Question

Determine rational numbers P and Q if (7+√5)/(7-√5)-(7-√5)/(7+√5)-p=-7√5).

Answer 1

$\bold{\bold{\Large{ \blue {\underline {\underline{given:-}}}}}}$

• $\sf \: \frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = p - 7 \sqrt{5} q$

$\bold{\bold{\Large{ \blue {\underline {\underline{formula:-}}}}}}$

$\sf {(a + b) }^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ \sf {a}^{2} - {b}^{2} = (a + b)(a - b)$

$\bold{\bold{\Large{ \blue {\underline {\underline{Step \: By \: Step \: Explaination:-}}}}}}$

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.

$\bold{\bold{\Large{ \blue {\underline {\underline{solution:-}}}}}}$

The right side of the expression is

$\sf \longmapsto \: p - 7 \sqrt{5} q \qquad \: - - - (1)$

Now take the first term of the left side ,

$\qquad \qquad\sf \longmapsto \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} }$

Rationalize it by multiplying and dividing of the denominator

$\sf \longmapsto \qquad \: \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} } \times \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} }$

using the formula ( a + b )² = a² + b² + 2ab in the numerator and a² -b² = (a + b) ( a - b ) in the denominator.

$\sf \longmapsto \qquad \: \dfrac{ {(7) }^{2} + ( \cancel{\sqrt{5}} )^{ \cancel2} + 2 \times 7 \times \sqrt{5} }{ {(7)}^{2} - ( \cancel{ \sqrt{5}} {)}^{ \cancel2} }$

simplify the terms.

$\sf \longmapsto \qquad \: \dfrac{49 + 5 + 14 \sqrt{5} }{49 - 5}$

cancel out the common factors

$\sf \longmapsto \qquad \: \dfrac{27 + 7 \sqrt{5} }{22} \qquad \: - - - (2)$

Now take second term of the left side ,

$\sf \longmapsto \qquad \: \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} }$

Rationalize it by multiplying and dividing the term by the conjugate of the denominator ,

$\sf \longmapsto\qquad \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} } \times \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} }$

using the formula ( a + b )² = a² + b² + 2ab in the numerator and a² -b² = (a + b) ( a - b ) in the denominator.

$\sf \longmapsto \qquad \: \dfrac{ {(7) }^{2} + ( \cancel{\sqrt{5}} )^{ \cancel2} + 2 \times 7 \times \sqrt{5} }{ {(7)}^{2} - ( \cancel{ \sqrt{5}} {)}^{ \cancel2} }$

cancel out common factors,

$\sf\longmapsto \qquad \: \dfrac{27 - 7 \sqrt{5} }{22} \qquad \: - - - (3)$

Now add the terms (2) and (3)

$\sf\longmapsto \qquad \: \dfrac{27 + 2 \sqrt{5} }{22} - \dfrac{27 - 7 \sqrt{5} }{22}$

change the sings ,

$\sf\longmapsto \qquad \: \dfrac{27 + 7 \sqrt{5} - 27 + 7 \sqrt{5} }{22}$

Add and subtract the like terms,

$\sf\longmapsto \qquad \: \dfrac{27 - 27 +( 7 + 7) \sqrt{5} }{22}$

$\sf\longmapsto \qquad \: \dfrac{14 \sqrt{5} }{22}$

cancel out the common terms,

$\sf\longmapsto \qquad \: \: \dfrac{ \cancel {14} \sqrt{5} }{ \cancel{22}}$

$\sf\longmapsto \qquad \: \dfrac{7 \sqrt{5} }{11}$

Equate the value with the equation (1)

p + 7 √5q = 0 + 7√5/11

hence the value of p is 0 and q is 1/11

Answer 2

Answer:

given:−

\begin{gathered} \sf \: \frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = p - 7 \sqrt{5} q \\ \end{gathered}

7−

5

7+

5

−

7+

5

7−

5

=p−7

5

q

\bold{\bold{\Large{ \blue {\underline {\underline{formula:-}}}}}}

formula:−

\begin{gathered} \sf {(a + b) }^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\ \sf {a}^{2} - {b}^{2} = (a + b)(a - b)\end{gathered}

(a+b)

2

=a

2

+b

2

+2ab

a

2

−b

2

=(a+b)(a−b)

\bold{\bold{\Large{ \blue {\underline {\underline{Step \: By \: Step \: Explaination:-}}}}}}

StepByStepExplaination:−

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.

