Q1(ii):

Find the roots of the following quadratic equations by factorisation: (ii) $2x^2 + x – 6 = 0$

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

To find: The roots of the given quadratic equation using the method of factorisation (splitting the middle term).

Step 1: Identify the coefficients of the quadratic equation.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Comparing $2x^2 + x - 6 = 0$ with the standard form:
$a = 2$
$b = 1$
$c = -6$

Step 2: Determine the product and the sum for splitting the middle term.
To factorise by splitting the middle term, we need to find two numbers such that:
1. Their product is equal to $a \times c = 2 \times (-6) = -12$.
2. Their sum is equal to $b = 1$.

Step 3: Find the two numbers.
We look for factors of $-12$ that add up to $1$:
Factors of $-12$: $(-1, 12), (1, -12), (-2, 6), (2, -6), (-3, 4), (3, -4)$.
Checking the sums:
$-3 + 4 = 1$.
Thus, the two numbers are $4$ and $-3$.

Step 4: Rewrite the middle term and factorise by grouping.
Rewrite the equation $2x^2 + x - 6 = 0$ as:
$2x^2 + 4x - 3x - 6 = 0$ [Splitting the middle term $x$ into $4x - 3x$]
Group the terms:
$(2x^2 + 4x) - (3x + 6) = 0$
Factor out the common terms from each group:
$2x(x + 2) - 3(x + 2) = 0$ [Factoring out $2x$ from the first group and $3$ from the second group]
$(2x - 3)(x + 2) = 0$ [Taking $(x + 2)$ as a common factor]

Step 5: Find the roots by setting each factor to zero.
According to the Zero Product Property, if the product of two factors is zero, at least one of the factors must be zero.
Case 1: $2x - 3 = 0$
$2x = 3$
$x = \frac{3}{2}$

Case 2: $x + 2 = 0$
$x = -2$

Final Answer: The roots of the quadratic equation $2x^2 + x - 6 = 0$ are $x = \frac{3}{2}$ and $x = -2$.