Question

Find the values of \[a\] and \[b\] if \[\dfrac{{5 + \sqrt 6 }}{{5 - \sqrt 6 }} = a + b\sqrt 6 \].

Answer 1

Hint: The given problem is based on the fraction here we are going to find the value of \[a\] and \[b\]. We have L.H.S\[ = \]R.H.S. By solving this we can easily find out \[a\] and \[b\]. Since it is a fraction and the denominator has surds (square root \[\sqrt 6 \]) we can rationalize the denominator, it is a special method used to simplify the problems. Rationalizing the denominator means conjugating the denominator and multiplying it with both the numerator and denominator.

Complete step by step answer:
In this problem, we are given that \[\dfrac{{5 + \sqrt 6 }}{{5 - \sqrt 6 }} = a + b\sqrt 6 \].Here the denominator is \[5 - \sqrt 6 \], the conjugate of \[5 - \sqrt 6 \] is \[5 + \sqrt 6 \].
Now by rationalize,
\[\dfrac{{5 + \sqrt 6 }}{{5 - \sqrt 6 }} \times \dfrac{{5 + \sqrt 6 }}{{5 + \sqrt 6 }} = a + b\sqrt 6 \]
By multiplying numerator by numerator and denominator by denominator,
\[\dfrac{{\left( {5 + \sqrt 6 } \right)\left( {5 + \sqrt 6 } \right)}}{{\left( {5 - \sqrt 6 } \right)\left( {5 + \sqrt 6 } \right)}} = a + b\sqrt 6 \]
We can see that the numerator is of the form \[\left( {a + b} \right) \times \left( {a + b} \right)\], which is equal to \[{\left( {a + b} \right)^2}\], we have the formula \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\], and denominator is of the form \[\left( {a - b} \right) \times \left( {a + b} \right)\], we have the formula \[\left( {a - b} \right) \times \left( {a + b} \right) = {a^2} - {b^2}\], where \[5 = a,\sqrt 6 = b\].

By applying formulas,
\[\dfrac{{{{\left( {5 + \sqrt 6 } \right)}^2}}}{{{5^2} - {{\left( {\sqrt 6 } \right)}^2}}} = a + b\sqrt 6 \]
We know that, \[{\left( {\sqrt 6 } \right)^2} = \sqrt 6 \times \sqrt 6 = 6\]
\[\dfrac{{{5^2} + 2(5)\left( {\sqrt 6 } \right) + {{\left( {\sqrt 6 } \right)}^2}}}{{25 - 6}} = a + b\sqrt 6 \]
Now subtracting in the denominator, \[25 - 6 = 19\]
\[\dfrac{{25 + 10\sqrt 6 + 6}}{{19}} = a + b\sqrt 6 \]
Performing addition in the numerator, \[25 + 6 = 31\]
\[\dfrac{{31 + 10\sqrt 6 }}{{19}} = a + b\sqrt 6 \]
Here the denominator \[19\] is common for both the values in the numerator, we can write this as,
\[\dfrac{{31}}{{19}} + \dfrac{{10\sqrt 6 }}{{19}} = a + b\sqrt 6 \]
Here L.H.S. \[ = \]R.H.S. Now in L.H.S. there are only numerals and in R.H.S variables. Clearly we can observe that the form variables in the R.H.S. match with the numerals in the L.H.S. Thus we can say that, \[a = \dfrac{{31}}{{19}}\] and \[b = \dfrac{{10\sqrt 6 }}{{19}}\].

Hence we have find that the value of \[a = \dfrac{{31}}{{19}}\] and \[b = \dfrac{{10\sqrt 6 }}{{19}}\].

Note:In case of simplifying the denominators with surds, rationalizing the denominator is a very useful method. In this method we simply multiply both the numerator and denominator with the conjugate of the denominator.
If the denominator only has a single term we can simply multiply the numerator and denominator by that denominator.
Example: \[\dfrac{{2\sqrt 5 + 3}}{{\sqrt 5 }}\]
\[ \dfrac{{2\sqrt 5 + 3}}{{\sqrt 5 }} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }}\] \[ = \dfrac{{2(5) + 3\sqrt 5 }}{5}\]
\[\dfrac{{10 + 3\sqrt 5 }}{5}\] \[ = \dfrac{{10}}{5} + \dfrac{{3\sqrt 5 }}{5}\]
\[ 2 + \dfrac{{3\sqrt 5 }}{5}\]
If the denominator only has two terms then we need to multiply the numerator and denominator by the conjugate of the denominator. Conjugate means change the sign in the middle of two numbers, that is, the conjugate of \[ + \] is \[ - \] and vice-versa.Example: \[\dfrac{{\sqrt 7 }}{{5 - \sqrt 7 }}\]
Conjugate of \[5 - \sqrt 7 \] is\[5 + \sqrt 7 \], the \[ - \] in the middle of two numbers changed into \[ + \].
\[\dfrac{{\sqrt 7 }}{{5 - \sqrt 7 }} \times \dfrac{{5 + \sqrt 7 }}{{5 + \sqrt 7 }}\] \[ = \dfrac{{5\sqrt 7 + 7}}{{{5^2} - {{\left( {\sqrt 7 } \right)}^2}}}\]
\[ \dfrac{{5\sqrt 7 + 7}}{{25 - 7}}\] \[ = \dfrac{{5\sqrt 7 + 7}}{{18}}\].
Any integer is the square of its surds, that is ,\[a = \sqrt a \times \sqrt a \]. Example: \[9 = \sqrt 9 \times \sqrt 9 \].