The equation given in the image is:

$3q^2 + 4q + \frac{4}{3} = 0$

You mentioned using the formula $a^2 + 2ab + b^2$, which is the expansion of a perfect square trinomial $(a + b)^2$. Let's check if this quadratic equation can be factored using that approach or another method.

Step 1: Simplify the Equation

The quadratic equation is: $3q^2 + 4q + \frac{4}{3} = 0$

To make the factoring easier, let's eliminate the fraction by multiplying the entire equation by 3:

$9q^2 + 12q + 4 = 0$

Step 2: Check for a Perfect Square Trinomial

We need to see if this trinomial can be written in the form $(aq + b)^2$, which expands to $a^2q^2 + 2abq + b^2$.

1. Identify the components:

   • $a^2q^2$ term: $a^2 = 9$, so $a = 3$.
   • $b^2$ term: $b^2 = 4$, so $b = 2$.
   • Middle term $2abq = 2(3)(2)q = 12q$.

2. Verification:

   • $9q^2 + 12q + 4$ does indeed match the form $(3q + 2)^2$.

Step 3: Factor the Equation

The quadratic can be factored as: $(3q + 2)^2 = 0$

Step 4: Solve for $q$

Setting the factor equal to zero: $3q + 2 = 0$

Solve for $q$: $q = -\frac{2}{3}$

Final Answer:

The solution to the equation is: $q = -\frac{2}{3}$

Would you like more details or have any questions?

Related Questions:

1. How can we verify if a quadratic equation is a perfect square trinomial?
2. What are the general methods for solving quadratic equations?
3. Can every quadratic equation be factored into perfect squares?
4. What does the discriminant tell us about the roots of a quadratic equation?
5. How would you solve a quadratic equation that doesn't factor easily?

Tip:

When dealing with quadratic equations, always check if they can be factored into a perfect square trinomial—it simplifies the solving process significantly.