Question

State with reason which of the following are surds.
\[(i)\sqrt[3]{{81}}\] \[(ii)\sqrt {140} \]

Answer 1

Hint:We solve both the parts separately using the information about surds that any number which cannot be simplified to remove it’s under root then it is called a surd.
* A surd always has its decimal representation as non-terminating and non-repeating which means the digits after the decimal are never ending and do not repeat to form approximation to any number.
* Every irrational number (a number whose under root cannot be simplified) is a surd.

Complete step-by-step answer:
From the definition of Surd, we know the number whose under root cannot be simplified is a surd, so we try to simplify both the terms in order to find if they are surd or not.
\[(i)\sqrt[3]{{81}}\]
Here we have the cube root of a number. So we try to simplify the number inside the root.
We can write \[81\] in multiples of other numbers as it is not a prime number.
We know \[81 = 9 \times 9 = 3 \times 3 \times 3 \times 3\]
Substitute the value inside the root and solve.
So we can write \[\sqrt[3]{{81}} = \sqrt[3]{{3 \times 3 \times 3 \times 3}}\]
Also, we know under root opens up as power of inverse of the number, say \[\sqrt[n]{x} = {(x)^{\dfrac{1}{n}}}\]
Substituting the value of \[n = 3\], \[\sqrt[3]{{81}} = {\left( {81} \right)^{\dfrac{1}{3}}}\]
So, we can write \[{\left( {81} \right)^{\dfrac{1}{3}}} = {\left( {3 \times 3 \times 3 \times 3} \right)^{\dfrac{1}{3}}}\]
Now we collect the powers on RHS of the equation using the formula \[\underbrace {x \times x \times x...... \times x}_n = {x^n}\]
So, we can write \[\left( {3 \times 3 \times 3 \times 3} \right) = {3^4}\]
Now the equation becomes
\[{\left( {81} \right)^{\dfrac{1}{3}}} = {\left( {{3^4}} \right)^{\dfrac{1}{3}}}\]
Also, from the property of exponents if base is same powers can be added, i.e. \[{x^m} \times {x^n} = {x^{m + n}}\]
Using this we can write \[{3^4} = {3^3} \times 3\]
Now substitute the value in equation \[{\left( {81} \right)^{\dfrac{1}{3}}} = {\left( {{3^4}} \right)^{\dfrac{1}{3}}}\]
\[{\left( {81} \right)^{\dfrac{1}{3}}} = {\left( {{3^3} \times 3} \right)^{\dfrac{1}{3}}}\]
Using the property of exponents, \[{\left( {ab} \right)^m} = {a^m} \times {b^m}\]
\[
  {\left( {81} \right)^{\dfrac{1}{3}}} = {\left( {{3^3}} \right)^{\dfrac{1}{3}}} \times {\left( 3 \right)^{\dfrac{1}{3}}} \\
  {\left( {81} \right)^{\dfrac{1}{3}}} = {3^{3 \times \dfrac{1}{3}}} \times {\left( 3 \right)^{\dfrac{1}{3}}} \\
  {\left( {81} \right)^{\dfrac{1}{3}}} = 3 \times {\left( 3 \right)^{\dfrac{1}{3}}} \\
 \]
Now writing in the form of under root
\[\sqrt[3]{{81}} = 3 \times \sqrt[3]{3}\]
And we know cube root of \[3 = 1.44224957\]
So, the value \[\sqrt[3]{{81}} = 3 \times 1.44224957 = 4.326748\] which is non-terminating and non-repeating.
So, \[\sqrt[3]{{81}}\] is a surd.
\[(ii)\sqrt {140} \]
Here we have the square root of a number. So we try to simplify the number inside the root.
We can write \[140\] in multiples of other numbers as it is not a prime number.
We know \[140 = 10 \times 14 = 2 \times 2 \times 5 \times 7\]
Substitute the value inside the root and solve.
So we can write \[\sqrt {140} = \sqrt {2 \times 2 \times 5 \times 7} \]
Also, we know under root opens up as power of inverse of the number, say \[\sqrt[n]{x} = {(x)^{\dfrac{1}{n}}}\]
Substituting the value of \[n = 2\], \[\sqrt[2]{{140}} = {\left( {140} \right)^{\dfrac{1}{2}}}\] { since we can write square root in both ways}
So, we can write \[{\left( {140} \right)^{\dfrac{1}{2}}} = {\left( {2 \times 2 \times 5 \times 7} \right)^{\dfrac{1}{2}}}\]
Now we collect the powers on RHS of the equation using the formula \[\underbrace {x \times x \times x...... \times x}_n = {x^n}\]
So, we can write \[\left( {2 \times 2} \right) = {2^2}\]
Now the equation becomes
\[{\left( {140} \right)^{\dfrac{1}{2}}} = {\left( {{2^2} \times 5 \times 7} \right)^{\dfrac{1}{2}}}\]
Using the property of exponents, \[{\left( {ab} \right)^m} = {a^m} \times {b^m}\]
\[
  {\left( {140} \right)^{\dfrac{1}{2}}} = {\left( {{2^2}} \right)^{\dfrac{1}{2}}} \times {\left( {5 \times 7} \right)^{\dfrac{1}{2}}} \\
  {\left( {140} \right)^{\dfrac{1}{2}}} = {2^{2 \times \dfrac{1}{2}}} \times {\left( {35} \right)^{\dfrac{1}{2}}} \\
  {\left( {140} \right)^{\dfrac{1}{2}}} = 2 \times {\left( {35} \right)^{\dfrac{1}{2}}} \\
 \]
Now writing in the form of under root
\[\sqrt {140} = 2 \times \sqrt {35} \]
And we know square root of \[35 = 5.916079\]
So, the value \[\sqrt {140} = 2 \times 5.916079 = 11.832158\] which is non-terminating and non-repeating.
So, \[\sqrt {140} \] is a surd.

Note: Students are likely to make mistake while calculating the under root without writing the value of \[n\] and without making use of the formula \[\sqrt[n]{x} = {(x)^{\dfrac{1}{n}}}\] which helps to see the calculations easily.