
What is the fourth root of $1296$?
Answer
512.4k+ views
Hint: To simplify this question , we need to solve it step by step . In order to solve and write the expression into the simplest form . . The square root is related to figuring out what should be the number which when multiplied by itself is equal to the number under the square root symbol $\sqrt {} $. This symbol is known as radical . Since in our case we have given the question in which we have to solve and find the fourth root value of $1296$, we will first get rid of the radical and remove the square root as we want the original value of $1296$ .
Complete step by step solution:
The term radical or root says that it is the mathematical opposite of an exponent, in the same manner that addition is the opposite of subtraction. The smallest radical is the square root, represented with the symbol $\sqrt {} $. And the next radical is the cube root and fourth root and so on , represented by the symbol $\sqrt[3]{{}}$, $\sqrt[4]{{}},\sqrt[n]{{}}$. Also the small number in front of the radical is its index number.
Here in our question we will work on index number 4 .
We can use prime factorisation for the number $1296$ inside the radical and pull out non- radical terms or perfect squares from the inside of square root to make the solution easier .
We get-
1296 is perfect square of 36 =
$
\sqrt {1296} = \sqrt {2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3} \\
= \sqrt {(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)} \\
= 4 \times 9 \\
= 36 \\
$
Cube root of $1296$=
\[
= \sqrt[3]{{(2 \times 2 \times 2) \times 2 \times (3 \times 3 \times 3) \times 3}} \\
= 2 \times 3 \times \sqrt[3]{6} \\
= 6 \times \sqrt[3]{6} \\
= 10.903 \\
\]
Fourth root of $1296$=
\[
= \sqrt[4]{{(2 \times 2 \times 2 \times 2) \times (3 \times 3 \times 3 \times 3)}} \\
= 2 \times 3 \\
= 6 \\
\]
Therefore, the required answer is 6 .
Note:
> Always keep in mind that we can use prime factorisation for the number inside the radical and pull out non- radical terms or perfect squares from the inside of the square root to make the solution easier .
> Always remember that the result of a radical operation is positive if the number under the radical is positive.
> Keep in mind , the result is negative if the number under the radical is negative and the index number is odd.
> Learn that , a negative number under the radical with an even index number produces an irrational number
Complete step by step solution:
The term radical or root says that it is the mathematical opposite of an exponent, in the same manner that addition is the opposite of subtraction. The smallest radical is the square root, represented with the symbol $\sqrt {} $. And the next radical is the cube root and fourth root and so on , represented by the symbol $\sqrt[3]{{}}$, $\sqrt[4]{{}},\sqrt[n]{{}}$. Also the small number in front of the radical is its index number.
Here in our question we will work on index number 4 .
We can use prime factorisation for the number $1296$ inside the radical and pull out non- radical terms or perfect squares from the inside of square root to make the solution easier .
We get-
1296 is perfect square of 36 =
$
\sqrt {1296} = \sqrt {2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3} \\
= \sqrt {(2 \times 2) \times (2 \times 2) \times (3 \times 3) \times (3 \times 3)} \\
= 4 \times 9 \\
= 36 \\
$
Cube root of $1296$=
\[
= \sqrt[3]{{(2 \times 2 \times 2) \times 2 \times (3 \times 3 \times 3) \times 3}} \\
= 2 \times 3 \times \sqrt[3]{6} \\
= 6 \times \sqrt[3]{6} \\
= 10.903 \\
\]
Fourth root of $1296$=
\[
= \sqrt[4]{{(2 \times 2 \times 2 \times 2) \times (3 \times 3 \times 3 \times 3)}} \\
= 2 \times 3 \\
= 6 \\
\]
Therefore, the required answer is 6 .
Note:
> Always keep in mind that we can use prime factorisation for the number inside the radical and pull out non- radical terms or perfect squares from the inside of the square root to make the solution easier .
> Always remember that the result of a radical operation is positive if the number under the radical is positive.
> Keep in mind , the result is negative if the number under the radical is negative and the index number is odd.
> Learn that , a negative number under the radical with an even index number produces an irrational number
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