Question

$${ 4 }^{ 2 } + { 5 }^{ 2 } + { z }^{ 2 } = { 21 }^{ 2 }$$

Answer

z=20,-20

Solution


Simplify  \({4}^{2}\)  to  \(16\).
\[16+{5}^{2}+{z}^{2}={21}^{2}\]
Simplify  \({5}^{2}\)  to  \(25\).
\[16+25+{z}^{2}={21}^{2}\]
Simplify  \({21}^{2}\)  to  \(441\).
\[16+25+{z}^{2}=441\]
Simplify  \(16+25+{z}^{2}\)  to  \({z}^{2}+41\).
\[{z}^{2}+41=441\]
Subtract \(41\) from both sides.
\[{z}^{2}=441-41\]
Simplify  \(441-41\)  to  \(400\).
\[{z}^{2}=400\]
Take the square root of both sides.
\[z=\pm \sqrt{400}\]
Since \(20\times 20=400\), the square root of \(400\) is \(20\).
\[z=\pm 20\]