Question

How do you solve \[{x^2} + 6x + 9 = 0\] by factoring?

Answer 1

Hint: Here, we will compare the given equation with the standard form of a quadratic equation to find the coefficients of the equation. We will split the middle term into two terms and factorize the given quadratic equation. Then, we will apply the principle of zero product and simplify the equations to get the required solutions of the quadratic equation.

Formula Used:
The principle of zero products states that if \[ab = 0\], then \[a = 0\] or \[b = 0\].

Complete step-by-step solution:
First, we will rewrite the given equation.
Rewriting the given equation, we get
\[1 \cdot {x^2} + 6x + 9 = 0\]
Now, we will find the coefficients and constants of the given quadratic equation.
Comparing the equation \[1 \cdot {x^2} + 6x + 9 = 0\] with the standard form of a quadratic equation \[a{x^2} + bx + c = 0\], we get
\[a = 1\], \[b = 6\], and \[c = 9\]
Thus, we get the coefficient of \[{x^2}\] as 1, the coefficient of \[x\] as 6, and the constant as 9.
The product of the coefficient of \[{x^2}\] and the constant 9 is \[1 \times 9 = 9\].
We need to find two factors of 9 such that their product is 9, and their sum is equal to the coefficient of \[x\], that is 6.
The sum of the factors 3 and 3 of 9 is equal to the coefficient of \[x\], that is 6.
Rewriting the expression \[6x\], we get
\[6x = 3x + 3x\]
Substituting \[6x = 3x + 3x\] in the quadratic equation, we get
\[ \Rightarrow {x^2} + 3x + 3x + 9 = 0\]
Factoring the equation using grouping, we get
\[\begin{array}{l} \Rightarrow x\left( {x + 3} \right) + 3\left( {x + 3} \right) = 0\\ \Rightarrow \left( {x + 3} \right)\left( {x + 3} \right) = 0\end{array}\]
The principle of zero products states that if \[ab = 0\], then \[a = 0\] or \[b = 0\].
Applying the principle of zero products, we get
\[ \Rightarrow x + 3 = 0\] or \[x + 3 = 0\].
Simplifying the expressions, we get
\[ \Rightarrow x = - 3, - 3\].

Therefore, the required solutions of the given quadratic equation are \[ - 3\] and \[ - 3\].

Note:
We used the term “quadratic equation” in our solution. A quadratic equation is an equation of degree 2. It is of the form \[a{x^2} + bx + c = 0\], where \[a\] is not equal to 0. A quadratic equation has 2 solutions.
The given equation can also be factored using the algebraic identity for the square of the sum of two numbers. The square of the sum of two numbers \[a\] and \[b\] is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
The given quadratic equation can be factored as
\[\begin{array}{l} \Rightarrow {x^2} + 2\left( 3 \right)\left( x \right) + {3^2} = 0\\ \Rightarrow {\left( {x + 3} \right)^2} = 0\end{array}\]
We can rewrite the above equation as:
\[ \Rightarrow \left( {x + 3} \right)\left( {x + 3} \right) = 0\]
Applying the principle of zero products, we get
\[ \Rightarrow x + 3 = 0\] or \[x + 3 = 0\].
Simplifying the expressions, we get
\[ \Rightarrow x = - 3, - 3\].