(x-7)^(2)-25=0 What are the solutions to the given equation? Choose 1 answer: (A) x=12,x=2 (B) x=12,x=-2 (C) x=-12,x=2 (D) x=-12,x=-2

Recognize difference of squares: Recognize that the equation is in the form of a difference of squares, which can be factored as (a^2 - b^2) = (a + b)(a - b). In this case, (x-7)^2 is a^2 and 25 is b^2.Factor the equation: Factor the equation using the difference of squares formula. (x-7)^2 - 25 = 0 (x-7 + 5)(x-7 - 5) = 0 (x-2)(x-12) = 0Solve for x: Set each factor equal to zero and solve for x. x - 2 = 0 or x - 12 = 0 x = 2 or x = 12

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(x-7)^(2)-25=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
x=12,x=2
(B)
x=12,x=-2
(C)
x=-12,x=2
(D)
x=-12,x=-2

$(x-7)^{2}-25=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=12, x=2$ $\newline$ (B) $x=12, x=-2$ $\newline$ (C) $x=-12, x=2$ $\newline$ (D) $x=-12, x=-2$

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Grade 6

Multiply two decimals: where does the decimal point go?

Full solution

(x-7)^{2}-25=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=12, x=2

\newline

(B)

x=12, x=-2

\newline

(C)

x=-12, x=2

\newline

(D)

x=-12, x=-2

Recognize difference of squares: Recognize that the equation is in the form of a difference of squares, which can be factored as $(a^2 - b^2) = (a + b)(a - b)$ . In this case, $(x-7)^2$ is $a^2$ and $25$ is $b^2$ .
Factor the equation: Factor the equation using the difference of squares formula. $\newline$ $(x-7)^2 - 25 = 0$ $\newline$ $(x-7 + 5)(x-7 - 5) = 0$ $\newline$ $(x-2)(x-12) = 0$
Solve for x: Set each factor equal to zero and solve for x. $\newline$ $x - 2 = 0$ or $x - 12 = 0$ $\newline$ $x = 2$ or $x = 12$