Question

to two decimal places, \\(\sqrt{97}\\) must lie between \\(\square\\) and \\(\square\\)

Explanation:

Step1: Find integers around 97

We know that \(9^2 = 81\) and \(10^2 = 100\). But we need a closer look. Let's check \(9.8^2=96.04\) and \(9.9^2 = 98.01\). Wait, but we need to find two numbers with two decimal places. Wait, actually, let's calculate \(9.84^2\) and \(9.85^2\). Wait, first, let's see: \(9.8^2 = 96.04\), \(9.85^2=(9.8 + 0.05)^2=9.8^2+2\times9.8\times0.05 + 0.05^2=96.04 + 0.98+0.0025 = 97.0225\). And \(9.84^2=(9.8 + 0.04)^2=9.8^2+2\times9.8\times0.04+0.04^2=96.04+0.784 + 0.0016=96.8256\)? Wait, no, that's wrong. Wait, \(9.84^2\): let's calculate \(984^2 = 968256\), so \(9.84^2=\frac{968256}{10000}=96.8256\)? No, that's not right. Wait, I made a mistake. Wait, \(9.8^2 = 96.04\), \(9.9^2=98.01\). Now, \(9.84^2\): let's do \(9.84\times9.84\). \(9\times9 = 81\), \(9\times0.84=7.56\), \(0.84\times9 = 7.56\), \(0.84\times0.84 = 0.7056\). Wait, no, that's the wrong way. The correct way: \((a + b)^2=a^2+2ab + b^2\), where \(a = 9.8\), \(b = 0.04\). So \(a^2=96.04\), \(2ab=2\times9.8\times0.04 = 0.784\), \(b^2=0.0016\). So \(9.84^2=96.04 + 0.784+0.0016 = 96.8256\)? No, that can't be. Wait, no, \(9.84\) is 9 + 0.84? No, \(9.84 = 9 + 0.84\)? No, \(9.84 = 9.8 + 0.04\). Wait, \(9.8\) is 98 tenths, \(9.84\) is 984 hundredths. So \(984\times984\): let's calculate \(1000 - 16 = 984\), so \((1000 - 16)^2=1000^2-2\times1000\times16+16^2=1000000 - 32000 + 256 = 968256\). So \(984^2 = 968256\), so \(9.84^2=\frac{968256}{10000}=96.8256\). Then \(9.85^2=(9.84 + 0.01)^2=9.84^2+2\times9.84\times0.01+0.01^2=96.8256+0.1968 + 0.0001=97.0225\). Wait, but \(97\) is between \(96.8256\) (which is \(9.84^2\)) and \(97.0225\) (which is \(9.85^2\))? Wait, no, \(9.84^2 = 96.8256\), \(9.85^2=97.0225\). But \(97\) is between \(96.8256\) and \(97.0225\)? Wait, no, that's not correct. Wait, I think I messed up the decimal places. Wait, let's calculate \(9.84^2\) again. Wait, \(9.84\times9.84\):

\(9.84\times9.84\)
= \((10 - 0.16)\times(10 - 0.16)\)
= \(10^2-2\times10\times0.16 + 0.16^2\)
= \(100 - 3.2 + 0.0256\)
= \(96.8256\). Yes, that's correct. Then \(9.85\times9.85\):

\(9.85\times9.85\)
= \((10 - 0.15)\times(10 - 0.15)\)
= \(100 - 3 + 0.0225\)
= \(97.0225\). Ah, there we go. So \(9.84^2 = 96.8256\), \(9.85^2 = 97.0225\). But we need to find two numbers with two decimal places such that their squares are around 97. Wait, maybe I should check \(9.84^2 = 96.8256\), \(9.85^2 = 97.0225\), \(9.86^2=(9.85 + 0.01)^2=9.85^2+2\times9.85\times0.01+0.01^2=97.0225+0.197+0.0001=97.2196\). Wait, no, that's not right. Wait, \(9.84^2 = 96.8256\), \(9.85^2 = 97.0225\), \(9.83^2\): let's calculate \(9.83^2=(9.84 - 0.01)^2=9.84^2-2\times9.84\times0.01+0.01^2=96.8256 - 0.1968+0.0001=96.6289\). No, that's not helpful. Wait, maybe I made a mistake in the initial approach. Let's think differently. We know that \(9^2 = 81\), \(10^2 = 100\), so \(\sqrt{97}\) is between 9 and 10. Now, let's find the tenths place. \(9.8^2 = 96.04\), \(9.9^2 = 98.01\), so it's between 9.8 and 9.9. Now, the hundredths place: \(9.84^2 = 96.8256\), \(9.85^2 = 97.0225\). Wait, \(9.84^2 = 96.8256\) (too low), \(9.85^2 = 97.0225\) (too high). So \(\sqrt{97}\) is between 9.84 and 9.85? Wait, no, because \(9.84^2 = 96.8256\), \(9.85^2 = 97.0225\), and 97 is between 96.8256 and 97.0225? Wait, no, 97 is greater than 96.8256 and less than 97.0225, so \(\sqrt{97}\) is between 9.84 and 9.85? Wait, but \(9.84^2 = 96.8256\), \(9.85^2 = 97.0225\), so 97 is between \(9.84^2\) and \(9.85^2\)? Wait, no, 96.8256 < 97 < 97.0225, so yes, \(\sqrt{97}\) is between 9.84 and 9.85? Wait, but let's check \(9.849^2\): let's calculate \(9.849^2\). But maybe a better way: use linear approximation. Let \(f(x)=\sqrt{x}\), \(x = 97\), \(a = 96.04\) (which is \(9.8^2\)), \(f(a)=9.8\), \(f'(x)=\frac{1}{2\sqrt{x}}\), so \(f(a + h)\approx f(a)+\frac{h}{2\sqrt{a}}\), where \(h = 97 - 96.04 = 0.96\). So \(f(97)\approx9.8+\frac{0.96}{2\times9.8}=9.8+\frac{0.96}{19.6}\approx9.8 + 0.04897\approx9.84897\). So \(\sqrt{97}\approx9.84897\), which is approximately 9.85 when rounded to two decimal places, but the question is "To two decimal places, \(\sqrt{97}\) must lie between [ ] and [ ]". Wait, maybe the question is asking for two consecutive numbers with two decimal places such that the lower one squared is less than 97 and the upper one squared is greater than 97. Let's calculate \(9.84^2 = 96.8256\), \(9.85^2 = 97.0225\). Since \(96.8256 < 97 < 97.0225\), so \(\sqrt{97}\) is between 9.84 and 9.85. Wait, but let's confirm: \(9.84^2 = 96.8256\) (less than 97), \(9.85^2 = 97.0225\) (greater than 97). So \(\sqrt{97}\) is between 9.84 and 9.85.

Step2: Verify the bounds

We calculate \(9.84^2 = 96.8256\) (which is less than 97) and \(9.85^2 = 97.0225\) (which is greater than 97). Since \((\sqrt{97})^2 = 97\), we have \(9.84^2<97<9.85^2\), so by the property of square roots (if \(a^2 < b^2\) and \(a,b>0\), then \(a < b\)), we get \(9.84 < \sqrt{97} < 9.85\).

Answer:

9.84 and 9.85