Question

approximate \\(\sqrt{82}\\) by following the steps below.\\(\sqrt{82}\\) must lie between the whole numbers 9 and 10 because \\(9^2 = 81\\) and \\(10^2 = 100\\), and 82 lies between these values.drag \\(\sqrt{82}\\) based on your estimate above:to one decimal place, \\(\sqrt{82}\\) must lie between 9.0 and 9.1 because \\(9^2 = 81\\) and \\(9.1^2 = 82.81\\), and 82 lies between these values.drag \\(\sqrt{82}\\) based on your estimate above:to two decimal places, \\(\sqrt{82}\\) must lie between \\(\square\\) and \\(\square\\)

Explanation:

Step1: Find the lower bound

We know that \(9.0^2 = 81\) and we need to check \(9.09^2\) and \(9.10^2\) (but we start with \(9.09\) maybe? Wait, actually, let's calculate \(9.09^2\) and \(9.10^2\) no, wait, let's do it step by step. First, we know that \(9.0^2 = 81\), \(9.1^2=82.81\). Now we need to find two numbers with two decimal places, say \(a\) and \(a + 0.01\) such that \(a^2<82<(a + 0.01)^2\). Let's start with \(9.05^2=9.05\times9.05 = 81.9025\). Then \(9.06^2=(9.05 + 0.01)^2=9.05^2+2\times9.05\times0.01+0.01^2=81.9025 + 0.181+0.0001 = 82.0836\). Wait, but \(82\) is between \(9.05^2 = 81.9025\) and \(9.06^2=82.0836\)? Wait no, wait \(9.05^2 = 81.9025\), \(9.06^2=82.0836\), but \(82\) is between them? Wait, \(81.9025<82<82.0836\), so actually, let's check \(9.09^2\)? Wait no, maybe I made a mistake. Wait, let's do it properly. Let's let \(x = 9.0 + d\), where \(d\) is between \(0\) and \(0.1\) (since we are looking at two decimal places, \(d\) is between \(0.00\) and \(0.01\) for the next step? Wait, no, the question is to two decimal places, so we need to find two numbers \(a.bc\) and \(a.bc + 0.01\) such that \((a.bc)^2<82<(a.bc + 0.01)^2\). We know that \(9.09^2=(9 + 0.09)^2=9^2+2\times9\times0.09+0.09^2=81 + 1.62+0.0081 = 82.6281\)? No, that's wrong. Wait, no, \(9.09\) is \(9 + 0.09\), so \( (9 + 0.09)^2=81 + 2*9*0.09 + 0.09^2=81 + 1.62 + 0.0081=82.6281\), which is more than 82. Wait, let's try \(9.05^2 = 81.9025\), \(9.06^2=82.0836\). Wait, but \(82\) is between \(9.05^2 = 81.9025\) and \(9.06^2=82.0836\)? Wait, no, \(81.9025<82<82.0836\), so actually, let's check \(9.09\) no, wait, maybe I messed up. Wait, the previous step was to one decimal place, between 9.0 and 9.1, because \(9.0^2 = 81\) and \(9.1^2=82.81\). Now, for two decimal places, we need to find \(x\) such that \(x\) is between \(9.00\) and \(9.10\), and find two consecutive two - decimal - place numbers \(a\) and \(a + 0.01\) where \(a^2<82<(a + 0.01)^2\).

Let's calculate \(9.09^2\): Wait, no, let's start with \(9.05^2 = 81.9025\), \(9.06^2=82.0836\). Wait, but \(82\) is between \(9.05^2\) and \(9.06^2\)? Wait, \(81.9025<82<82.0836\), so \(9.05^2 = 81.9025\), \(9.06^2=82.0836\). But wait, maybe a better way: we know that \(9.0^2 = 81\), \(9.1^2=82.81\). Let's let \(f(x)=x^2\), we want to find \(x\) such that \(x^2 = 82\), \(x\) is in \([9.0,9.1]\). Let's use linear approximation or just test values. Let's try \(9.05^2=81.9025\), \(9.06^2 = 9.05^2+2\times9.05\times0.01+0.01^2=81.9025 + 0.181+0.0001 = 82.0836\). Wait, but \(82\) is between \(9.05^2 = 81.9025\) and \(9.06^2=82.0836\)? Wait, \(81.9025<82<82.0836\), so actually, the lower bound is \(9.05\) and upper bound is \(9.06\)? Wait, no, that can't be. Wait, maybe I made a mistake in calculation. Wait, \(9.09^2\): no, wait, \(9.09\times9.09\): \(9\times9 = 81\), \(9\times0.09=0.81\), \(0.09\times9 = 0.81\), \(0.09\times0.09 = 0.0081\), so \((9 + 0.09)^2=81+0.81 + 0.81+0.0081=82.6281\), which is more than 82. Wait, let's try \(9.04^2=(9 + 0.04)^2=81+0.72 + 0.0016=81.7216\), \(9.05^2=81.9025\), \(9.06^2=82.0836\), \(9.07^2=82.2649\), \(9.08^2=82.4464\), \(9.09^2=82.6281\), \(9.10^2=82.81\). Wait, but \(82\) is between \(9.05^2 = 81.9025\) and \(9.06^2=82.0836\)? Wait, \(81.9025<82<82.0836\), so the two - decimal - place numbers are \(9.05\) and \(9.06\)? Wait, no, that seems off. Wait, maybe the question is to find between which two consecutive two - decimal - place numbers \(\sqrt{82}\) lies. Let's calculate \(9.09^2\) no, wait, let's do it again. Wait, \(9.0^2 = 81\), \(9.1^2=82.81\). Let's find \(x\) such that \(x^2 = 82\), \(x\) is in \([9.0,9.1]\). Let's use the formula for linear approximation. Let \(f(x)=x^2\), \(f^\prime(x)=2x\). We know that \(f(9)=81\), we want to find \(\Delta x\) such that \(f(9+\Delta x)=82\). So \(81 + 2\times9\times\Delta x+( \Delta x)^2=82\). Ignoring \((\Delta x)^2\) (since \(\Delta x\) is small), we have \(81+18\Delta x\approx82\), so \(18\Delta x\approx1\), \(\Delta x\approx\frac{1}{18}\approx0.0556\). So \(x\approx9.0556\). So \(9.05^2 = 81.9025\), \(9.06^2=82.0836\). So \(82\) is between \(9.05^2\) and \(9.06^2\). So to two decimal places, \(\sqrt{82}\) lies between \(9.05\) and \(9.06\) because \(9.05^2 = 81.9025\) and \(9.06^2=82.0836\), and \(82\) lies between these values.

Step1: Calculate the square of 9.05

\(9.05^2=(9 + 0.05)^2=9^2+2\times9\times0.05 + 0.05^2=81+0.9+0.0025 = 81.9025\)

Step2: Calculate the square of 9.06

\(9.06^2=(9 + 0.06)^2=9^2+2\times9\times0.06+0.06^2=81 + 1.08+0.0036=82.0836\)

Step3: Compare with 82

Since \(81.9025<82<82.0836\), \(\sqrt{82}\) lies between \(9.05\) and \(9.06\) when rounded to two decimal places.

Answer:

\(9.05\) and \(9.06\)