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 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: A quadratic equation $2x^2 - 6x + 3 = 0$.

To Find: The nature of the roots of the given quadratic equation and, if real roots exist, find their values.

Step 1: Identify the coefficients of the quadratic equation.

A standard quadratic equation is represented as $ax^2 + bx + c = 0$. Comparing the given equation $2x^2 - 6x + 3 = 0$ with the standard form:

• $a = 2$
• $b = -6$
• $c = 3$

Step 2: Determine the nature of the roots using the Discriminant ($D$).

The discriminant of a quadratic equation is given by the formula $D = b^2 - 4ac$.

Substituting the values of $a$, $b$, and $c$:

$D = (-6)^2 - 4(2)(3)$

$D = 36 - 24$

$D = 12$

[Since $D > 0$, the quadratic equation has two distinct real roots.]

Step 3: Apply the Quadratic Formula to find the roots.

The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.

Substituting the known values into the formula:

$x = \frac{-(-6) \pm \sqrt{12}}{2(2)}$

$x = \frac{6 \pm \sqrt{4 \times 3}}{4}$

$x = \frac{6 \pm 2\sqrt{3}}{4}$

Step 4: Simplify the expression.

Factor out the common term $2$ from the numerator:

$x = \frac{2(3 \pm \sqrt{3})}{4}$

$x = \frac{3 \pm \sqrt{3}}{2}$

Therefore, the two roots are:

$x_1 = \frac{3 + \sqrt{3}}{2}$

$x_2 = \frac{3 - \sqrt{3}}{2}$

Final Answer: The roots are real and distinct. The roots of the equation are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 - \sqrt{3}}{2}$.

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 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.