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$Solve for x. Enter the solutions from least to greatest. {:[(x-3)^(2)-81=0],[" lesser "x=◻],[" greater "x=◻]:}$

Solve for $x$ . Enter the solutions from least to greatest. $\newline$ $\begin{array}{l} (x-3)^{2}-81=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}$

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Evaluate exponential functions

Full solution

Q. Solve for

x

. Enter the solutions from least to greatest.

\newline

\begin{array}{l} (x-3)^{2}-81=0 \\ \text { lesser } x=\square \\ \text { greater } x=\square \end{array}

Identify the equation: Identify the equation to solve. $\newline$ The given equation is $(x-3)^2 - 81 = 0$ .
Rewrite as difference of squares: Rewrite the equation as a difference of squares. $\newline$ The equation $(x-3)^2 - 81 = 0$ can be written as $(x-3)^2 - 9^2 = 0$ , since $81$ is $9$ squared.
Factor the difference of squares: Factor the difference of squares. $\newline$ The equation $(x-3)^2 - 9^2 = 0$ can be factored into $((x-3) - 9)((x-3) + 9) = 0$ .
Solve for x (first factor): Set each factor equal to zero and solve for x. $\newline$ First factor: $(x-3) - 9 = 0$ $\newline$ $x - 3 - 9 = 0$ $\newline$ $x - 12 = 0$ $\newline$ $x = 12$
Solve for x (second factor): Solve the second factor for x. $\newline$ Second factor: $(x-3) + 9 = 0$ $\newline$ $x - 3 + 9 = 0$ $\newline$ $x + 6 = 0$ $\newline$ $x = -6$
Order the solutions: Order the solutions from least to greatest. $\newline$ The lesser value of $x$ is $-6$ , and the greater value of $x$ is $12$ .

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