Let's solve the quadratic equation [tex]\(4x^2 + 12x + 9 = 0\)[/tex] using the quadratic formula and determine the correct substitution of the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex].
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, identify the coefficients [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] from the given equation:
[tex]\[ a = 4 \][/tex]
[tex]\[ b = 12 \][/tex]
[tex]\[ c = 9 \][/tex]
Let's apply these values to the quadratic formula step-by-step.
1. Substitute [tex]\(a = 4\)[/tex], [tex]\(b = 12\)[/tex], and [tex]\(c = 9\)[/tex] into the formula:
[tex]\[ x = \frac{-12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} \][/tex]
2. Calculate the discriminant ([tex]\(b^2 - 4ac\)[/tex]):
[tex]\[ b^2 - 4ac = 12^2 - 4(4)(9) \][/tex]
[tex]\[ = 144 - 144 \][/tex]
[tex]\[ = 0 \][/tex]
3. Substitute the discriminant back into the formula:
[tex]\[ x = \frac{-12 \pm \sqrt{0}}{8} \][/tex]
4. Simplify the square root and the expression:
[tex]\[ x = \frac{-12 \pm 0}{8} \][/tex]
[tex]\[ x = \frac{-12}{8} \][/tex]
[tex]\[ x = -1.5 \][/tex]
Since the discriminant is zero, there is exactly one real solution:
[tex]\[ x = -1.5 \][/tex]
Now let's match the correct substitution from the given options:
1. [tex]\( x = \frac{12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} \)[/tex]
2. [tex]\( x = \frac{-12 \pm \sqrt{12^2 + 4(4)(9)}}{2(4)} \)[/tex]
3. [tex]\( x = \frac{-12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} \)[/tex]
Comparing these options to our step-by-step substitution, we find that the correct substitution is given by:
[tex]\[ x = \frac{-12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} \][/tex]
Therefore, the correct answer is:
[tex]\[ x = \frac{-12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} \][/tex]
Thus, the correct option is:
[tex]\[ (3) \][/tex]
So, the solution to the quadratic equation [tex]\(4x^2 + 12x + 9 = 0\)[/tex] is correctly substituted in:
[tex]\[ x = \frac{-12 \pm \sqrt{12^2 - 4(4)(9)}}{2(4)} \][/tex]
The correct option that shows the correct substitution is [tex]\(\boxed{3}\)[/tex].
