Compute approximately: 1° sqrt(4.3); 2°
Compute approximately: 1° $\sqrt{4.3}$; 2° $\sqrt{\dfrac{2.037^{2}-1}{2.037^{2}+1}}$.
Explanation of Method
The root formula applied, then a compound expression linearised.
Step 1: Part 1°: choose the nearest square
Write $4.3 = 2^{2}+0.3$ and apply the formula of Problem 259 with $n = 2,\ a = 2,\ x = 0.3$: \[\sqrt{a^{2}+x} \approx a+\frac{x}{2a} \;\Longrightarrow\; \sqrt{4.3} \approx 2+\frac{0.3}{4} = 2+0.075.\]
\[\sqrt{4.3} \approx 2.075\]
The true value is $2.0736441$, so the estimate is high by $0.00136$ — about $0.07\%$. Squaring back gives $2.075^{2} = 4.305625$, which overshoots $4.3$ by $0.0056$, as expected for a concave function.
Step 2: Part 2°: set up the small parameter
Put $t = 2.037 = 2+\varepsilon$ with $\varepsilon = 0.037$, and
\[ F(\varepsilon) = \sqrt{\frac{t^{2}-1}{t^{2}+1}}. \]
Since $t^{2} = 4+4\varepsilon+\varepsilon^{2}$ and $\varepsilon^{2} = 0.001369$ is negligible beside $4\varepsilon = 0.148$, \[t^{2}-1 \approx 3+4\varepsilon, \qquad t^{2}+1 \approx 5+4\varepsilon, \qquad F(\varepsilon) \approx \left(\frac{3+4\varepsilon}{5+4\varepsilon}\right)^{1/2}.\]
Step 3: Linearise at $\varepsilon = 0$
Write $u = 3+4\varepsilon,\ v = 5+4\varepsilon$, so $F = \bigl(u/v\bigr)^{1/2}$. At $\varepsilon = 0$: $F(0) = \sqrt{\frac35} = \frac{\sqrt{15}}{5} \approx 0.7745967$.
Differentiating, and noting $u' = v' = 4$ so that $u'v-uv' = 4\bigl(v-u\bigr) = 4\cdot2 = 8$ identically: \[F'(\varepsilon) = \frac12\sqrt{\frac{v}{u}}\cdot\frac{u'v-uv'}{v^{2}} = \frac{u'v-uv'}{2\,v^{3/2}u^{1/2}}, \qquad F'(0) = \frac{8}{2\cdot5^{3/2}\cdot3^{1/2}} = \frac{4}{5\sqrt{15}}.\]
Step 4: Assemble
\[ F(\varepsilon) \approx \frac{\sqrt{15}}{5}+\frac{4\varepsilon}{5\sqrt{15}} = \frac{15+4\varepsilon}{5\sqrt{15}}. \]
With $\varepsilon = 0.037$, so $15+4\varepsilon = 15.148$ and $5\sqrt{15} \approx 19.36492$: $F \approx \frac{15.148}{19.36492} \approx 0.78224$.
Step 5: Verification
\[\sqrt{4.3} = 2.0736441 \quad\text{vs}\quad 2.075; \qquad \text{error } +0.00136 \;\;\checkmark\]
\[2.037^{2} = 4.149369, \qquad \frac{3.149369}{5.149369} = 0.6116033, \qquad \sqrt{0.6116033} = 0.7820504\]
\[\text{Estimate } 0.7822394 \text{ differs by } 0.00019, \text{ i.e. } 0.024\% \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\sqrt{4.3} \approx 2.075\,}\]
Final answer:
\[\boxed{\,\sqrt{4.3} \approx 2.075, \qquad \sqrt{\dfrac{2.037^{2}-1}{2.037^{2}+1}} \approx 0.7822\,}\]
Remark
Remark. Part 2° is worth comparing with the brute-force route: squaring $2.037$, forming the quotient and extracting a square root takes three awkward operations, whereas the linearisation needs only $15+4\varepsilon$ over the fixed constant $5\sqrt{15}$. The reason the result is so accurate — four correct digits — is that $\varepsilon = 0.037$ enters the answer divided by roughly $4$, and the quadratic term is smaller still.