Answer :
To determine which of the given choices are solutions to the equation [tex]\(4x^2 - 12x + 9 = 5\)[/tex], let's follow these steps:
1. Rewriting the Equation:
First, we need to bring the constant term on the right-hand side to the left-hand side to form a standard quadratic equation:
[tex]\[ 4x^2 - 12x + 9 - 5 = 0 \implies 4x^2 - 12x + 4 = 0 \][/tex]
2. Simplifying the Equation:
Next, we can simplify the quadratic equation by dividing all terms by 4:
[tex]\[ x^2 - 3x + 1 = 0 \][/tex]
Now, we have a simpler quadratic equation [tex]\(x^2 - 3x + 1 = 0\)[/tex].
3. Solving the Quadratic Equation:
To solve this equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\(x^2 - 3x + 1 = 0\)[/tex], [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex]. Plugging in these values, we get:
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Therefore, the solutions to the equation are:
[tex]\[ x = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{5}}{2} \][/tex]
4. Comparing with Given Choices:
Let's check the given choices against these solutions:
- A. [tex]\(x = -\sqrt{4} - 3 = -2 - 3 = -5\)[/tex] (Not a solution)
- B. [tex]\(x = -\sqrt{5} + \frac{3}{2}\)[/tex] (Not a solution)
- C. [tex]\(x = \sqrt{5} + \frac{3}{2}\)[/tex] (Not a solution)
- D. [tex]\(x = \frac{\sqrt{5} + 3}{2} = \frac{3 + \sqrt{5}}{2}\)[/tex] (Solution)
- E. [tex]\(x = \sqrt{4} - 3 = 2 - 3 = -1\)[/tex] (Not a solution)
- F. [tex]\(x = \frac{-\sqrt{5} + 3}{2} = \frac{3 - \sqrt{5}}{2}\)[/tex] (Solution)
So, the correct answers are:
[tex]\[ \boxed{D \text{ and } F} \][/tex]
1. Rewriting the Equation:
First, we need to bring the constant term on the right-hand side to the left-hand side to form a standard quadratic equation:
[tex]\[ 4x^2 - 12x + 9 - 5 = 0 \implies 4x^2 - 12x + 4 = 0 \][/tex]
2. Simplifying the Equation:
Next, we can simplify the quadratic equation by dividing all terms by 4:
[tex]\[ x^2 - 3x + 1 = 0 \][/tex]
Now, we have a simpler quadratic equation [tex]\(x^2 - 3x + 1 = 0\)[/tex].
3. Solving the Quadratic Equation:
To solve this equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\(x^2 - 3x + 1 = 0\)[/tex], [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex]. Plugging in these values, we get:
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Therefore, the solutions to the equation are:
[tex]\[ x = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{5}}{2} \][/tex]
4. Comparing with Given Choices:
Let's check the given choices against these solutions:
- A. [tex]\(x = -\sqrt{4} - 3 = -2 - 3 = -5\)[/tex] (Not a solution)
- B. [tex]\(x = -\sqrt{5} + \frac{3}{2}\)[/tex] (Not a solution)
- C. [tex]\(x = \sqrt{5} + \frac{3}{2}\)[/tex] (Not a solution)
- D. [tex]\(x = \frac{\sqrt{5} + 3}{2} = \frac{3 + \sqrt{5}}{2}\)[/tex] (Solution)
- E. [tex]\(x = \sqrt{4} - 3 = 2 - 3 = -1\)[/tex] (Not a solution)
- F. [tex]\(x = \frac{-\sqrt{5} + 3}{2} = \frac{3 - \sqrt{5}}{2}\)[/tex] (Solution)
So, the correct answers are:
[tex]\[ \boxed{D \text{ and } F} \][/tex]
