Quadratic Equations: Question 1

Worked solution

Part (a): Factorise $2x^2 + 5x - 12$

Because the coefficient of $x^2$ is not $1$ (a non-monic quadratic), use the “split the middle term” method.

Step 1: Find the key product and sum. Multiply the leading coefficient by the constant term: $a \times c = 2 \times (-12) = -24.$ We need two numbers whose product is $-24$ and whose sum is $+5$ (the coefficient of $x$).

Testing factor pairs of $-24$:

The required numbers are $+8$ and $-3$.

Step 2: Rewrite the middle term and group. $2x^2 + 5x - 12 = 2x^2 + 8x - 3x - 12$ $= 2x(x + 4) - 3(x + 4)$ $= (x + 4)(2x - 3).$

Check by expanding: $(2x-3)(x+4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$ ✓

$\boxed{2x^2 + 5x - 12 = (2x - 3)(x + 4)}$

Part (b): Solve $2x^2 + 5x = 12$

Step 1: Rearrange to the form $=0$. The zero-product rule (“if a product is $0$, one of the factors is $0$”) only applies when one side is zero, so move everything to the left: $2x^2 + 5x - 12 = 0.$

Step 2: Use the factorisation from part (a). $(2x - 3)(x + 4) = 0.$

Step 3: Set each factor equal to zero. $2x - 3 = 0 \quad\Rightarrow\quad 2x = 3 \quad\Rightarrow\quad x = \tfrac{3}{2}$ $x + 4 = 0 \quad\Rightarrow\quad x = -4$

$\boxed{x = \tfrac{3}{2} \quad \text{or} \quad x = -4}$

Alternative method (quadratic formula, useful as a check)

For $2x^2 + 5x - 12 = 0$ with $a=2,\ b=5,\ c=-12$: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4(2)(-12)}}{2(2)} = \frac{-5 \pm \sqrt{25 + 96}}{4} = \frac{-5 \pm \sqrt{121}}{4} = \frac{-5 \pm 11}{4}.$ This gives $x = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2} \qquad \text{or} \qquad x = \frac{-5 - 11}{4} = \frac{-16}{4} = -4,$ which confirms the factorising answer.

Verification by substitution

• $x = \tfrac{3}{2}$: $\ 2\left(\tfrac{3}{2}\right)^2 + 5\left(\tfrac{3}{2}\right) = 2\left(\tfrac{9}{4}\right) + \tfrac{15}{2} = \tfrac{9}{2} + \tfrac{15}{2} = \tfrac{24}{2} = 12$ ✓
• $x = -4$: $\ 2(-4)^2 + 5(-4) = 32 - 20 = 12$ ✓

Both satisfy the original equation $2x^2 + 5x = 12$.