Enter the values for the variables that give the correct simplified expressions, [tex]\( x \geq 0 \)[/tex].

[tex]\[
\begin{array}{l}
\sqrt{50 x^2} = \sqrt{25 \cdot 2 \cdot x^2} = 5x \sqrt{b} \\
b = \square \\
\sqrt{32 x} = \sqrt{16 \cdot 2 \cdot x} = c \sqrt{2x} \\
c = \square \\
\sqrt{18 n} = \sqrt{9 \cdot 2 \cdot n} = e \sqrt{2n} \\
e = \square \\
\sqrt{72 x^2} = \sqrt{36 \cdot 2 \cdot x^2} = gx \sqrt{2} \\
g = \square
\end{array}
\][/tex]



Answer :

Let's start solving these expressions one by one, by simplifying the square roots step-by-step.

1. Expression: [tex]\(\sqrt{50 x^2}\)[/tex]
- We start with [tex]\(\sqrt{50 x^2}\)[/tex].
- We can factorize [tex]\(50 x^2\)[/tex] as [tex]\(25 \cdot 2 \cdot x^2\)[/tex].
- Taking the square root of each factor:
[tex]\[ \sqrt{25 \cdot 2 \cdot x^2} = \sqrt{25} \cdot \sqrt{2} \cdot \sqrt{x^2} \][/tex]
- Knowing that [tex]\(\sqrt{25} = 5\)[/tex] and [tex]\(\sqrt{x^2} = x\)[/tex] (for [tex]\(x \geq 0\)[/tex]):
[tex]\[ \sqrt{50 x^2} = 5x \sqrt{2} \][/tex]
- Thus, [tex]\(b = 2\)[/tex].

2. Expression: [tex]\(\sqrt{32 x}\)[/tex]
- We start with [tex]\(\sqrt{32 x}\)[/tex].
- We can factorize [tex]\(32 x\)[/tex] as [tex]\(16 \cdot 2 \cdot x\)[/tex].
- Taking the square root of each factor:
[tex]\[ \sqrt{16 \cdot 2 \cdot x} = \sqrt{16} \cdot \sqrt{2x} \][/tex]
- Knowing that [tex]\(\sqrt{16} = 4\)[/tex]:
[tex]\[ \sqrt{32 x} = 4 \sqrt{2x} \][/tex]
- Thus, [tex]\(c = 4\)[/tex].

3. Expression: [tex]\(\sqrt{18 n}\)[/tex]
- We start with [tex]\(\sqrt{18 n}\)[/tex].
- We can factorize [tex]\(18 n\)[/tex] as [tex]\(9 \cdot 2 \cdot n\)[/tex].
- Taking the square root of each factor:
[tex]\[ \sqrt{9 \cdot 2 \cdot n} = \sqrt{9} \cdot \sqrt{2n} \][/tex]
- Knowing that [tex]\(\sqrt{9} = 3\)[/tex]:
[tex]\[ \sqrt{18 n} = 3 \sqrt{2n} \][/tex]
- Thus, [tex]\(e = 3\)[/tex].

4. Expression: [tex]\(\sqrt{72 x^2}\)[/tex]
- We start with [tex]\(\sqrt{72 x^2}\)[/tex].
- We can factorize [tex]\(72 x^2\)[/tex] as [tex]\(36 \cdot 2 \cdot x^2\)[/tex].
- Taking the square root of each factor:
[tex]\[ \sqrt{36 \cdot 2 \cdot x^2} = \sqrt{36} \cdot \sqrt{2} \cdot \sqrt{x^2} \][/tex]
- Knowing that [tex]\(\sqrt{36} = 6\)[/tex] and [tex]\(\sqrt{x^2} = x\)[/tex] (for [tex]\(x \geq 0\)[/tex]):
[tex]\[ \sqrt{72 x^2} = 6x \sqrt{2} \][/tex]
- Thus, [tex]\(g = 6\)[/tex].

In summary, the values are:
[tex]\[ \begin{array}{l} b = 2 \\ c = 4 \\ e = 3 \\ g = 6 \end{array} \][/tex]

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