Question

If p=3-root 5/3+root 5 and q=3+root5/3-root 5,find the value of p^2+q^2.​

Answer 1

${ \bf{ \underline{ \underline \orange{Answer}}}}$

Given that

${ \bf \purple{ \implies p = \frac{3 - \sqrt{5} }{3 + \sqrt{5} } }} \\ { \bf \red{ \implies q = \frac{3 + \sqrt{5} }{3 - \sqrt{5} } }}$

Assuming the value of p as equation ( 1 )

& Assuming the value of q as equation ( 2 )

Taking equation ( 1 ) & S.O.B.S ( squaring on both sides )

${ \bf \purple{ \to {p}^{2} = ( \frac{3 - \sqrt{5} }{3 + \sqrt{5} })^{2} = \frac{ {(3 - \sqrt{5}) }^{2} }{ {(3 + \sqrt{5}) }^{2} } }}$

Using (a - b)² = a² - 2ab + b²

(a + b)² = a² + 2ab + b²

${ \bf \purple{ \to {p}^{2} = \frac{ {3}^{2} -2( 3 )(\sqrt{5}) + {( \sqrt{5} )}^{2} }{ {3}^{2} + 2(3)( \sqrt{5}) + {( \sqrt{5} )}^{2} } }}$

${ \bf \purple{ \to {p}^{2} = \frac{9 - 6 \sqrt{5} + 5 }{9 + 6 \sqrt{5} + 5} = \frac{14 - 6 \sqrt{5} }{14 + 6 \sqrt{5} } }}$

Assuming as equation ( 3 )

Taking equation ( 2 ) S.O.B.S

${ \bf \red{ \to {q}^{2} = (\frac{3 + \sqrt{5} }{3 - \sqrt{5} } )^{2} }}$

${ \bf \red{ \to {q}^{2} = \frac{ {3}^{2} + 2(3){( \sqrt{5} ) } + { (\sqrt{5}) }^{2} }{ {3}^{2} - 2(3)(\sqrt{5}) + {( \sqrt{5} )}^{2} } }}$

${ \bf \red{ \to {q}^{2} = \frac{9 +6 \sqrt{5} + 5 }{9 - 6\sqrt{5} + 5} = \frac{14 + 6 \sqrt{5} }{14 - 6 \sqrt{5} } }}$

Assuming as equation ( 4 )

Now,

equation ( 3 ) + equation ( 4 )

${ \bf { \green{ {p}^{2}} + \orange {{q}^{2}} = \green{\frac{14 - 6 \sqrt{5} }{14 + 6 \sqrt{5} } } + \orange{ \frac{14 + 6 \sqrt{5} }{14 - 6 \sqrt{5} } }}}$

Using

→ (a - b)(a + b) = a² - b²

→ (a + b)² = a² + 2ab + b²

→ (a - b)² = a² - 2ab + b²

${ \bf { \green{ {p}^{2} } + \orange{ {q}^{2} } = \pink{\frac{ {(14 - 6 \sqrt{5} )}^{2} + {(14 + 6 \sqrt{5} )}^{2} }{ {14}^{2} - ({6 \sqrt{5} )}^{2} } } }}$

${ \bf{ \green{ {p}^{2} } + \orange{ {q}^{2} } = \pink{ \frac{ {14}^{2} - 2(14)(6 \sqrt{5}) + {(6 \sqrt{5}) }^{2} + {14}^{2} + 2(14)(6 \sqrt{5}) + {(6 \sqrt{5}) }^{2} }{196 - 180 } }}}$

${\bf{ \green{ {p}^{2} } + \orange{ {q}^{2} } = \pink{\frac{196 + 180 + 196 + 180}{36}} }}$

${ \bf{ \green{ {p}^{2} } + \orange{ {q}^{2} }= \pink{\dfrac{752}{36} } }}$

p² + q² = 20.8