
Find the cube root of 24839 by estimation method.
Answer
483.9k+ views
Hint:For finding the case root by estimation method, the groups of 3 digits have to be made. Then the unit digit of the right most group will be the unit digit of the actual case root of the given number.
Then check that the second group is falling in which number, cube root in falling in which number, cube root in between. Then the smallest number will be attached to the left-hand side of the above found unit digit.
Complete step-by- step solution:
Given the number is 24389 let us divide the number into 2 groups.
\[\underbrace {24}_{}\,\underbrace {389}_{}\]
Let \[(389)\] be the first group and \[(24)\]is second group
The unit digit of the first group i.e. \[389\] is \[9\] . So, it will be the unit place digit of the actual answer.
For remaining part i.e. ten’s place
\[{2^3} < 24 < {3^3}\]
As 24 is greater than 8 less than \[27\].
To get the ten’s place value the condition is that the first group number has to lie between any 2 cube root values,in that the smallest term will be the ten’s place of the required answer.
Here 2 is less than 3
So, the number at ten’s place will be \[2\]
i.e. \[\sqrt[3]{{24839}} = 29\]
Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from \[I{I^{nd}}\].
In mathematics, a cube root of a number x is a number y such that \[\;{y^3}\; = \;x\]. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8, denoted \[\sqrt[3]{8}\], is 2, because \[{2^3}\; = \;8\]
Then check that the second group is falling in which number, cube root in falling in which number, cube root in between. Then the smallest number will be attached to the left-hand side of the above found unit digit.
Complete step-by- step solution:
Given the number is 24389 let us divide the number into 2 groups.
\[\underbrace {24}_{}\,\underbrace {389}_{}\]
Let \[(389)\] be the first group and \[(24)\]is second group
The unit digit of the first group i.e. \[389\] is \[9\] . So, it will be the unit place digit of the actual answer.
For remaining part i.e. ten’s place
\[{2^3} < 24 < {3^3}\]
As 24 is greater than 8 less than \[27\].
To get the ten’s place value the condition is that the first group number has to lie between any 2 cube root values,in that the smallest term will be the ten’s place of the required answer.
Here 2 is less than 3
So, the number at ten’s place will be \[2\]
i.e. \[\sqrt[3]{{24839}} = 29\]
Note: Cube root need to group digit in no. 3 and taking the unit place digit of the first group & ten’s from \[I{I^{nd}}\].
In mathematics, a cube root of a number x is a number y such that \[\;{y^3}\; = \;x\]. All nonzero real numbers have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8, denoted \[\sqrt[3]{8}\], is 2, because \[{2^3}\; = \;8\]
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