b = -10$, and $c = -3$.
Since $a \neq 0$ and the highest degree of the variable $x$ is 2, the equation satisfies the definition of a quadratic equation.

Final Answer: Yes, the given equation $(x - 3)(2x + 1) = x(x + 5)$ is a quadratic equation because it simplifies to the form $x^2 - 10x - 3 = 0$.

Given: The product of two consecutive positive integers is $306$.

To find: Represent the given situation in the form of a quadratic equation.

Step 1: Defining the variables

Let the first positive integer be $x$.

Since the integers are consecutive, the next integer must be $x + 1$.

[Assumption: $x$ is a positive integer, therefore $x > 0$].

Step 2: Formulating the equation based on the given condition

According to the problem, the product of these two consecutive integers is $306$.

Mathematically, this is expressed as:

$x(x + 1) = 306$

Step 3: Expanding the expression

Distribute $x$ into the parentheses:

$x^2 + x = 306$

Step 4: Rearranging into standard quadratic form

A quadratic equation is represented in the standard form $ax^2 + bx + c = 0$.

Subtract $306$ from both sides of the equation to set the right side to zero:

$x^2 + x - 306 = 0$

Step 5: Verification of the form

Comparing $x^2 + x - 306 = 0$ with the standard form $ax^2 + bx + c = 0$:

Here, $a = 1$, $b = 1$, and $c = -306$.

Since the highest power of the variable $x$ is $2$, this is a quadratic equation.

Final Answer: The quadratic equation representing the given situation is $x^2 + x - 306 = 0$, where $x$ is the smaller integer.

Given: The algebraic equation $(x + 2)^3 = 2x(x^2 - 1)$.

To Find: Determine whether the given equation is a quadratic equation.

Definition: A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$. The highest power (degree) of the variable $x$ must be exactly 2.

Step 1: Expanding the Left-Hand Side (LHS)
The LHS is $(x + 2)^3$. We use the algebraic identity for the cube of a binomial: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.
Here, $a = x$ and $b = 2$.
$(x + 2)^3 = x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3$
$= x^3 + 6x^2 + 3(x)(4) + 8$
$= x^3 + 6x^2 + 12x + 8$ [Expanding the terms]

Step 2: Expanding the Right-Hand Side (RHS)
The RHS is $2x(x^2 - 1)$. We use the distributive property of multiplication over subtraction: $a(b - c) = ab - ac$.
$2x(x^2 - 1) = (2x \cdot x^2) - (2x \cdot 1)$
$= 2x^3 - 2x$ [Distributing $2x$ into the parentheses]

Step 3: Equating LHS and RHS and Simplifying
Now, set the expanded LHS equal to the expanded RHS:
$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$
To bring all terms to one side, subtract $(2x^3 - 2x)$ from both sides:
$x^3 - 2x^3 + 6x^2 + 12x + 2x + 8 = 0$
$-x^3 + 6x^2 + 14x + 8 = 0$ [Combining like terms]

Step 4: Analyzing the Degree of the Equation
The resulting equation is $-x^3 + 6x^2 + 14x + 8 = 0$.
The highest power (degree) of the variable $x$ in this equation is 3. [Since the term $-x^3$ exists and its coefficient is non-zero].

Conclusion:
Since the degree of the equation is 3, it is a cubic equation, not a quadratic equation. A quadratic equation must have a maximum degree of 2.

Final Answer: The given equation $(x + 2)^3 = 2x(x^2 - 1)$ is not a quadratic equation.

Given: The equation $x^2 - 2x = (-2)(3 - x)$.

To Find: Determine whether the given equation is a quadratic equation.

Definition: A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

Step 1: Simplify the right-hand side (RHS) of the equation.

The given equation is:

$x^2 - 2x = (-2)(3 - x)$

Applying the distributive property of multiplication over subtraction, $a(b - c) = ab - ac$:

$x^2 - 2x = (-2 \times 3) - (-2 \times x)$

$x^2 - 2x = -6 + 2x$

Step 2: Rearrange the equation into the standard form $ax^2 + bx + c = 0$.

To bring all terms to the left-hand side, subtract $2x$ and add $6$ to both sides of the equation:

$x^2 - 2x - 2x + 6 = 0$

Combine the like terms ($-2x$ and $-2x$):

$x^2 - 4x + 6 = 0$

Step 3: Compare with the standard form.

Comparing $x^2 - 4x + 6 = 0$ with the standard quadratic form $ax^2 + bx + c = 0$:

Here, $a = 1$, $b = -4$, and $c = 6$.

[Since $a = 1 \neq 0$, the equation satisfies the condition for being a quadratic equation.]

Conclusion:

Since the highest power of the variable $x$ in the simplified equation is $2$, and the coefficient of $x^2$ is non-zero, the given equation is a quadratic equation.

Final Answer: Yes, the given equation $x^2 - 2x = (-2)(3 - x)$ is a quadratic equation.

Given: The equation $x^2 + 3x + 1 = (x - 2)^2$.

To Find: Determine whether the given equation is a quadratic equation.

Definition: A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

Step 1: Expand the right-hand side of the equation.

