Using Heron's method, what is the square root of:

-120

- 85

- 20

Please include steps

Guest Aug 1, 2017

edited by
Guest
Aug 1, 2017

#1**+2 **

Heron's method of finding the square root of a number is interesting, but this is what you do

1. Take a reasonable guess of what the answer would be (to the integer)

2. Divide the radicand by your guess.

3. Find the average of your guess and the radicand divided by your guess.

4. Repeat

Ok, I will show how to estimate the values above \(\sqrt{120}\).

1. **Take a Reasonable Guess**

The first step to estimating the value of \(\sqrt{120}\) is to take a guess. I would suggest leaving it as an integer.

I happen to know that \(11^2=121\), so 11 should be a very good guess to what the actual decimal expansion is of \(\sqrt{120}\) is. You may overestimate or underestimate; it should not make a difference. However, the closer your original guess is, the more accurate your result will be.

2. **Divide the Radicand by Your Guess**

The radicand, if you are unaware, is the number inside of the radical. For example, in this case, the radicand of \(\sqrt{120}\) is 120. The radicand of \(\sqrt{2}\) is 2. We want to take 120 and divide it by our guess, 11.

\(\frac{120}{11}=10.\overline{9090}\), so I will leave 120/11 as it is because the fraction cannot be simplified any further.

3. **Find the Average of Your Guess and the Radicand Divided by Your Guess**

In this step, simply take average of your guess, 11, and the radicand divided by your guess, which was calculated to be \(\frac{120}{11}\). Let's do that:

\(\frac{11+\frac{120}{11}}{2}\) | Now, just simplify. I will convert 11 into an improper fraction. \(11*\frac{11}{11}=\frac{121}{11}\) |

\(\frac{\frac{121}{11}+\frac{120}{11}}{2}\) | Find the sum of the numerator by adding the fractions. |

\(\frac{\frac{241}{11}}{2}=\frac{241}{11*2}=\frac{241}{22}\) | Unfortunately, 241/11 cannot be simplified further. Now, I will utilize a fraction rule that states that \(\frac{\frac{a}{b}}{c}=\frac{a}{b*c}\) |

You can continue this process to get a more accurate decimal approximation, too, but this fraction is OK for just an approximation. Here's how close this approximation was:

\(\frac{241}{22}=10.95\overline{45}\)

\(\sqrt{120}\approx10.95445\)

That's really accurate, wouldn't you agree?

Ok, I will do the next one, too:

1. **Take a Reasonable Guess**

Yet again, we must make a guess for the \(\sqrt{85}\).

I happen to know that \(9^2=81\), so this seems like a good estimate.

2. **Divide the Radicand by Your Guess**

**\(\frac{85}{9}\)** cannot be simplified further because the numerator and denominator are co-prime, so we will have to leave it as is. Also, \(\frac{85}{9}=9.\overline{4}\), so it would be difficult to calculate it in decimal form.

3. **Find the Average of Your Guess and the Radicand Divided by Your Guess**

Just like the previous problem, we are finding the average of 81 and \(\frac{85}{9}\):

\(\frac{9+\frac{85}{9}}{2}\) | First, convert 9 into an improper fraction such that the denominator is 9. |

\(\frac{\frac{81}{9}+\frac{85}{9}}{2}\) | Simplify the numerator. |

\(\frac{\frac{166}{9}}{2}\) | Just like the previous problem, we will utilize the fraction rule \(\frac{\frac{a}{b}}{c}=\frac{a}{b*c}\) |

\(\frac{\frac{166}{9}}{2}=\frac{166}{2*9}=\frac{166}{18}\) | In this case, the numbers 166 and 18 have a GCF of 2, so we can simplify this faction further. |

\(\frac{166}{18}\div\frac{2}{2}=\frac{83}{9}\) | |

Let's see how close our approximation is:

\(\frac{83}{9}=9.\overline{2}\)

\(\sqrt{85}\approx9.2195\)

I would say that that is a good approximation.

--------------------------------------------------------------------------------

And finally 20.

My guess will be 4 as \(4^2=16\), so that seems like a good approximation.

\(\frac{20}{4}=5\). Now that I have divided the radicand by my guess. We can do the next step:

\(\frac{4+5}{2}\) | Find the sum in the numerator. |

\(\frac{9}{2}\) | |

As a check, let's see how close our approximation truly is:

\(\frac{9}{2}=4.5 \)

\(\sqrt{20}\approx4.47214\)

That seems pretty good!

TheXSquaredFactor
Aug 1, 2017