Question

Solve 2(x-3)^2 + 5 = 29 using square roots?

A. By using the square root property: 2(x-3)^2 + 5 = 29 → 2(x-3)^2 = 29 - 5 → 2(x-3)^2 = 24 → (x-3)^2 = 12
B. By using the square root property: 2(x-3)^2 + 5 = 29 → 2(x-3)^2 = 29 - 5 → 2(x-3)^2 = 24 → x-3 = ±√12
C. By using the square root property: 2(x-3)^2 + 5 = 29 → 2(x-3)^2 = 29 - 5 → 2(x-3)^2 = 24 → x = 3 ± √12
D. By using the square root property: 2(x-3)^2 + 5 = 29 → 2(x-3)^2 = 29 - 5 → 2(x-3)^2 = 24 → x = 3 ± 12

Answer 1

To solve the quadratic equation 2(x-3)^2 + 5 = 29, we isolate the term (x-3)^2, divide by 2, take the square root of both sides, and solve for x, yielding x = 3 ± 2√3.

Step-by-step explanation:

To solve the equation 2(x-3)^2 + 5 = 29 using square roots we follow these steps:

1. Isolate the quadratic term: 2(x-3)^2 = 29 - 5, which simplifies to 2(x-3)^2 = 24.
2. Divide both sides by 2 to further isolate the squared term: (x-3)^2 = 12.
3. Use the square root property to solve for x: Take the square root of both sides, yielding x - 3 = ±√12, which simplifies to x - 3 = ±2√3 since 12 is a perfect square multiplied by 3 (12=4×3).
4. Finally, solve for x: x = 3 ± 2√3.

In conclusion, the correct form that uses the square root property is option B, although it needs a slight adjustment:
2(x-3)^2 + 5 = 29 → 2(x-3)^2 = 29 - 5 → 2(x-3)^2 = 24 → x-3 = ±√12 → x = 3 ± 2√3.