Conical bottle

When a conical bottle rests on its flat base, the water in the bottle is 8 cm from it vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the height of the bottle?

Correct result:

h =  10.2195 cm

Solution:

$$\begin{gathered} h_1=8\doteq 10.2195\text{ cm } \\ {h}_2=2\doteq -8.2195\text{ cm } \\ V={\displaystyle \frac{1}{3}}\pi r^{2}h \\ {V}_3={\displaystyle \frac{1}{3}}\pi r_1^2{h}_1 \\ {V}_1=V-V_3 \\ {V}_1={\displaystyle \frac{1}{3}}\pi r^{2}h-{\displaystyle \frac{1}{3}}\pi r_1^2{h}_1 \\ {V}_1={\displaystyle \frac{1}{3}}\pi (r^{2}h-r_1^2{h}_1) \\ {r}_1<r_2<r \\ {r}_1:h_1=r:h \\ {r}_1=r\cdot h_1\mathrm{/}h \\ {r}_2:(h-h_2)=r:h \\ {r}_2=r\cdot (h-h_2)\mathrm{/}h \\ {V}_2={\displaystyle \frac{1}{3}}\pi r_2^2(h-h_2) \\ {V}_1=V_2 \\ {\displaystyle \frac{1}{3}}\pi (r^{2}h-r_1^2{h}_1)={\displaystyle \frac{1}{3}}\pi r_2^2(h-h_2) \\ (r^{2}h-r_1^2{h}_1)=r_2^2(h-h_2) \\ (r^{2}h-(r\cdot h_1\mathrm{/}h{)}^2{h}_1)=(r(h-h_2)\mathrm{/}h{)}^2(h-h_2) \\ (h-(h_1\mathrm{/}h{)}^2\cdot h_1)=((h-h_2)\mathrm{/}h{)}^2(h-h_2) \\ (h-(8\mathrm{/}h{)}^2\cdot 8)=((h-2)\mathrm{/}h{)}^2(h-2) \\ h(h-2)=84 \\ {h}^2-2h-84=0 \\ a=1;b=-2;c=-84 \\ D=b^2-4ac=2^2-4\cdot 1\cdot (-84)=340 \\ D>0 \\ {h}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{2\pm \sqrt{340}}{2}}={\displaystyle \frac{2\pm 2\sqrt{85}}{2}} \\ {h}_{1,2}=1\pm 9.21954445729 \\ {h}_1=10.2195444573 \\ {h}_2=-8.21954445729 \\ \text{ Factored form of the equation: } \\ (h-10.2195444573)(h+8.21954445729)=0 \\ h>0 \\ h=h_1=10.2195=10.2195\text{ cm} \end{gathered}$$

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