Step-by-Step Solution
Key Concept: A quadratic equation of the form $ax^2+bx+c=0$ can be solved by (i) factorisation when the product $ac$ can be expressed as a sum of two numbers whose sum is $b$, (ii) completing the square, or (iii) using the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. The appropriate method is chosen based on the coefficients.
Example 1 consists of four quadratic equations.\
1. Equation (i): $x^{2}-5x+6=0$\
*Factorisation:* Find two numbers whose product is $6$ and sum is $-5$. They are $-2$ and $-3$.\
$$x^{2}-5x+6=(x-2)(x-3)=0$$\
Hence $x-2=0$ or $x-3=0$ \=> $x=2$ or $x=3$.\
2. Equation (ii): $2x^{2}+3x-2=0$\
*Factorisation:* Multiply $a$ and $c$: $2\times(-2)=-4$. Find two numbers whose product is $-4$ and sum is $3$: $4$ and $-1$.\
Rewrite the middle term:\
$$2x^{2}+4x-x-2=0$$\
Grouping:\
$$(2x^{2}+4x)-(x+2)=0$$\
$$2x(x+2)-1(x+2)=0$$\
$$(2x-1)(x+2)=0$$\
Hence $2x-1=0$ or $x+2=0$ \=> $x=\frac{1}{2}$ or $x=-2$.\
3. Equation (iii): $x^{2}+4x+4=0$\
*Perfect square:* Recognise $(x+2)^{2}=x^{2}+4x+4$.\
$$ (x+2)^{2}=0$$\
Hence $x+2=0$ \=> $x=-2$ (double root).\
4. Equation (iv): $3x^{2}-2x-8=0$\
*Quadratic formula:* $a=3$, $b=-2$, $c=-8$.\
$$\Delta = b^{2}-4ac = (-2)^{2}-4\times3\times(-8)=4+96=100$$\
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}=\frac{2 \pm 10}{6}$$\
Therefore, \$x = \frac{2+10}{6}=\frac{12}{6}=2\$ or \$x = \frac{2-10}{6}=\frac{-8}{6}= -\frac{4}{3}\$.\
Thus all four equations are solved.
Correct Answer: 1) $x=2$ or $x=3$; 2) $x=\frac{1}{2}$ or $x=-2$; 3) $x=-2$ (double root); 4) $x=2$ or $x=-\frac{4}{3}$.