Question

to one decimal place, \\(\sqrt{96}\\) must lie between 9.7 and 9.8 because \\(9.7^2 = 94.09\\) and \\(9.8^2 = 96.04\\), and 96 lies between these values. drag \\(\sqrt{96}\\) based on your estimate above: number line image to two decimal places, \\(\sqrt{96}\\) must lie between \\(\square\\) and \\(\square\\)

Explanation:

Step1: Analyze the difference between 96 and 9.7²

First, we know that \(9.7^2 = 94.09\) and \(9.8^2 = 96.04\). The value 96 is closer to \(9.8^2 = 96.04\) than to \(9.7^2 = 94.09\). Let's find the midpoint between 9.7 and 9.8, which is \(9.75\). Calculate \(9.75^2\): \(9.75\times9.75=\frac{39}{4}\times\frac{39}{4}=\frac{1521}{16} = 95.0625\). Since \(95.0625<96\), we know that \(\sqrt{96}\) is greater than 9.75. Now let's try 9.79: \(9.79^2=(9.8 - 0.01)^2=9.8^2-2\times9.8\times0.01 + 0.01^2=96.04 - 0.196+0.0001 = 95.8441\). Then 9.80² is 96.04. Wait, no, wait, 9.79² is 95.8441, 9.80² is 96.04. Wait, we need to find two consecutive decimals to the hundredth place (two decimal places) such that the square of the first is less than 96 and the square of the second is greater than 96. Let's start from 9.79: \(9.79^2 = 95.8441\), \(9.80^2=96.04\). Wait, but 95.8441 < 96 < 96.04, but we need a closer one. Wait, let's calculate 9.795²? No, we need two decimal places. Wait, let's do it step by step. Let's let \(x = 9.7 + 0.01n\), where \(n\) is an integer from 0 to 10 (since 9.7 to 9.8 is 0.1, which is 10 hundredths). We know that \(9.7^2 = 94.09\), \(9.71^2=9.7^2 + 2\times9.7\times0.01+0.01^2=94.09 + 0.194+0.0001 = 94.2841\), \(9.72^2=94.2841+2\times9.71\times0.01 + 0.0001=94.2841 + 0.1942+0.0001 = 94.4784\), this is too slow. Alternatively, since we know that \(9.79^2 = 95.8441\), \(9.80^2 = 96.04\), but 96 is between 9.79² and 9.80²? Wait, no, 9.79² is 95.8441, 9.80² is 96.04, but 96 is between them? Wait, 95.8441 < 96 < 96.04, but we need a better approximation. Wait, actually, the correct way is: we know that \(\sqrt{96}\) is between 9.7 and 9.8 (to one decimal place). To two decimal places, we need to find \(a\) and \(a + 0.01\) such that \(a^2<96<(a + 0.01)^2\). Let's calculate \(9.79^2 = (9.8 - 0.01)^2=9.8^2-2\times9.8\times0.01 + 0.01^2=96.04 - 0.196+0.0001 = 95.8441\), \(9.80^2 = 96.04\). Wait, but 95.8441 < 96 < 96.04, but that's a big gap. Wait, no, I made a mistake. Wait, 9.79²: 9.79 * 9.79. Let's calculate 979 * 979 = (1000 - 21)^2 = 10000 - 42000 + 441 = 10000 - 42000 is -32000 + 441 = -31559? No, that's wrong. Wait, 9.79 * 9.79: 9 * 9 = 81, 9 * 0.79 = 7.11, 0.79 * 9 = 7.11, 0.79 * 0.79 = 0.6241. Then (9 + 0.79)^2 = 9² + 2*9*0.79 + 0.79² = 81 + 14.22 + 0.6241 = 95.8441. Correct. Then 9.80² = 96.04. But 96 is between 95.8441 and 96.04, but we need a closer pair. Wait, maybe I messed up the initial approach. Let's use linear approximation. The function \(f(x)=x^2\), we know that \(f(9.7)=94.09\), \(f(9.8)=96.04\). We want to find \(x\) such that \(f(x)=96\). The derivative \(f’(x)=2x\), so the linear approximation at \(x = 9.7\) is \(f(9.7)+f’(9.7)(x - 9.7)=94.09 + 19.4(x - 9.7)\). Set this equal to 96: \(94.09+19.4(x - 9.7)=96\) → \(19.4(x - 9.7)=1.91\) → \(x - 9.7=\frac{1.91}{19.4}\approx0.09845\) → \(x\approx9.7 + 0.09845\approx9.79845\). So to two decimal places, that's approximately 9.80, but wait, no, because \(9.79^2 = 95.8441\), \(9.80^2=96.04\), and 96 is between them. But we need two consecutive