(x-3)^(2)-81=0 What are the solutions to the given equation? Choose 1 answer: (A) x=12,x=6 (B) x=-12,x=-6 (c) x=12,x=-6 (D) x=-12,x=6

Setting the equation and identifying terms: First, we need to set the equation equal to zero and identify the terms. (x-3)^(2) - 81 = 0 This is a quadratic equation in the form of a difference of squares.Applying the difference of squares formula: Next, we apply the difference of squares formula, which is a^2 - b^2 = (a + b)(a - b), where a = (x-3) and b = 9. (x-3 + 9)(x-3 - 9) = 0Simplifying the factors: Now, we simplify the factors. (x-3 + 9) = (x + 6) (x-3 - 9) = (x - 12) So, the equation becomes: (x + 6)(x - 12) = 0Setting each factor equal to zero: To find the solutions, we set each factor equal to zero. x + 6 = 0 or x - 12 = 0Solving for x in the first equation: Solve for x in the first equation. x + 6 = 0 x = -6Solving for x in the second equation: Solve for x in the second equation. x - 12 = 0 x = 12

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(x-3)^(2)-81=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
x=12,x=6
(B)
x=-12,x=-6
(c)
x=12,x=-6
(D)
x=-12,x=6

$(x-3)^{2}-81=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=12, x=6$ $\newline$ (B) $x=-12, x=-6$ $\newline$ (C) $x=12, x=-6$ $\newline$ (D) $x=-12, x=6$

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Algebra 1

Solve linear equations: mixed review

Full solution

(x-3)^{2}-81=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=12, x=6

\newline

(B)

x=-12, x=-6

\newline

(C)

x=12, x=-6

\newline

(D)

x=-12, x=6

Setting the equation and identifying terms: First, we need to set the equation equal to zero and identify the terms. $\newline$ $(x-3)^{2} - 81 = 0$ $\newline$ This is a quadratic equation in the form of a difference of squares.
Applying the difference of squares formula: Next, we apply the difference of squares formula, which is $a^2 - b^2 = (a + b)(a - b)$ , where $a = (x-3)$ and $b = 9$ . $\newline$ $(x-3 + 9)(x-3 - 9) = 0$
Simplifying the factors: Now, we simplify the factors. $\newline$ $(x-3 + 9) = (x + 6)$ $\newline$ $(x-3 - 9) = (x - 12)$ $\newline$ So, the equation becomes: $\newline$ $(x + 6)(x - 12) = 0$
Setting each factor equal to zero: To find the solutions, we set each factor equal to zero. $\newline$ $x + 6 = 0$ or $x - 12 = 0$
Solving for x in the first equation: Solve for $x$ in the first equation. $x + 6 = 0$ $x = -6$
Solving for x in the second equation: Solve for x in the second equation. $\newline$ $x - 12 = 0$ $\newline$ $x = 12$