The given equation is:

$x^2 + 3x + 1 = (x - 2)^2$

We use the algebraic identity $(a - b)^2 = a^2 - 2ab + b^2$. Here, $a = x$ and $b = 2$.

$(x - 2)^2 = x^2 - 2(x)(2) + (2)^2$

$(x - 2)^2 = x^2 - 4x + 4$

Step 2: Substitute the expanded form back into the original equation.

$x^2 + 3x + 1 = x^2 - 4x + 4$

Step 3: Rearrange the terms to one side to set the equation to zero.

Subtract $(x^2 - 4x + 4)$ from both sides of the equation:

$x^2 + 3x + 1 - (x^2 - 4x + 4) = 0$

$x^2 + 3x + 1 - x^2 + 4x - 4 = 0$

Step 4: Simplify the expression by combining like terms.

Group the $x^2$ terms, the $x$ terms, and the constant terms:

$(x^2 - x^2) + (3x + 4x) + (1 - 4) = 0$

$0x^2 + 7x - 3 = 0$

$7x - 3 = 0$

Step 5: Analyze the resulting equation.

The simplified equation is $7x - 3 = 0$. This is a linear equation because the highest power of the variable $x$ is 1. Comparing this to the standard form $ax^2 + bx + c = 0$, we see that the coefficient of $x^2$ (which is $a$) is $0$. Since a quadratic equation must have $a \neq 0$, this equation does not satisfy the definition.

Final Answer: The given equation $x^2 + 3x + 1 = (x - 2)^2$ is not a quadratic equation because the $x^2$ terms cancel out, leaving a linear equation.

Given: The equation $(x + 1)^2 = 2(x - 3)$.

To Find: Determine whether the given equation is a quadratic equation.

Definition: A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

Step 1: Expanding the left-hand side (LHS) of the equation.
The expression is $(x + 1)^2$. We use the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$.
Substituting $a = x$ and $b = 1$:
$(x + 1)^2 = x^2 + 2(x)(1) + (1)^2$
$(x + 1)^2 = x^2 + 2x + 1$

Step 2: Expanding the right-hand side (RHS) of the equation.
The expression is $2(x - 3)$. We use the distributive property $a(b - c) = ab - ac$.
$2(x - 3) = 2(x) - 2(3)$
$2(x - 3) = 2x - 6$

Step 3: Equating the expanded sides and simplifying.
Equating the results from Step 1 and Step 2:
$x^2 + 2x + 1 = 2x - 6$

Step 4: Bringing all terms to one side to form the standard quadratic form.
Subtract $2x$ from both sides:
$x^2 + 2x - 2x + 1 = -6$
$x^2 + 1 = -6$

Add $6$ to both sides:
$x^2 + 1 + 6 = 0$
$x^2 + 7 = 0$

Step 5: Comparing with the standard form $ax^2 + bx + c = 0$.
The equation $x^2 + 7 = 0$ can be written as $1x^2 + 0x + 7 = 0$.
Here, $a = 1$, $b = 0$, and $c = 7$.
Since $a \neq 0$ (as $1 \neq 0$), the equation satisfies the condition for being a quadratic equation.

Final Answer: Yes, the given equation $(x + 1)^2 = 2(x - 3)$ is a quadratic equation because it can be simplified to the form $ax^2 + bx + c = 0$ where $a \neq 0$.

Given: The equation $(2x - 1)(x - 3) = (x + 5)(x - 1)$.

To Find: Determine whether the given equation is a quadratic equation.

Definition: A quadratic equation in the variable $x$ is an equation of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

Step 1: Expanding the Left-Hand Side (LHS)

The LHS is $(2x - 1)(x - 3)$. We apply the distributive property (FOIL method):

$(2x - 1)(x - 3) = 2x(x) + 2x(-3) - 1(x) - 1(-3)$

$= 2x^2 - 6x - x + 3$

$= 2x^2 - 7x + 3$ [Combining like terms $-6x$ and $-x$]

Step 2: Expanding the Right-Hand Side (RHS)

The RHS is $(x + 5)(x - 1)$. We apply the distributive property:

$(x + 5)(x - 1) = x(x) + x(-1) + 5(x) + 5(-1)$

$= x^2 - x + 5x - 5$

$= x^2 + 4x - 5$ [Combining like terms $-x$ and $5x$]

Step 3: Equating LHS and RHS and Simplifying

Now, set the expanded LHS equal to the expanded RHS:

$2x^2 - 7x + 3 = x^2 + 4x - 5$

To bring the equation into the standard form $ax^2 + bx + c = 0$, subtract $(x^2 + 4x - 5)$ from both sides:

$2x^2 - x^2 - 7x - 4x + 3 + 5 = 0$

$x^2 - 11x + 8 = 0$

Step 4: Verification against the Standard Form

The resulting equation is $x^2 - 11x + 8 = 0$.

Comparing this with the standard form $ax^2 + bx + c = 0$:

Here, $a = 1$, $b = -11$, and $c = 8$.

Since $a = 1 \neq 0$, the equation satisfies the condition for being a quadratic equation.

Final Answer: Yes, the given equation $(2x - 1)(x - 3) = (x + 5)(x - 1)$ is a quadratic equation.