Question

The number obtained by rationalizing the denominator of $\dfrac{1}{{\sqrt 7 - 2}}$is
A.$\dfrac{{\sqrt 7 + 2}}{3}$
B.$\dfrac{{\sqrt 7 - 2}}{3}$
C.$\dfrac{{\sqrt 7 + 2}}{5}$
D.$\dfrac{{\sqrt 7 + 2}}{{45}}$

Answer 1

Hint: For rationalizing the number, we have to multiply both numerator and denominator by the conjugate of denominator. Use (a – b) (a + b) = ${a}^{2}$ – ${b}^{2}$ to simplify the product.

Complete step-by-step answer:
$\dfrac{1}{{\sqrt 7 - 2}}$
If the denominator is a monomial in some radical, say with k < n, rationalisation consists of multiplying the numerator and the denominator by and replacing by if n is even or by x if n is odd (if k ≥ n, the same replacement allows us to reduce k until it becomes lower than n.
If the denominator is linear in some square root, say rationalisation consists of multiplying the numerator and the denominator by and expanding the product in the denominator.
Here, conjugate of $\dfrac{1}{{\sqrt 7 - 2}}$ is $\dfrac{1}{{\sqrt 7 + 2}}$
Now, by multiplying numerator and denominator by this conjugate. We get
$ = \dfrac{1}{{\sqrt 7 - 2}} \times \dfrac{{\sqrt 7 + 2}}{{\sqrt 7 + 2}}$
\[ = \dfrac{{\sqrt 7 + 2}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( 2 \right)}^2}}}\]{∵ (a – b) (a + b) = ${a}^{2} – {b}^{2}$}
$ = \dfrac{{\sqrt 7 + 2}}{{7 - 4}}$
$ = \dfrac{{\sqrt 7 + 2}}{3}$
∴ Answer is option A $\dfrac{{\sqrt 7 + 2}}{3}$.

Note: There is another special way to move a square root from the bottom of a fraction to the top, and this is done in a very simple way by multiplying numerator and denominator with conjugate.
Rationalisation can be extended to all algebraic numbers and algebraic functions (as an application of norm forms). For example, to rationalise a cube root, two linear factors involving cube roots of unity should be used, or equivalently a quadratic factor.