Answer :
To find the zeros of the quadratic equation [tex]\( y = x^2 + 6x - 4 \)[/tex] by completing the square, follow these steps:
1. Start with the given quadratic equation:
[tex]\[ x^2 + 6x - 4 = 0 \][/tex]
2. Move the constant term to the other side of the equation:
[tex]\[ x^2 + 6x = 4 \][/tex]
3. To complete the square on the left side, we need to add and subtract the square of half the coefficient of [tex]\(x\)[/tex]. The coefficient of [tex]\(x\)[/tex] is 6, so half of it is 3, and its square is 9. Thus, we add and subtract 9:
[tex]\[ x^2 + 6x + 9 - 9 = 4 \][/tex]
4. Rewrite the equation by grouping the perfect square trinomial and the constant terms:
[tex]\[ (x + 3)^2 - 9 = 4 \][/tex]
5. Simplify by isolating the perfect square:
[tex]\[ (x + 3)^2 = 4 + 9 \implies (x + 3)^2 = 13 \][/tex]
6. Take the square root of both sides:
[tex]\[ x + 3 = \pm \sqrt{13} \][/tex]
7. Solve for [tex]\(x\)[/tex] by isolating it on one side:
[tex]\[ x = -3 \pm \sqrt{13} \][/tex]
Therefore, the zeros of the quadratic equation [tex]\( y = x^2 + 6x - 4 \)[/tex] are:
[tex]\[ x = -3 \pm \sqrt{13} \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{A. \; x=-3 \pm \sqrt{13}} \][/tex]
The result matches the choice A in the provided options.
1. Start with the given quadratic equation:
[tex]\[ x^2 + 6x - 4 = 0 \][/tex]
2. Move the constant term to the other side of the equation:
[tex]\[ x^2 + 6x = 4 \][/tex]
3. To complete the square on the left side, we need to add and subtract the square of half the coefficient of [tex]\(x\)[/tex]. The coefficient of [tex]\(x\)[/tex] is 6, so half of it is 3, and its square is 9. Thus, we add and subtract 9:
[tex]\[ x^2 + 6x + 9 - 9 = 4 \][/tex]
4. Rewrite the equation by grouping the perfect square trinomial and the constant terms:
[tex]\[ (x + 3)^2 - 9 = 4 \][/tex]
5. Simplify by isolating the perfect square:
[tex]\[ (x + 3)^2 = 4 + 9 \implies (x + 3)^2 = 13 \][/tex]
6. Take the square root of both sides:
[tex]\[ x + 3 = \pm \sqrt{13} \][/tex]
7. Solve for [tex]\(x\)[/tex] by isolating it on one side:
[tex]\[ x = -3 \pm \sqrt{13} \][/tex]
Therefore, the zeros of the quadratic equation [tex]\( y = x^2 + 6x - 4 \)[/tex] are:
[tex]\[ x = -3 \pm \sqrt{13} \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{A. \; x=-3 \pm \sqrt{13}} \][/tex]
The result matches the choice A in the provided options.
