Q1(i):

Check whether the following are quadratic equations : (i) $(x + 1)^2 = 2(x – 3)$

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 $a \neq 0$.