Get the answers you need from a community of experts on IDNLearn.com. Ask your questions and receive reliable and comprehensive answers from our dedicated community of professionals.
Sagot :
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.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.