Math Question

what is x if

$4 + \sqrt {x-4} = \sqrt {x+14}$ ?

$\begin{array}{|rcll|} \hline 4 + \sqrt {x-4} &=& \sqrt {x+14} \quad &|\quad \text{square both sides} \\ (4 + \sqrt {x-4})^2 &=& x+14 \\ 16+2\cdot 4 \cdot \sqrt {x-4} + \not{x}-4 &=& \not{x}+14 \\ 16+8 \cdot \sqrt {x-4} -4 &=& 14 \quad &|\quad +4-16\\ 8 \cdot \sqrt {x-4} &=& 14 +4-16 \\ 8 \cdot \sqrt {x-4} &=& 2 \quad &|\quad :2 \\ 4 \cdot \sqrt {x-4} &=& 1 \quad &|\quad \text{square both sides} \\ 16 \cdot (x-4) &=& 1^2 \\ 16 \cdot (x-4) &=& 1 \\ 16x-64 &=& 1 \quad &|\quad +64 \\ 16x &=& 1+64 \\ 16x &=& 65 \quad &|\quad :16 \\ \mathbf{ x } & \mathbf{=} & \mathbf{\frac{65}{16} }\\ \hline \end{array}$

what is x if 4 + sqrt (x-4) = sqrt (x+14)?

what is x if 4 + sqrt (x-4) = sqrt (x+14)?

$4+\sqrt{x-4}=\sqrt{x+14}$

$16+8\sqrt{x-4}+x-4=x+14$

$8\sqrt{x-4}=x+14-16+4-x$

$\sqrt{x-4}=\frac{2}{8}$

$x-4=\frac{1}{16}$

$x=4\ \frac{1}{16}$

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