decimals (to two decimal places) where the first squared is less than 96 and the second squared is greater than 96. Wait, 9.79² = 95.8441 < 96, 9.80² = 96.04 > 96. But wait, is there a number between 9.79 and 9.80 whose square is 96? Let's check 9.798²: (9.8 - 0.002)² = 9.8² - 2*9.8*0.002 + 0.002² = 96.04 - 0.0392 + 0.000004 = 96.000804. Oh! Wait, that's very close to 96. So 9.798² ≈ 96.0008, which is just over 96. Then 9.797²: (9.8 - 0.003)² = 96.04 - 2*9.8*0.003 + 0.003² = 96.04 - 0.0588 + 0.000009 = 95.981209. So 95.981209 < 96 < 96.000804. So to two decimal places, the number is between 9.79 and 9.80? Wait, no, 9.797 is 9.79 when rounded to two decimal places? No, 9.797 is 9.80 when rounded to two decimal places? Wait, no: 9.795 rounds to 9.80, 9.794 rounds to 9.79. Wait, but we need the exact bounds to two decimal places, i.e., two numbers with two decimal places, say \(a.bc\) and \(a.bc + 0.01\), such that \((a.bc)^2 < 96 < (a.bc + 0.01)^2\). Let's calculate 9.79² = 95.8441, 9.80² = 96.04. But 96 is between them. But wait, 9.798² ≈ 96.0008, which is 9.80 when rounded to two decimal places? No, 9.798 is 9.80 when rounded to two decimal places? Wait, no: 9.798, the third decimal is 8, so when rounding to two decimal places, it's 9.80. But the actual value of \(\sqrt{96}\) is approximately 9.79795897113, which is approximately 9.80 when rounded to two decimal places. But we need to find two numbers with two decimal places where the first is less than \(\sqrt{96}\) and the second is greater. So let's take 9.79 and 9.80? Wait, no, because 9.79² = 95.8441 < 96, and 9.80² = 96.04 > 96. But is there a number between 9.79 and 9.80 with two decimal places? No, because two decimal places means the numbers are 9.79, 9.80, etc. Wait, no, two decimal places are numbers like 9.70, 9.71, ..., 9.79, 9.80. So the consecutive hundredths between 9.7 and 9.8 are 9.70, 9.71, ..., 9.79, 9.80. We need to find the largest 9.7x (x is a digit from 0 to 9) such that (9.7x)² < 96, and the next one (9.7(x + 1))² > 96. Let's calculate 9.79² = 95.8441 < 96, 9.80² = 96.04 > 96. So the two decimal places bounds are 9.79 and 9.80? Wait, but 9.79² is 95.8441, which is less than 96, and 9.80² is 96.04, which is greater than 96. But wait, 9.798² is approximately 96.0008, which is very close to 96, so the actual bounds to two decimal places should be 9.79 and 9.80? Wait, no, let's check 9.79² = 95.8441, 9.80² = 96.04. So 96 is between 95.8441 and 96.04, so the square root of 96 is between 9.79 and 9.80? Wait, but 9.798 is approximately 9.80 when rounded to two decimal places, but the exact bounds for two decimal places (i.e., the greatest number with two decimal places less than \(\sqrt{96}\) and the smallest number with two decimal places greater than \(\sqrt{96}\)) are 9.79 and 9.80? Wait, no, let's calculate 9.79² = 95.8441, 9.80² = 96.04. So 96 is between 95.8441 and 96.04, so \(\sqrt{96}\) is between 9.79 and 9.80? Wait, but 9.798 is about 9.80, but the two decimal places numbers are 9.79, 9.80. So the answer should be 9.79 and 9.80. Wait, but let's verify: 9.79² = 95.8441 < 96, 9.80² = 96.04 > 96, so \(\sqrt{96}\) is between 9.79 and 9.80.

Step2: Confirm the bounds

We calculated that \(9.79^2 = 95.8441\) which is less than 96, and \(9.80^2 = 96.04\) which is greater than 96. Therefore, to two decimal places, \(\sqrt{96}\) lies between 9.79 and 9.80.

Answer:

9.79 and 9.80