Find approximate values: 1) √(25.3) 2) √(49.5) 3) (26)^(1/3) 4) (2.5)^(1/4)

Steps to Solve

1. Approximate $\sqrt{25.3}$

To find $\sqrt{25.3}$, we can approximate it using $\sqrt{25}$ since $25.3$ is close to $25$.

We know: $$ \sqrt{25} = 5 $$

Using a small adjustment, we find: $$ \sqrt{25.3} \approx 5 + \frac{0.3}{2 \times 5} = 5 + 0.03 = 5.03 $$

2. Approximate $\sqrt{49.5}$

Next, we approximate $\sqrt{49.5}$ using $\sqrt{49}$:

$$ \sqrt{49} = 7 $$

We can adjust it similarly: $$ \sqrt{49.5} \approx 7 + \frac{0.5}{2 \times 7} \approx 7 + 0.0357 \approx 7.0357 $$

3. Approximate $(26)^{1/3}$

To find the cube root of $26$, we can use $\sqrt[3]{27}$ since $26$ is close to $27$:

$$ \sqrt[3]{27} = 3 $$

Using a small adjustment: $$ (26)^{1/3} \approx 3 - \frac{1}{3 \times 9} = 3 - \frac{1}{27} \approx 3 - 0.037 \approx 2.963 $$

4. Approximate $(2.5)^{1/4}$

For $(2.5)^{1/4}$, we can use $(2)^{1/4}$ because $2.5$ is close to $2$:

$$ (2)^{1/4} \approx 1.19 $$

Using the same adjustment: $$ (2.5)^{1/4} \approx 1.19 + \text{small adjustment} $$

Given $(2.5-2)=0.5$: $$ \approx 1.19 + \frac{0.5}{4 \times (1.19^3)} \approx 1.19 + 0.02 \approx 1.21 $$