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3n^(2)=27
What are the solutions to the given equation?
Choose 1 answer:
(A)
n=sqrt3
(B)
n=3
(C)
n=-sqrt3 and
n=sqrt3
(D)
n=-3 and
n=3

$3 n^{2}=27$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $n=\sqrt{3}$ $\newline$ (B) $n=3$ $\newline$ (C) $n=-\sqrt{3}$ and $n=\sqrt{3}$ $\newline$ (D) $n=-3$ and $n=3$

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Solve multi-variable equations

Full solution

Q.

3 n^{2}=27

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

n=\sqrt{3}

\newline

(B)

n=3

\newline

(C)

n=-\sqrt{3}

and

n=\sqrt{3}

\newline

(D)

n=-3

and

n=3

Isolate $n^2$ : We have the equation $3n^2 = 27$ . To solve for $n$ , we first need to isolate $n^2$ . We do this by dividing both sides of the equation by $3$ . $\newline$ Calculation: $(3n^2) / 3 = 27 / 3$ $\newline$ $n^2 = 9$
Take square root: Now that we have $n^2 = 9$ , we take the square root of both sides to solve for $n$ . Remember that taking the square root of a number gives us two solutions: one positive and one negative. $\newline$ Calculation: $\sqrt{n^2} = \pm\sqrt{9}$ $\newline$ n = $\pm3$

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