Using the Babylonian method find √600 ???????????? √235.how?

Answer:With the Babylonian method [tex]\sqrt{600} \approx 24.494897[/tex] and [tex]\sqrt{235} \approx 15.329709 [/tex].Step-by-step explanation:To find the square root of 600, do the following:1. Make an initial guess:  Because [tex]\sqrt{576}=24[/tex] and [tex]\sqrt{625} =25[/tex] you can start with [tex]x_{0}=24[/tex] as your initial guess.2. Apply the formula: [tex]x_{1}=\frac{(x_{0}+\frac{x}{x_{0}})}{2}[/tex] where [tex]x=600[/tex][tex]x_{1}=\frac{(24+\frac{600}{24})}{2}\\x_{1}=24.5[/tex]The number [tex]x_{1}[/tex] is a better approximation to [tex]\sqrt{600}[/tex]3. Iterate until convergence:For this apply the formula:[tex]x_{n+1}=\frac{(x_{n}+\frac{x}{x_{n}})}{2}[/tex] Convergence is achieved when the digits of [tex]x_{n+1}[/tex] and [tex]x_{n}[/tex] agree to as many decimal places as you desire.[tex]x_{2}=\frac{(x_{1}+\frac{x}{x_{1}})}{2}\\x_{2}=\frac{(24.5+\frac{600}{24.5})}{2}\\x_{2}=24.494897[/tex][tex]x_{3}=\frac{(x_{2}+\frac{x}{x_{2}})}{2}\\x_{3}=\frac{(24.494897+\frac{600}{24.494897})}{2}\\x_{3}=24.494897\\[/tex]Because [tex]x_{2}[/tex] and [tex]x_{3}[/tex] agree to six decimal places. we can say that an estimate for [tex]\sqrt{600} \approx 24.494897[/tex].You can compare with the value that WolframAlpha gives you which is [tex]\sqrt{600} \approx 24.49489742[/tex] you can see that it agrees to six decimal places.To find the square root of 235, do the following:1. Make an initial guess:  Because [tex]\sqrt{225}=15[/tex] and [tex]\sqrt{256} =16[/tex] you can start with [tex]x_{0}=15[/tex] as your initial guess.2. Apply the formula: [tex]x_{1}=\frac{(x_{0}+\frac{x}{x_{0}})}{2}[/tex] where [tex]x=235[/tex][tex]x_{1}=\frac{(15+\frac{235}{15})}{2}\\x_{1}=\frac{46}{3}[/tex]3. Iterate until convergence:Apply the formula:[tex]x_{n+1}=\frac{(x_{n}+\frac{x}{x_{n}})}{2}[/tex] [tex]x_{2}=\frac{(x_{1}+\frac{235}{x_{1}})}{2}\\x_{2}=\frac{(\frac{46}{3}+\frac{235}{\frac{46}{3}})}{2} \\x_{2}=15.329710[/tex][tex]x_{3}=\frac{(x_{2}+\frac{235}{x_{2}})}{2}\\x_{3}=\frac{(15.329710+\frac{235}{15.329710})}{2} \\x_{3}=15.329709[/tex][tex]x_{4}=\frac{(x_{3}+\frac{235}{x_{3}})}{2}\\x_{4}=\frac{(15.329709+\frac{235}{15.329709})}{2} \\x_{3}=15.329709[/tex]Because [tex]x_{3}[/tex] and [tex]x_{4}[/tex] agree to six decimal places. We can say that an estimate for [tex]\sqrt{235} \approx 15.329709 [/tex].WolframAlpha gives you [tex]\sqrt{235} \approx 15.329709716 [/tex]