Question

If \[x = 7 + \sqrt {40} \], find the value of \[\sqrt x + \dfrac{1}{{\sqrt x }}\]

Answer 1

Hint: We use the method of prime factorization to bring out the value from square root and write the given under root in simpler form. Write the terms on the right hand side of the equation in such a way that they can be related with the identity of the square of the sum of two numbers. Calculate the value of \[\sqrt x \] and substitute in the required equation. Rationalize the fraction obtained if required.
* If ‘a’ and ‘b’ are two numbers then the square of their sum is given by \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
* Rationalization is a process where we multiply both numerator and denominator with the conjugate of the denominator of the given number.
* Prime factorization is a process of writing a number in multiple of its factors where all factors are prime numbers.

Complete step-by-step solution:
We are given \[x = 7 + \sqrt {40} \]................… (1)
We solve the value of under root by writing the number in terms of its prime factorization
\[ \Rightarrow \]Prime factorization of \[40 = 2 \times 2 \times 2 \times 5\]
We can collect the powers of a number when the base is same
\[ \Rightarrow \]Prime factorization of \[40 = {2^2} \times 2 \times 5\]
Substitute the value of 40 from prime factorization in equation (1)
\[ \Rightarrow x = 7 + \sqrt {{2^2} \times 2 \times 5} \]
Cancel square root by square power and bring out that term from the square root
\[ \Rightarrow x = 7 + 2\sqrt {2 \times 5} \]
Now we can break the term \[7 = 5 + 2\]. Substitute this value of 7 in RHS of the equation
\[ \Rightarrow x = 5 + 2 + 2\sqrt {2 \times 5} \]
Break the values under the square root using the property \[\sqrt {m \times n} = \sqrt m \times \sqrt n \]
\[ \Rightarrow x = 5 + 2 + 2\sqrt 5 \times \sqrt 2 \]
Since we know square root cancels the square power, we can write \[5 = {(\sqrt 5 )^2}\]and\[2 = {(\sqrt 2 )^2}\] in RHS
\[ \Rightarrow x = {\left( {\sqrt 5 } \right)^2} + {\left( {\sqrt 2 } \right)^2} + 2 \times \sqrt 5 \times \sqrt 2 \]
On comparing to the identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get \[a = \sqrt 5 ;b = \sqrt 2 \]
\[ \Rightarrow x = {\left( {\sqrt 5 + \sqrt 2 } \right)^2}\]
Take under root on both sides of the equation
\[ \Rightarrow \sqrt x = \sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} \]
Cancel square root by square power on right hand side of the equation
\[ \Rightarrow \sqrt x = \sqrt 5 + \sqrt 2 \]..................… (2)
We have to find the value of \[\sqrt x + \dfrac{1}{{\sqrt x }}\]
Substitute the required value from equation (2) in \[\sqrt x + \dfrac{1}{{\sqrt x }}\]
\[ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \sqrt 5 + \sqrt 2 + \dfrac{1}{{\sqrt 5 + \sqrt 2 }}\]
Rationalize the term with values in square root in the denominator
\[ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \sqrt 5 + \sqrt 2 + \dfrac{1}{{\sqrt 5 + \sqrt 2 }} \times \dfrac{{\sqrt 5 - \sqrt 2 }}{{\sqrt 5 - \sqrt 2 }}\]
Apply the property \[(a + b)(a - b) = {a^2} - {b^2}\]in denominator of RHS
\[ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \sqrt 5 + \sqrt 2 + \dfrac{{\sqrt 5 - \sqrt 2 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}\]
Cancel square root by square power in denominator
\[ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \sqrt 5 + \sqrt 2 + \dfrac{{\sqrt 5 - \sqrt 2 }}{{5 - 2}}\]
\[ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \sqrt 5 + \sqrt 2 + \dfrac{{\sqrt 5 - \sqrt 2 }}{3}\]
Take LCM in RHS of the equation
\[ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \dfrac{{3\sqrt 5 + 3\sqrt 2 + \sqrt 5 - \sqrt 2 }}{3}\]
Add like terms in numerator in RHS
\[ \Rightarrow \sqrt x + \dfrac{1}{{\sqrt x }} = \dfrac{{4\sqrt 5 + 2\sqrt 2 }}{3}\]

\[\therefore \]The value of \[\sqrt x + \dfrac{1}{{\sqrt x }}\] is \[\dfrac{{4\sqrt 5 + 2\sqrt 2 }}{3}\]

Note: Students many times make mistake of solving this problem by substituting the value of \[\sqrt {40} \] using the calculator which is not appropriate as after finding the value of ‘x’ we will have to calculate the square root again which will be a complex process. While rationalizing any fraction keep in mind we always try to form the denominator such that we can apply the identity \[(a + b)(a - b) = {a^2} - {b^2}\] which helps us to remove square roots.