
If $\sqrt 2 = 1.414$, find the value of $\dfrac{1}{{\sqrt 2 + 1}}$.
Answer
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Hint: Here we will find the value of the given expression by using the rationalizing method. Whenever there is a square root in the denominator, then we rationalize it to remove the radicals from the fraction. First, we will multiply and divide the conjugate of the denominator by the value and then we will solve it by using the suitable algebraic identity. Finally, we will substitute the value of $\sqrt 2 $ in the simplified form to get the required answer.
Complete step-by-step answer:
We have been given that $\sqrt 2 = 1.414$.
As the value is in decimal and also it is an approx value, we won’t directly put it in the expression, rather we will rationalize the expression and then substitute the value in it.
We have to find the value of $\dfrac{1}{{\sqrt 2 + 1}}$.
Rationalizing the expression, we get
$\Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{1}{{\sqrt 2 + 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \\
\Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} \\ $
We know by algebraic identities $\left( {a + b} \right) \times \left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$.
So by using the identity in the above equation, we get
$ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{{\left( {{{\sqrt 2 }^2} - {1^2}} \right)}}$
Squaring the terms in the denominator, we get
$ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{{2 - 1}}$
Subtracting the terms in the denominator, we get
$ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{1}$
Now substituting $\sqrt 2 = 1.414$ in thee above equation, we get
$\ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = 1.414 - 1 \\
\Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = 0.414 \\ $
So the value of $\dfrac{1}{{\sqrt 2 + 1}}$ is $0.414$.
Note: Rationalization is a method to remove radical’s value in the denominator of a function which is algebraic in nature. We multiply a radical that will remove the other radicals in the denominator. This process is used to simplify the question and thus reduce the heavy calculation. A radical is used to represent any root of a number. We can say that if we have a radical which is added or subtracted by some integer in the denominator we will multiply and divide the conjugate of that denominator with the whole value.
Complete step-by-step answer:
We have been given that $\sqrt 2 = 1.414$.
As the value is in decimal and also it is an approx value, we won’t directly put it in the expression, rather we will rationalize the expression and then substitute the value in it.
We have to find the value of $\dfrac{1}{{\sqrt 2 + 1}}$.
Rationalizing the expression, we get
$\Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{1}{{\sqrt 2 + 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \\
\Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} \\ $
We know by algebraic identities $\left( {a + b} \right) \times \left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$.
So by using the identity in the above equation, we get
$ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{{\left( {{{\sqrt 2 }^2} - {1^2}} \right)}}$
Squaring the terms in the denominator, we get
$ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{{2 - 1}}$
Subtracting the terms in the denominator, we get
$ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = \dfrac{{\sqrt 2 - 1}}{1}$
Now substituting $\sqrt 2 = 1.414$ in thee above equation, we get
$\ \Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = 1.414 - 1 \\
\Rightarrow \dfrac{1}{{\sqrt 2 + 1}} = 0.414 \\ $
So the value of $\dfrac{1}{{\sqrt 2 + 1}}$ is $0.414$.
Note: Rationalization is a method to remove radical’s value in the denominator of a function which is algebraic in nature. We multiply a radical that will remove the other radicals in the denominator. This process is used to simplify the question and thus reduce the heavy calculation. A radical is used to represent any root of a number. We can say that if we have a radical which is added or subtracted by some integer in the denominator we will multiply and divide the conjugate of that denominator with the whole value.
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