Q1(iv):

Find the roots of the following quadratic equations by factorisation: (iv) $2x^2 – x + \frac{1}{8} = 0$

Given: The quadratic equation $2x^2 - x + \frac{1}{8} = 0$.

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

Step 1: Simplifying the Equation
To make the factorisation process easier, we first eliminate the fraction by multiplying the entire equation by $8$.

$8 \times (2x^2 - x + \frac{1}{8}) = 8 \times 0$

$16x^2 - 8x + 1 = 0$

Step 2: Splitting the Middle Term
We need to factorise the quadratic expression $16x^2 - 8x + 1$. We look for two numbers such that:
1. Their product is equal to the product of the coefficient of $x^2$ and the constant term: $16 \times 1 = 16$.
2. Their sum is equal to the coefficient of $x$: $-8$.

The two numbers that satisfy these conditions are $-4$ and $-4$, since $(-4) \times (-4) = 16$ and $(-4) + (-4) = -8$.

Step 3: Factorising by Grouping
Rewrite the middle term $-8x$ as $-4x - 4x$:

$16x^2 - 4x - 4x + 1 = 0$

Group the terms into two pairs:

$(16x^2 - 4x) - (4x - 1) = 0$

Factor out the common terms from each group:

$4x(4x - 1) - 1(4x - 1) = 0$

Step 4: Extracting the Factors
Now, factor out the common binomial $(4x - 1)$:

$(4x - 1)(4x - 1) = 0$

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

Step 5: Finding the Roots
To find the roots, set each factor equal to zero [By the Zero Product Property]:

$4x - 1 = 0$

$4x = 1$

$x = \frac{1}{4}$

Since both factors are identical, the equation has two equal real roots.

Final Answer: The roots of the quadratic equation are $x = \frac{1}{4}$ and $x = \frac{1}{4}$.