Find the square root: $$8 + 2\sqrt 7 $$.
Answer
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Hint: Here in this question, we need to find the square root of a given number which is a nested radical. For this, first we need to start simplifying the radical from the inner side. To simplify a radical should know the square numbers and further simplify by using an algebraic identity to get the required value.
Complete step-by-step solution:
The square root of a natural number is a value, which can be written in the form of $$y = \sqrt a $$. It means ‘y’ is equal to the square root of a, where ‘a’ is any natural number. We can also express it as ${y^2} = a$. Thus, it is concluded here that square root is a value which when multiplied by itself gives the original number, i.e., $$a = y \times y$$.
The symbol or sign to represent a square root is ‘$$\sqrt {} $$’. This symbol is also called a radical. Also, the number under the root is called a radicand.
Consider the given question:
We need to find the square root value of $$8 + 2\sqrt 7 $$
Or
Find the value of
$$ \Rightarrow \,\,\,\,\sqrt {8 + 2\sqrt 7 } $$
We need to simplify from inner side of radicals
8 can be written as $$1 + 7$$, then
$$ \Rightarrow \,\,\,\,\sqrt {1 + 7 + 2\sqrt 7 } $$
As we know, the square number of $$\sqrt 7 $$ is $$7$$ i.e., $${\left( {\sqrt 7 } \right)^2} = 7$$ and square number of 1 is always 1 i.e., $${\left( 1 \right)^2} = 1$$
$$ \Rightarrow \,\,\,\,\sqrt {{{\left( 1 \right)}^2} + {{\left( {\sqrt 7 } \right)}^2} + 2 \times 1 \times \sqrt 7 } $$
By using the algebraic identity $${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$$, then
$$ \Rightarrow \,\,\,\,\sqrt {{{\left( {1 + \sqrt 7 } \right)}^2}} $$
On cancelling the square and root, we get
$$\therefore \,\,\,\,1 + \sqrt 7 $$
Therefore, the value of $$\sqrt {8 + 2\sqrt 7 } = 1 + \sqrt 7 $$.
Note: Remember, when simplifying the nested radicand, solve the radical one by one from the inner radical. And should know the square and square root numbers at least from 1 to 100. We can also simplify radicand by converting to the exponential numbers then by law of indices we can solve the given number.
Complete step-by-step solution:
The square root of a natural number is a value, which can be written in the form of $$y = \sqrt a $$. It means ‘y’ is equal to the square root of a, where ‘a’ is any natural number. We can also express it as ${y^2} = a$. Thus, it is concluded here that square root is a value which when multiplied by itself gives the original number, i.e., $$a = y \times y$$.
The symbol or sign to represent a square root is ‘$$\sqrt {} $$’. This symbol is also called a radical. Also, the number under the root is called a radicand.
Consider the given question:
We need to find the square root value of $$8 + 2\sqrt 7 $$
Or
Find the value of
$$ \Rightarrow \,\,\,\,\sqrt {8 + 2\sqrt 7 } $$
We need to simplify from inner side of radicals
8 can be written as $$1 + 7$$, then
$$ \Rightarrow \,\,\,\,\sqrt {1 + 7 + 2\sqrt 7 } $$
As we know, the square number of $$\sqrt 7 $$ is $$7$$ i.e., $${\left( {\sqrt 7 } \right)^2} = 7$$ and square number of 1 is always 1 i.e., $${\left( 1 \right)^2} = 1$$
$$ \Rightarrow \,\,\,\,\sqrt {{{\left( 1 \right)}^2} + {{\left( {\sqrt 7 } \right)}^2} + 2 \times 1 \times \sqrt 7 } $$
By using the algebraic identity $${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$$, then
$$ \Rightarrow \,\,\,\,\sqrt {{{\left( {1 + \sqrt 7 } \right)}^2}} $$
On cancelling the square and root, we get
$$\therefore \,\,\,\,1 + \sqrt 7 $$
Therefore, the value of $$\sqrt {8 + 2\sqrt 7 } = 1 + \sqrt 7 $$.
Note: Remember, when simplifying the nested radicand, solve the radical one by one from the inner radical. And should know the square and square root numbers at least from 1 to 100. We can also simplify radicand by converting to the exponential numbers then by law of indices we can solve the given number.
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