2. Solve [tex]x^2-6x+4=0[/tex] by completing the square.

A. [tex]x=3+\sqrt{13}, 3-\sqrt{13}[/tex]

B. [tex]x=-3+\sqrt{5},-3-\sqrt{5}[/tex]

C. [tex]x=3+\sqrt{5}, 3-\sqrt{5}[/tex]

D. [tex]x=3,-3[/tex]



Answer :

To solve the quadratic equation [tex]\(x^2 - 6x + 4 = 0\)[/tex] by completing the square, follow these steps:

1. Start with the given equation:
[tex]\[ x^2 - 6x + 4 = 0 \][/tex]

2. Move the constant term to the right side:
[tex]\[ x^2 - 6x = -4 \][/tex]

3. Complete the square on the left side:
- Take half of the coefficient of [tex]\(x\)[/tex] (which is [tex]\(-6\)[/tex]), divide by 2, and square it:
[tex]\[ \left(\frac{-6}{2}\right)^2 = 9 \][/tex]
- Add and subtract this value to/from the left side:
[tex]\[ x^2 - 6x + 9 - 9 = -4 \][/tex]
- Rearrange the equation:
[tex]\[ (x - 3)^2 - 9 = -4 \][/tex]

4. Move the constant term [tex]\(-9\)[/tex] to the right side to isolate the square term:
[tex]\[ (x - 3)^2 = 5 \][/tex]

5. Take the square root of both sides:
[tex]\[ x - 3 = \pm \sqrt{5} \][/tex]

6. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = 3 + \sqrt{5} \quad \text{or} \quad x = 3 - \sqrt{5} \][/tex]

So, the solutions for the quadratic equation [tex]\(x^2 - 6x + 4 = 0\)[/tex] are:
[tex]\[ x = 3 + \sqrt{5} \quad \text{and} \quad x = 3 - \sqrt{5} \][/tex]