Let's solve each of these quadratic equations step-by-step.

a) $(6x - 18)(3x + 2) = 0$

For a product to equal zero, at least one of the factors must be zero:

1. $6x - 18 = 0$
   • Solve for $x$:
     $x = \frac{18}{6} = 3$
2. $3x + 2 = 0$
   • Solve for $x$:
     $x = -\frac{2}{3}$

Solutions: $x = 3$ and $x = -\frac{2}{3}$

b) $x^2 - 7x + 10 = 0$

This is a quadratic equation in standard form. We can factor it as: $x^2 - 7x + 10 = (x - 5)(x - 2) = 0$ Set each factor equal to zero:

1. $x - 5 = 0 \Rightarrow x = 5$
2. $x - 2 = 0 \Rightarrow x = 2$

Solutions: $x = 5$ and $x = 2$

c) $2x^2 + 2x - 12 = 0$

First, simplify by dividing all terms by 2: $x^2 + x - 6 = 0$ Now, factor it: $(x + 3)(x - 2) = 0$ Set each factor equal to zero:

1. $x + 3 = 0 \Rightarrow x = -3$
2. $x - 2 = 0 \Rightarrow x = 2$

Solutions: $x = -3$ and $x = 2$

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

This is a difference of squares: $(2x - 1)(2x + 1) = 0$ Set each factor equal to zero:

1. $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$
2. $2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$

Solutions: $x = \frac{1}{2}$ and $x = -\frac{1}{2}$

e) $x^2 = 9$

Take the square root of both sides: $x = \pm \sqrt{9} = \pm 3$

Solutions: $x = 3$ and $x = -3$

f) $x^2 - 2x + 1 = 0$

This is a perfect square trinomial: $(x - 1)^2 = 0$ Set the factor equal to zero: $x - 1 = 0 \Rightarrow x = 1$

Solution: $x = 1$ (double root)

Would you like more details on any step?

Related Questions:

1. How do you factor trinomials where the leading coefficient is not 1?
2. How can you solve quadratic equations using the quadratic formula?
3. What are the different methods for solving quadratic equations?
4. How do you identify perfect square trinomials?
5. How does the difference of squares formula work?

Tip:

When solving quadratics, check if the equation can be simplified or factored before resorting to the quadratic formula.