The problem asks us to solve the quadratic equation $x^2 + 6x + 9 = 0$ by factorization and describe the nature of the roots of the equation.

1. Problem Description

The problem asks us to solve the quadratic equation $x^2 + 6x + 9 = 0$ by factorization and describe the nature of the roots of the equation.

2. Solution Steps

The given equation is $x^2 + 6x + 9 = 0$. We need to factorize the quadratic expression $x^2 + 6x + 9$.

We are looking for two numbers that multiply to 9 and add up to

6. These numbers are 3 and

3. So, we can write the expression as:

$x^2 + 3x + 3x + 9 = 0$

Now, we can factor by grouping:

$x(x + 3) + 3(x + 3) = 0$

$(x + 3)(x + 3) = 0$

$(x + 3)^2 = 0$

Taking the square root of both sides:

$x + 3 = 0$

$x = -3$

Since $(x+3)^2=0$, there is only one solution $x=-3$.

The roots of the equation are real and equal. The discriminant of the quadratic equation $ax^2 + bx + c = 0$ is given by $D = b^2 - 4ac$. If $D = 0$, then the roots are real and equal. In this case, $a=1$, $b=6$, and $c=9$, so the discriminant is $D = 6^2 - 4(1)(9) = 36 - 36 = 0$. Therefore, the roots are real and equal.

3. Final Answer

The solution to the equation $x^2 + 6x + 9 = 0$ is $x = -3$. The roots are real and equal.