\bold{\bold{\Large{ \blue {\underline {\underline{solution:-}}}}}}

solution:−

The right side of the expression is

\sf \longmapsto \: p - 7 \sqrt{5} q \qquad \: - - - (1)⟼p−7

5

q−−−(1)

Now take the first term of the left side ,

\qquad \qquad\sf \longmapsto \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} }⟼

7−

5

7+

5

Rationalize it by multiplying and dividing of the denominator

\sf \longmapsto \qquad \: \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} } \times \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} }⟼

7−

5

7+

5

×

7−

5

7+

5

using the formula ( a + b )² = a² + b² + 2ab in the numerator and a² -b² = (a + b) ( a - b ) in the denominator.

\sf \longmapsto \qquad \: \dfrac{ {(7) }^{2} + ( \cancel{\sqrt{5}} )^{ \cancel2} + 2 \times 7 \times \sqrt{5} }{ {(7)}^{2} - ( \cancel{ \sqrt{5}} {)}^{ \cancel2} }⟼

(7)

2

−(

5

)

2

(7)

2

+(

5

)

2

+2×7×

5

simplify the terms.

\sf \longmapsto \qquad \: \dfrac{49 + 5 + 14 \sqrt{5} }{49 - 5}⟼

49−5

49+5+14

5

cancel out the common factors

\sf \longmapsto \qquad \: \dfrac{27 + 7 \sqrt{5} }{22} \qquad \: - - - (2)⟼

22

27+7

5

−−−(2)

Now take second term of the left side ,

\sf \longmapsto \qquad \: \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} }⟼

7+

5

7−

5

Rationalize it by multiplying and dividing the term by the conjugate of the denominator ,

\sf \longmapsto\qquad \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} } \times \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} }⟼

7+

5

7−

5

×

7−

5

7+

5

using the formula ( a + b )² = a² + b² + 2ab in the numerator and a² -b² = (a + b) ( a - b ) in the denominator.

\sf \longmapsto \qquad \: \dfrac{ {(7) }^{2} + ( \cancel{\sqrt{5}} )^{ \cancel2} + 2 \times 7 \times \sqrt{5} }{ {(7)}^{2} - ( \cancel{ \sqrt{5}} {)}^{ \cancel2} }⟼

(7)

2

−(

5

)

2

(7)

2

+(

5

)

2

+2×7×

5

cancel out common factors,

\sf\longmapsto \qquad \: \dfrac{27 - 7 \sqrt{5} }{22} \qquad \: - - - (3)⟼

22

27−7

5

−−−(3)

Now add the terms (2) and (3)

\sf\longmapsto \qquad \: \dfrac{27 + 2 \sqrt{5} }{22} - \dfrac{27 - 7 \sqrt{5} }{22}⟼

22

27+2

5

−

22

27−7

5

change the sings ,

\sf\longmapsto \qquad \: \dfrac{27 + 7 \sqrt{5} - 27 + 7 \sqrt{5} }{22}⟼

22

27+7

5

−27+7

5

Add and subtract the like terms,

\sf\longmapsto \qquad \: \dfrac{27 - 27 +( 7 + 7) \sqrt{5} }{22}⟼

22

27−27+(7+7)

5

\sf\longmapsto \qquad \: \dfrac{14 \sqrt{5} }{22}⟼

22

14

5

cancel out the common terms,

\sf\longmapsto \qquad \: \: \dfrac{ \cancel {14} \sqrt{5} }{ \cancel{22}}⟼

22

14

5

\sf\longmapsto \qquad \: \dfrac{7 \sqrt{5} }{11}⟼

11

7

5

Equate the value with the equation (1)

p + 7 √5q = 0 + 7√5/11

hence the value of p is 0 and q is 1/11