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comparing the difference consider the number line with the plotted squa…

Question

comparing the difference
consider the number line with the plotted square roots.
inspect the number line. which square roots have a
difference of about 0.5?
○ \\(\sqrt{11}\\) and \\(\sqrt{14}\\)
○ \\(\sqrt{11}\\) and \\(\sqrt{19}\\)
○ \\(\sqrt{14}\\) and \\(\sqrt{19}\\)
○ \\(\sqrt{19}\\) and \\(\sqrt{24}\\)

Explanation:

Step1: Estimate each square root

First, recall the perfect squares around the radicands:

  • For $\sqrt{11}$: $3^2 = 9$, $4^2 = 16$. So $3 < \sqrt{11} < 4$. Let's approximate: $3.3^2 = 10.89$, so $\sqrt{11} \approx 3.31$.
  • For $\sqrt{14}$: $3.7^2 = 13.69$, $3.8^2 = 14.44$. So $\sqrt{14} \approx 3.74$.
  • For $\sqrt{19}$: $4.3^2 = 18.49$, $4.4^2 = 19.36$. So $\sqrt{19} \approx 4.36$.
  • For $\sqrt{24}$: $4.8^2 = 23.04$, $4.9^2 = 24.01$. So $\sqrt{24} \approx 4.90$.

Step2: Calculate the differences

  • Difference between $\sqrt{11}$ and $\sqrt{14}$: $3.74 - 3.31 = 0.43 \approx 0.5$ (close enough).
  • Difference between $\sqrt{11}$ and $\sqrt{19}$: $4.36 - 3.31 = 1.05$ (not about 0.5).
  • Difference between $\sqrt{14}$ and $\sqrt{19}$: $4.36 - 3.74 = 0.62$ (not about 0.5).
  • Difference between $\sqrt{19}$ and $\sqrt{24}$: $4.90 - 4.36 = 0.54 \approx 0.5$? Wait, but let's check the first pair again. Wait, maybe my approximation for $\sqrt{11}$: $3.3^2 = 10.89$, $3.4^2 = 11.56$, so $\sqrt{11} \approx 3.32$. $\sqrt{14}$: $3.74^2 = 13.9876$, so $\sqrt{14} \approx 3.74$. Then $3.74 - 3.32 = 0.42 \approx 0.5$. For $\sqrt{19}$ and $\sqrt{24}$: $4.9 - 4.36 = 0.54 \approx 0.5$. Wait, but let's check the number line. The number line has marks: from 3 to 4, how many ticks? From 3 to 4, there are 10 ticks (since between 3 and 4, each tick is 0.1? Wait, the number line: 3, then $\sqrt{11}$, then $\sqrt{14}$ near 4, then $\sqrt{19}$, then $\sqrt{24}$ near 5. Wait, maybe my initial approximation was off. Let's re - approximate using the number line.

Looking at the number line:

  • $\sqrt{11}$ is between 3 and 4, closer to 3.3 (since 3.3^2 = 10.89, 3.4^2 = 11.56).
  • $\sqrt{14}$ is between 3 and 4, closer to 3.7 (3.7^2 = 13.69, 3.8^2 = 14.44).
  • $\sqrt{19}$ is between 4 and 5, closer to 4.3 (4.3^2 = 18.49, 4.4^2 = 19.36).
  • $\sqrt{24}$ is between 4 and 5, closer to 4.9 (4.9^2 = 24.01).

Now, calculate the differences:

  • $\sqrt{11}$ and $\sqrt{14}$: $3.7 - 3.3 = 0.4 \approx 0.5$ (considering the number line ticks, maybe each tick is 0.1).
  • $\sqrt{19}$ and $\sqrt{24}$: $4.9 - 4.3 = 0.6$? Wait, no. Wait the options: the first option is $\sqrt{11}$ and $\sqrt{14}$, the last is $\sqrt{19}$ and $\sqrt{24}$. Wait, let's calculate the actual square roots:

$\sqrt{11} \approx 3.3166$, $\sqrt{14} \approx 3.7417$, difference is $3.7417 - 3.3166 = 0.4251 \approx 0.43 \approx 0.5$.

$\sqrt{19} \approx 4.3589$, $\sqrt{24} \approx 4.8990$, difference is $4.8990 - 4.3589 = 0.5401 \approx 0.54 \approx 0.5$. But let's check the number line again. The number line has $\sqrt{11}$, $\sqrt{14}$, $\sqrt{19}$, $\sqrt{24}$. Let's count the ticks between them. From $\sqrt{11}$ to $\sqrt{14}$: how many ticks? If between 3 and 4, there are, say, 10 ticks (each 0.1), then $\sqrt{11}$ is at 3.3 (3 + 0.3), $\sqrt{14}$ is at 3.7 (3 + 0.7), so the difference is $0.7 - 0.3 = 0.4$ (in terms of 0.1 - unit ticks), which is 0.4, close to 0.5. From $\sqrt{19}$ to $\sqrt{24}$: $\sqrt{19}$ is at 4.3 (4 + 0.3), $\sqrt{24}$ is at 4.9 (4 + 0.9), difference is $0.9 - 0.3 = 0.6$, which is 0.6, not 0.5. Wait, maybe my tick - count is wrong. Alternatively, maybe the intended answer is $\sqrt{11}$ and $\sqrt{14}$ because the other differences are either too big or not as close. Wait, the first option's difference is about 0.4 - 0.5, and the last option's is about 0.5 - 0.6. But let's check the exact values:

$\sqrt{11} \approx 3.3166$, $\sqrt{14} \approx 3.7417$, difference $\approx 0.425 \approx 0.5$.

$\sqrt{19} \approx 4.3589$, $\sqrt{24} \approx 4.8990$, difference $\approx 0.54 \approx 0.5$. But…

Answer:

$\boldsymbol{\sqrt{11}}$ and $\boldsymbol{\sqrt{14}}$ (the first option)