How can you simplify \[\sqrt {42} \] ?
Answer
563.4k+ views
Hint: We can also solve it by method for factorisation. We can take the LCM of the number to find out the square root. If no square factors are present, then we have to find the square root with a simple continued fraction expansion.
Complete step by step solution:
We need to factorise the number within square root. Now, \[42\, = \,2 \times 3 \times 7\] has no square. All the factors occur once, so \[\sqrt {42} \] cannot be simplified further. The value of \[\sqrt {42} \] is an irrational number between 6 and 7. We know \[{6^2}\, = \,36\]and \[{7^2}\, = \,49\]. Therefore \[36 < 42 < 49\]
\[
\Rightarrow \,\,\sqrt {36} \, < \sqrt {42} < \,\sqrt {49} \\
\Rightarrow \,\,6 < \,\sqrt {42} \, < \,7 \\
\]
Hence, it cannot be simplified by factorisation method. Square root of this form has to be solved by simple continued fraction expansion. The fraction expansion for square root of product of two consecutive numbers is given by: \[\sqrt {n(n + 1)} = \,\,[n;\overline {2,2n} ]\,\, = \,\,n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \cdot \cdot \cdot }}}}}}}}}}\]
Now we know \[42\, = 6 \times 7\].
Therefore, for the above expansion n=6 and (n+1) =7. We can carry the expansion depending upon the number of decimal values. Putting n=6 in the above equation we get-
\[\sqrt {42} \,\, = \sqrt {6 \times 7} = \,\,[6;\overline {2,12} ]\,\, = \,\,6 + \dfrac{1}{{2 + \dfrac{1}{{12 + \dfrac{1}{2}}}}}\] = \[\dfrac{{337}}{{52}}\, = \,6.48\]
We have truncated the continued fraction early to get a good rational approximation for \[\sqrt {42} \].
Hence the correct answer for \[\sqrt {42} \] is 6.48.
Formula used:
The formula used in this problem is \[\sqrt {n(n + 1)} = \,\,[n;\overline {2,2n} ]\,\, = \,\,n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \cdot \cdot \cdot }}}}}}}}}}\]
Note:
It is to be noted that this approximation will have approximately as many significant digits as the sum of significant digits of the numerator and denominator. So, we should terminate after 7 decimal places in this problem. Square roots in case of those numbers which have squares of each factor can be easily simplified by factorisation method.
Complete step by step solution:
We need to factorise the number within square root. Now, \[42\, = \,2 \times 3 \times 7\] has no square. All the factors occur once, so \[\sqrt {42} \] cannot be simplified further. The value of \[\sqrt {42} \] is an irrational number between 6 and 7. We know \[{6^2}\, = \,36\]and \[{7^2}\, = \,49\]. Therefore \[36 < 42 < 49\]
\[
\Rightarrow \,\,\sqrt {36} \, < \sqrt {42} < \,\sqrt {49} \\
\Rightarrow \,\,6 < \,\sqrt {42} \, < \,7 \\
\]
Hence, it cannot be simplified by factorisation method. Square root of this form has to be solved by simple continued fraction expansion. The fraction expansion for square root of product of two consecutive numbers is given by: \[\sqrt {n(n + 1)} = \,\,[n;\overline {2,2n} ]\,\, = \,\,n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \cdot \cdot \cdot }}}}}}}}}}\]
Now we know \[42\, = 6 \times 7\].
Therefore, for the above expansion n=6 and (n+1) =7. We can carry the expansion depending upon the number of decimal values. Putting n=6 in the above equation we get-
\[\sqrt {42} \,\, = \sqrt {6 \times 7} = \,\,[6;\overline {2,12} ]\,\, = \,\,6 + \dfrac{1}{{2 + \dfrac{1}{{12 + \dfrac{1}{2}}}}}\] = \[\dfrac{{337}}{{52}}\, = \,6.48\]
We have truncated the continued fraction early to get a good rational approximation for \[\sqrt {42} \].
Hence the correct answer for \[\sqrt {42} \] is 6.48.
Formula used:
The formula used in this problem is \[\sqrt {n(n + 1)} = \,\,[n;\overline {2,2n} ]\,\, = \,\,n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \dfrac{1}{{2n + \dfrac{1}{{2 + \cdot \cdot \cdot }}}}}}}}}}\]
Note:
It is to be noted that this approximation will have approximately as many significant digits as the sum of significant digits of the numerator and denominator. So, we should terminate after 7 decimal places in this problem. Square roots in case of those numbers which have squares of each factor can be easily simplified by factorisation method.
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