Find the zeros of [tex]$y = x^2 - 6x - 4$[/tex] by completing the square.

A. [tex]$x = 3 \pm \sqrt{5}$[/tex]
B. [tex][tex]$x = -3 \pm \sqrt{13}$[/tex][/tex]
C. [tex]$x = 3 \pm \sqrt{13}$[/tex]
D. [tex]$x = \pm 3$[/tex]



Answer :

To find the zeros of the quadratic equation [tex]\( y = x^2 - 6x - 4 \)[/tex] by completing the square, we will transform the equation into a form that makes it easier to solve. Here are the detailed steps:

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

2. Set [tex]\( y \)[/tex] to 0 to find the zeros:
[tex]\[ 0 = x^2 - 6x - 4 \][/tex]

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

4. Complete the square on the left side. To do this, take half of the coefficient of [tex]\( x \)[/tex], square it, and add it to both sides of the equation. The coefficient of [tex]\( x \)[/tex] is [tex]\(-6\)[/tex], half of [tex]\(-6\)[/tex] is [tex]\(-3\)[/tex], and squaring [tex]\(-3\)[/tex] gives [tex]\(9\)[/tex]:
[tex]\[ x^2 - 6x + 9 = 4 + 9 \][/tex]
[tex]\[ (x - 3)^2 = 13 \][/tex]

5. Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x - 3 = \pm \sqrt{13} \][/tex]

6. Solve for [tex]\( x \)[/tex] by isolating it:
[tex]\[ x = 3 \pm \sqrt{13} \][/tex]

The zeros of the quadratic equation [tex]\( y = x^2 - 6x - 4 \)[/tex] are:
[tex]\[ x = 3 - \sqrt{13} \][/tex]
[tex]\[ x = 3 + \sqrt{13} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{C. \; x = 3 \pm \sqrt{13}} \][/tex]