Answer :
Let's solve the quadratic equation [tex]\( 4x^2 - 12x + 9 = 5 \)[/tex] for [tex]\( x \)[/tex].
1. Rewrite the equation in standard form:
[tex]\[ 4x^2 - 12x + 9 - 5 = 0 \implies 4x^2 - 12x + 4 = 0 \][/tex]
2. Factor the quadratic equation (if possible):
[tex]\[ 4x^2 - 12x + 4 = 0 \][/tex]
This can be immediately solved by recognizing it's a quadratic equation. To solve, we should find the roots using the quadratic formula, which is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 4 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 4 \)[/tex].
3. Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{144 - 64}}{8} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{80}}{8} \][/tex]
[tex]\[ x = \frac{12 \pm 4 \sqrt{5}}{8} \][/tex]
[tex]\[ x = \frac{12}{8} \pm \frac{4 \sqrt{5}}{8} \][/tex]
[tex]\[ x = \frac{3}{2} \pm \frac{\sqrt{5}}{2} \][/tex]
4. Simplify the solutions found:
[tex]\[ x = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{5}}{2} \][/tex]
Now, compare these solutions to the provided options:
- [tex]\( \boxed{\text{A.}\ x = \frac{\sqrt{5} + 3}{2}} \)[/tex]
- B. [tex]\( x = -\sqrt{4} - 3 = -2 - 3 = -5 \)[/tex]
- C. [tex]\( x = -\sqrt{5} + \frac{3}{2} \)[/tex]
- D. [tex]\( x = \sqrt{5} + \frac{3}{2} = \frac{\sqrt{5} + 3}{2} \)[/tex] (matches option A again)
- [tex]\( \boxed{\text{E.}\ x = \frac{-\sqrt{5} + 3}{2}} \)[/tex]
- F. [tex]\( x = \sqrt{4} - 3 = 2 - 3 = -1 \)[/tex]
From the given options, the solutions that apply are:
- [tex]\( \boxed{\text{A.}\ x = \frac{\sqrt{5} + 3}{2}} \)[/tex]
- [tex]\( \boxed{\text{E.}\ x = \frac{-\sqrt{5} + 3}{2}} \)[/tex]
Thus, the correct options are A and E.
1. Rewrite the equation in standard form:
[tex]\[ 4x^2 - 12x + 9 - 5 = 0 \implies 4x^2 - 12x + 4 = 0 \][/tex]
2. Factor the quadratic equation (if possible):
[tex]\[ 4x^2 - 12x + 4 = 0 \][/tex]
This can be immediately solved by recognizing it's a quadratic equation. To solve, we should find the roots using the quadratic formula, which is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 4 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 4 \)[/tex].
3. Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{144 - 64}}{8} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{80}}{8} \][/tex]
[tex]\[ x = \frac{12 \pm 4 \sqrt{5}}{8} \][/tex]
[tex]\[ x = \frac{12}{8} \pm \frac{4 \sqrt{5}}{8} \][/tex]
[tex]\[ x = \frac{3}{2} \pm \frac{\sqrt{5}}{2} \][/tex]
4. Simplify the solutions found:
[tex]\[ x = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x = \frac{3 - \sqrt{5}}{2} \][/tex]
Now, compare these solutions to the provided options:
- [tex]\( \boxed{\text{A.}\ x = \frac{\sqrt{5} + 3}{2}} \)[/tex]
- B. [tex]\( x = -\sqrt{4} - 3 = -2 - 3 = -5 \)[/tex]
- C. [tex]\( x = -\sqrt{5} + \frac{3}{2} \)[/tex]
- D. [tex]\( x = \sqrt{5} + \frac{3}{2} = \frac{\sqrt{5} + 3}{2} \)[/tex] (matches option A again)
- [tex]\( \boxed{\text{E.}\ x = \frac{-\sqrt{5} + 3}{2}} \)[/tex]
- F. [tex]\( x = \sqrt{4} - 3 = 2 - 3 = -1 \)[/tex]
From the given options, the solutions that apply are:
- [tex]\( \boxed{\text{A.}\ x = \frac{\sqrt{5} + 3}{2}} \)[/tex]
- [tex]\( \boxed{\text{E.}\ x = \frac{-\sqrt{5} + 3}{2}} \)[/tex]
Thus, the correct options are A and E.