Q1(v):

Check whether the following are quadratic equations : (v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$

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.