Question

Find the approximate value of $\sqrt{50}$.

Answer 1

Hint: We find the nearest perfect squares of 50 say $a$ and $c$. We use the property $a < b < c\Rightarrow \sqrt{a}<\sqrt{b}<\sqrt{c}$ where $b$ is 50. We find the square of the midpoint of $a$ and $c$ that is $\dfrac{a+c}{2}$ . If we have $b>\dfrac{a+c}{2}\Rightarrow \sqrt{b}>\sqrt{\dfrac{a+c}{2}}$ we replace $c$ with $\dfrac{a+c}{2}$ and if we have $b < \dfrac{a+c}{2}\Rightarrow \sqrt{b} < \sqrt{\dfrac{a+c}{2}}$ we replace $a$. We continue the process until the value of $\dfrac{a+c}{2}$ has the decimal points upto which we want to approximate.

Complete step-by-step answer:
Perfect square is a positive integer whose square root is also a positive integer, for example 1,4,9, 16,25,36,49, 64 etc.
We are asked to find the square root of 50. We first find the perfect square right above it which is 49 and right below it. We know for three real numbers $a,b,c$ greater than 1 the property $a < b < c\Rightarrow \sqrt{a} < \sqrt{b} < \sqrt{c}$.
 Then the midpoint of $a$ and $c$ is $\dfrac{a+c}{2}$. Now we have,
\[\begin{align}
  & b>\dfrac{a+c}{2}\Rightarrow \sqrt{b}>\sqrt{\dfrac{a+c}{2}} \\
 & b<\dfrac{a+c}{2}\Rightarrow \sqrt{b}<\sqrt{\dfrac{a+c}{2}} \\
\end{align}\]
We use this property for 49, 59, 64 and have,
\[49<50<64\Rightarrow 7<\sqrt{50}<8\]
We find the square of midpoint of 7 and 8 and check whether it is greater or lesser than 50 . So we have
\[{{\left( \dfrac{7+8}{2} \right)}^{2}}={{\left( 7.5 \right)}^{2}}=56.25>50\]
 We use the property again and have $49<50<56.25\Rightarrow 7<\sqrt{50}<7.5$. We find the square of midpoint of 7 and 7.5 to check whether it is greater or lesser than 50 .
\[{{\left( \dfrac{7+7.5}{2} \right)}^{2}}={{\left( 7.25 \right)}^{2}}=52.5625>50\]
We use the property and have $49<50<52.5625\Rightarrow 7<\sqrt{50}<7.25$. We find the square of midpoint of 7 and 7.25 and check whether it is greater or lesser than 50 .
\[{{\left( \dfrac{7+7.25}{2} \right)}^{2}}={{\left( 7.125 \right)}^{2}}=50.765625>50\]
We now have $49<50<50.765625\Rightarrow 7<\sqrt{50}<7.125$. We find the square of midpoint of 7 and 7.125 to check whether it is greater or lesser than 50 .
\[{{\left( \dfrac{7+7.125}{2} \right)}^{2}}={{\left( 7.0625 \right)}^{2}}=49.87890625<50\]
We now have $49.87890625<50<50.765625\Rightarrow 7.0625<\sqrt{50}<7.125$. We find the square of midpoint of 7.0625 and 7.125 to check whether it is greater or lesser than 50 .
\[{{\left( \dfrac{7.0625+7.125}{2} \right)}^{2}}={{\left( 7.09375 \right)}^{2}}=50.\text{321289}0\text{625}50\]
 So we have $49.87890625<50<50.\text{321289}0\text{625}\Rightarrow 7.0625<\sqrt{50}<7.09735$. So for $\sqrt{50}$ one digit after decimal point is 0. We can similarly calculate the next digit as 7. So the approximate value of $\sqrt{50}\simeq 7.07$ up to two decimal.

Note: The method we used is called bisection method. We can alternatively find the approximate of $\sqrt{50}$ if we know the approximate value of $\sqrt{2}\simeq 1.414$. So we have $\sqrt{50}=\sqrt{5\times 5\times 2}=5\sqrt{2}=5\times 1.141=7.07$ . We can also solve with the logarithm on both sides of $x=\sqrt{50}={{50}^{\dfrac{1}{2}}}$.