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${:[4],[4^(2)+4^(2)=x^(2),],[16+16=x^(2),x=sqrt(x^(2))=sqrt32],[32=x^(2)," or "4sqrt2],[," or "5.656 dots]:} 3$

$\begin{array}{rr} 4 \\ 4^{2}+4^{2}=x^{2} & \\ 16+16=x^{2} & x=\sqrt{x^{2}}=\sqrt{32} \\ 32=x^{2} & \text { or } 4 \sqrt{2} \\ & \text { or } 5.656 \ldots \end{array}$ $\newline$ $3$

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Domain and range of absolute value functions: equations

Full solution

Q.

\begin{array}{rr} 4 \\ 4^{2}+4^{2}=x^{2} & \\ 16+16=x^{2} & x=\sqrt{x^{2}}=\sqrt{32} \\ 32=x^{2} & \text { or } 4 \sqrt{2} \\ & \text { or } 5.656 \ldots \end{array}

\newline

3

Addition of $16$ s: Now, add the two $16$ s together to get the value on the left side of the equation. $\newline$ $16 + 16 = 32$ , so now we have $32 = x^{2}$ .
Finding $x$ : To find $x$ , we take the square root of both sides of the equation. $x = \sqrt{32}$ , but we can simplify $\sqrt{32}$ further.
Simplifying $\sqrt{32}$ : Simplify $\sqrt{32}$ by breaking it down into $\sqrt{16} \times \sqrt{2}$ . Since $\sqrt{16} = 4$ , we have $x = 4 \times \sqrt{2}$ .
Checking for further simplification: Now, let's check if we can simplify $4 \times \sqrt{2}$ further. $\newline$ Nope, that's as simple as it gets, so $x = 4\sqrt{2}$ .
Considering negative solution: But wait, we need to check if there's another possible value for $x$ . $\newline$ Since we took the square root, $x$ could also be negative, so $x = -4\sqrt{2}$ is also a solution.

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