Q1(v):

Find the roots of the following quadratic equations by factorisation: (v) $100x^2 – 20x + 1 = 0$

Given: A quadratic equation $100x^2 - 20x + 1 = 0$.

To find: The roots of the given quadratic equation using the method of factorisation.

Step 1: Identify the coefficients of the quadratic equation.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Comparing $100x^2 - 20x + 1 = 0$ with the standard form:
$a = 100$
$b = -20$
$c = 1$

Step 2: Determine the splitting of the middle term.
To factorise by splitting the middle term, we need to find two numbers such that:
1. Their product is equal to $a \times c = 100 \times 1 = 100$.
2. Their sum is equal to $b = -20$.

Let the two numbers be $p$ and $q$.
$p \times q = 100$
$p + q = -20$
Considering the factors of $100$: $(-10) \times (-10) = 100$ and $(-10) + (-10) = -20$.
Thus, the two numbers are $-10$ and $-10$.

Step 3: Rewrite the middle term and factorise by grouping.
Substitute $-20x$ with $-10x - 10x$ in the original equation:
$100x^2 - 10x - 10x + 1 = 0$

Group the terms into two pairs:
$(100x^2 - 10x) - (10x - 1) = 0$

Factor out the greatest common factor from each group:
$10x(10x - 1) - 1(10x - 1) = 0$

Step 4: Extract the common binomial factor.
$(10x - 1)(10x - 1) = 0$
This can be written as:
$(10x - 1)^2 = 0$

Step 5: Solve for $x$.
To find the roots, set the factor equal to zero:
$10x - 1 = 0$
$10x = 1$
$x = \frac{1}{10}$

Since the equation is a perfect square, both roots are identical:
$x = \frac{1}{10}, \frac{1}{10}$

Final Answer: The roots of the quadratic equation $100x^2 - 20x + 1 = 0$ are $\frac{1}{10}$ and $\frac{1}{10}$.