To solve these problems, let's take them step by step.
### Part 1: Solving Equations
#### b. Solve [tex]\(2x^2 + 12x + 9 = -8 + 9\)[/tex]
1. Simplify the equation:
[tex]\[
2x^2 + 12x + 9 = 1
\][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[
2x^2 + 12x + 8 = 0
\][/tex]
3. Divide through by 2 to simplify:
[tex]\[
x^2 + 6x + 4 = 0
\][/tex]
4. Solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 6\)[/tex], and [tex]\(c = 4\)[/tex]:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 - 16}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{20}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm 2\sqrt{5}}{2}
\][/tex]
[tex]\[
x = -3 \pm \sqrt{5}
\][/tex]
So, the solutions are:
[tex]\[
x = -3 + \sqrt{5}, \quad x = -3 - \sqrt{5}
\][/tex]
#### d. Solve [tex]\(x^2 + 6x + 9 = -4 + 9\)[/tex]
1. Simplify the equation:
[tex]\[
x^2 + 6x + 9 = 5
\][/tex]
2. Rearrange the equation to set it to zero:
[tex]\[
x^2 + 6x + 4 = 0
\][/tex]
3. Solve the quadratic equation in the same manner as above:
[tex]\[
x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{36 - 16}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm \sqrt{20}}{2}
\][/tex]
[tex]\[
x = \frac{-6 \pm 2\sqrt{5}}{2}
\][/tex]
[tex]\[
x = -3 \pm \sqrt{5}
\][/tex]
So, the solutions are:
[tex]\[
x = -3 + \sqrt{5}, \quad x = -3 - \sqrt{5}
\][/tex]
### Part 2: Area of the Right Triangle
One leg of a right triangle is 4 inches long and the hypotenuse is 5 inches long. What is the area of this triangle?
1. We know the hypotenuse ([tex]\(c\)[/tex]) is 5 inches, and one leg ([tex]\(a\)[/tex]) is 4 inches.
2. Using the Pythagorean theorem [tex]\(a^2 + b^2 = c^2\)[/tex], solve for the other leg ([tex]\(b\)[/tex]):
[tex]\[
4^2 + b^2 = 5^2
\][/tex]
[tex]\[
16 + b^2 = 25
\][/tex]
[tex]\[
b^2 = 25 - 16
\][/tex]
[tex]\[
b^2 = 9
\][/tex]
[tex]\[
b = 3
\][/tex]
3. Now, calculate the area of the triangle using the formula for the area of a right triangle [tex]\((\text{Area} = \frac{1}{2} \times \text{base} \times \text{height})\)[/tex]:
[tex]\[
\text{Area} = \frac{1}{2} \times 4 \times 3
\][/tex]
[tex]\[
\text{Area} = \frac{1}{2} \times 12
\][/tex]
[tex]\[
\text{Area} = 6 \, \text{square inches}
\][/tex]
Therefore, the correct answer is:
a. 6 square inches
