# Computing Square Roots

#### xyz_1965

See attachment. Can someone get me started here? It looks like I need to plug each square root given for n in the formula and simplify.

#### romsek

MHF Helper
pick $a_0$ by taking a rough guess at the square root.

you know $\sqrt{9}=3$ so try $a_0 = 2.7$

you know $\sqrt{81}=9, ~\sqrt{100}=10$ so try $a_0 = 9.5$

#### greg1313

This is a consequence of the binomial theorem.

$$\displaystyle (a^2 + b)^{1/2} \$$

$$\displaystyle = \ \bigg(a^2\bigg)^{\tfrac{1}{2}} \ + \ \dfrac{1}{2}\bigg(a^2\bigg)^{\tfrac{-1}{2}}b^1 \ + \ ...$$

$$\displaystyle = \ a \ + \ \dfrac{1}{2}a^{-1}b \ + \ ...$$

$$\displaystyle \approx \ a \ + \ \dfrac{b}{2a}$$

$$\displaystyle 2.8^2 \ = \ 7.84 \$$ is 0.16 less than 8.

$$\displaystyle 2.9^2 \ = \ 8.41 \$$ is 0.41 more than 8

Then someone might use 2.8 for a and 0.16 for b.

$$\displaystyle \sqrt{8} \ \approx \ 2.8 \ + \ \dfrac{0.16}{2*2.8}$$

$$\displaystyle \sqrt{8} \ \approx \ 2.8 \ + \ 0.0285714...$$

$$\displaystyle \sqrt{8} \ \approx \ 2.8285714$$

Compare this first approximation to $$\displaystyle \ \sqrt{8} \ = \ 2.828427...$$