Question

Simplify the following surd: ${\left( {\sqrt 5 } \right)^2}$

Answer 1

Hint:We have to simply solve the surd and the power which is placed over the number. After simplifying the surd (square root) we get an index of $\dfrac{1}{2}$and then we multiply it with the index placed over the bracket of the given expression.

Complete solution step by step:
Firstly, we write the given expression
${\left( {\sqrt 5 } \right)^2}$
Simplifying the bracket part first using the square root simplification i.e. when we remove the square root, we replace it by index (power) of $\dfrac{1}{2}$ in this manner
$
\sqrt p = {(p)^{\dfrac{1}{2}}} \\
\Rightarrow {\left( {{{(5)}^{\dfrac{1}{2}}}} \right)^2} \\
$
Now to solve this expression we use the index properties i.e.
${({p^q})^r} = {({p^r})^q} = {p^{q \times r}}$
Using this property and taking first two parts of it we solve our expression
$
{\left( {{5^{\dfrac{1}{2}}}} \right)^2} = {\left( {{5^2}} \right)^{\dfrac{1}{2}}} \\
= {(25)^{\dfrac{1}{2}}} \\
$
Converting the index into square root we have
$
{(25)^{\dfrac{1}{2}}} = \sqrt {25} \\
= 5 \\
$
Our answer comes out to be ‘5’. Now we check it by taking first and third part of the property
$
{\left( {{5^{\dfrac{1}{2}}}} \right)^2} = {(5)^{\dfrac{1}{2} \times 2}} \\
= 5 \\
$
Our answer comes out the same as the previous result i.e. ‘5’.

Additional information:We can solve the problem using basic index method also in which we solve the index part of the expression as explained below
${\left( {\sqrt 5 } \right)^2}$
The expression has an index of 2 which means the number inside the bracket will be multiplied twice with itself so we have -
${\left( {\sqrt 5 } \right)^2} = \sqrt 5 \times \sqrt 5 $
Now we use multiplication property of square roots i.e.
$\sqrt p \times \sqrt p = p$
Now using the property we have
${\left( {\sqrt 5 } \right)^2} = \sqrt 5 \times \sqrt 5 = 5$
Our answer comes out to be ‘5’, the same as when calculated by the previous method.

Note:In laymen terms index is the power raised to a number and surd is the ‘nth’ root of a number i.e. $\sqrt[n]{a} = {(a)^{\dfrac{1}{n}}}$ and we can say both are opposite in nature because after solving index (power) of a number will be the raised value of the number whereas surd will become the reduced root value of the number.