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]
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]