
Determine the rational numbers p and q if \[\dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} - \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} = p - 7\sqrt 5 q\].
Answer
546.9k+ views
Hint: First rationalize each term of the left side by multiplying and dividing the terms by the conjugate of their denominator and simplify them. Then subtract the second term from the first term. After that compare the rationalizing part with the right side to get the value of p and q.
Complete step-by-step solution:
The right side of the expression is,
$ \Rightarrow p - 7\sqrt 5 q$..................….. (1)
Now take the first term of the left side,
$ \Rightarrow \dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }}$
Rationalize it by multiplying and dividing the term by the conjugate of the denominator,
$ \Rightarrow \dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} \times \dfrac{{7 + \sqrt 5 }}{{7 + \sqrt 5 }}$
Use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the numerator and ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator,
$ \Rightarrow \dfrac{{{{\left( 7 \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} + 2 \times 7 \times \sqrt 5 }}{{{{\left( 7 \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}}}$
Simplify the terms,
$ \Rightarrow \dfrac{{49 + 5 + 14\sqrt 5 }}{{49 - 5}}$
Add the terms in the numerator and subtract the terms in the denominator,
$ \Rightarrow \dfrac{{54 + 14\sqrt 5 }}{{44}}$
Cancel out the common factors,
$ \Rightarrow \dfrac{{27 + 7\sqrt 5 }}{{22}}$...................….. (2)
Now take the second term of the left side,
$ \Rightarrow \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }}$
Rationalize it by multiplying and dividing the term by the conjugate of the denominator,
$ \Rightarrow \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} \times \dfrac{{7 - \sqrt 5 }}{{7 - \sqrt 5 }}$
Use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the numerator and ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator,
$ \Rightarrow \dfrac{{{{\left( 7 \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2 \times 7 \times \sqrt 5 }}{{{{\left( 7 \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}}}$
Simplify the terms,
$ \Rightarrow \dfrac{{49 + 5 - 14\sqrt 5 }}{{49 - 5}}$
Add the terms in the numerator and subtract the terms in the denominator,
$ \Rightarrow \dfrac{{54 - 14\sqrt 5 }}{{44}}$
Cancel out the common factors,
$ \Rightarrow \dfrac{{27 - 7\sqrt 5 }}{{22}}$...................….. (3)
Now add the terms (2) and (3),
$ \Rightarrow \dfrac{{27 + 7\sqrt 5 }}{{22}} - \dfrac{{27 - 7\sqrt 5 }}{{22}}$
Change the signs,
$ \Rightarrow \dfrac{{27 + 7\sqrt 5 - 27 + 7\sqrt 5 }}{{22}}$
Add and subtract the like terms,
$ \Rightarrow \dfrac{{14\sqrt 5 }}{{22}}$
Cancel out the common terms,
$ \Rightarrow \dfrac{{7\sqrt 5 }}{{11}}$
Equate the value with the equation (1),
$ \Rightarrow p + 7\sqrt 5 q = 0 + \dfrac{{7\sqrt 5 }}{{11}}$
Hence, the value of $p$ is 0 and $q$ is $\dfrac{1}{{11}}$.
Note: You might be wondering as it is a question's requirement so we have rationalized this expression. But in general, why is there a need to rationalize? The answer is as you can see that after rationalization meaning multiplying and dividing the whole expression by a conjugate, the denominator of the expression is reduced to integers by using basic algebraic identities. So, rationalization will simplify the denominator in such a way that it contains only rational numbers. So, in a calculation, if you find the denominator can be rationalized then go for it, as it will reduce the complexity of the problem.
Complete step-by-step solution:
The right side of the expression is,
$ \Rightarrow p - 7\sqrt 5 q$..................….. (1)
Now take the first term of the left side,
$ \Rightarrow \dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }}$
Rationalize it by multiplying and dividing the term by the conjugate of the denominator,
$ \Rightarrow \dfrac{{7 + \sqrt 5 }}{{7 - \sqrt 5 }} \times \dfrac{{7 + \sqrt 5 }}{{7 + \sqrt 5 }}$
Use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the numerator and ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator,
$ \Rightarrow \dfrac{{{{\left( 7 \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} + 2 \times 7 \times \sqrt 5 }}{{{{\left( 7 \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}}}$
Simplify the terms,
$ \Rightarrow \dfrac{{49 + 5 + 14\sqrt 5 }}{{49 - 5}}$
Add the terms in the numerator and subtract the terms in the denominator,
$ \Rightarrow \dfrac{{54 + 14\sqrt 5 }}{{44}}$
Cancel out the common factors,
$ \Rightarrow \dfrac{{27 + 7\sqrt 5 }}{{22}}$...................….. (2)
Now take the second term of the left side,
$ \Rightarrow \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }}$
Rationalize it by multiplying and dividing the term by the conjugate of the denominator,
$ \Rightarrow \dfrac{{7 - \sqrt 5 }}{{7 + \sqrt 5 }} \times \dfrac{{7 - \sqrt 5 }}{{7 - \sqrt 5 }}$
Use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the numerator and ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator,
$ \Rightarrow \dfrac{{{{\left( 7 \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2 \times 7 \times \sqrt 5 }}{{{{\left( 7 \right)}^2} - {{\left( {\sqrt 5 } \right)}^2}}}$
Simplify the terms,
$ \Rightarrow \dfrac{{49 + 5 - 14\sqrt 5 }}{{49 - 5}}$
Add the terms in the numerator and subtract the terms in the denominator,
$ \Rightarrow \dfrac{{54 - 14\sqrt 5 }}{{44}}$
Cancel out the common factors,
$ \Rightarrow \dfrac{{27 - 7\sqrt 5 }}{{22}}$...................….. (3)
Now add the terms (2) and (3),
$ \Rightarrow \dfrac{{27 + 7\sqrt 5 }}{{22}} - \dfrac{{27 - 7\sqrt 5 }}{{22}}$
Change the signs,
$ \Rightarrow \dfrac{{27 + 7\sqrt 5 - 27 + 7\sqrt 5 }}{{22}}$
Add and subtract the like terms,
$ \Rightarrow \dfrac{{14\sqrt 5 }}{{22}}$
Cancel out the common terms,
$ \Rightarrow \dfrac{{7\sqrt 5 }}{{11}}$
Equate the value with the equation (1),
$ \Rightarrow p + 7\sqrt 5 q = 0 + \dfrac{{7\sqrt 5 }}{{11}}$
Hence, the value of $p$ is 0 and $q$ is $\dfrac{1}{{11}}$.
Note: You might be wondering as it is a question's requirement so we have rationalized this expression. But in general, why is there a need to rationalize? The answer is as you can see that after rationalization meaning multiplying and dividing the whole expression by a conjugate, the denominator of the expression is reduced to integers by using basic algebraic identities. So, rationalization will simplify the denominator in such a way that it contains only rational numbers. So, in a calculation, if you find the denominator can be rationalized then go for it, as it will reduce the complexity of the problem.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

If overrightarrow a overrightarrow b overrightarrow class 12 maths CBSE

If a b and c are unit coplanar vectors then left 2a class 12 maths CBSE

Trending doubts
Write a letter to your class teacher asking for 2 days class 8 english CBSE

The past tense of Cut is Cutted A Yes B No class 8 english CBSE

What is the difference between rai and mustard see class 8 biology CBSE

Who commanded the Hector the first British trading class 8 social science CBSE

What is roughage Give two examples class 8 biology CBSE

State the differences between manure and fertilize class 8 biology CBSE

