Question

If alpha, beta are the roots of the equation 2x square -10x + 5 = 0, then find the value of
$\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$
​

Answer 1

2x² - 10x + 5 = 0

a = 2

b = - 10

c = 5

So, Here

D = b² - 4ac

D = (-10)² - 4(2)(5)

D = 100 - 40

D = 60

$\sf \sqrt{D} = \sqrt{60}$

$\sf \sqrt{D} = \sqrt{4×15}$

$\sf \sqrt{D} = 2 \sqrt{15}$

________________________

Now,

α = $\sf \frac{-b + \sqrt{D}}{2a}$

α = $\sf \frac{-(-10) + 2\sqrt{15}}{2(2)}$

α = $\sf \frac{10 + 2\sqrt{15}}{2(2)}$

α = $\sf \frac{ \cancel2(5+ \sqrt{15})}{\cancel2×2}$

α = $\sf \frac{5 + \sqrt{15}}{2}$

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Now,

ß = $\sf \frac{-b - \sqrt{D}}{2a}$

ß = $\sf \frac{-(-10) - 2\sqrt{15}}{2(2)}$

ß = $\sf \frac{10 - 2\sqrt{15}}{2(2)}$

ß = $\sf \frac{ \cancel2(5- \sqrt{15})}{\cancel2×2}$

ß = $\sf \frac{5 - \sqrt{15}}{2}$

_________________________

According to the Question,

We have to Find the value of $\sf \frac{α}{ß} + \frac{ß}{α}$

$→ \sf \frac{α}{ß} + \frac{ß}{α}$

$→ \sf \frac{\frac{5 + \sqrt{15}}{2} }{\frac{5 - \sqrt{15}}{2}} + \frac{\frac{5 - \sqrt{15}}{2}}{\frac{5 + \sqrt{15}}{2}}$

$→ \sf \frac{\frac{5 + \sqrt{15}}{ \cancel2} }{\frac{5 - \sqrt{15}}{ \cancel2}} +\frac{\frac{5 - \sqrt{15}}{ \cancel2}}{\frac{5 + \sqrt{15}}{ \cancel2}}$

$→ \sf \frac{5 + \sqrt{15} }{5 - \sqrt{15}} +\frac{5 - \sqrt{15}}{5 + \sqrt{15}}$

$→ \sf \frac{5 + \sqrt{15} (5 + \sqrt{15}) }{5 - \sqrt{15}} +\frac{5 - \sqrt{15}(5 - \sqrt{15}) }{5 + \sqrt{15}}$

$→ \sf \frac{(5 + \sqrt{15})² +( 5 - \sqrt{15})² }{ {(5)}^{2} - (\sqrt{15})^{2} }$

$→ \sf \frac{(5) ^{2} + 2(5)( \sqrt{15} ) + (\sqrt{15})² +( 5) ^{2} - 2(5)( \sqrt{15}) + ( \sqrt{15})² }{ 25 - 15 }$

$→ \sf \frac{25+ 10( \sqrt{15} ) + 15 +25 - 10( \sqrt{15}) + 15 }{ 25 - 15 }$

$→ \sf \frac{25 \cancel{ + 10( \sqrt{15} )} + 15 +25 \cancel{- 10( \sqrt{15})} + 15 }{ 10 }$

$→ \sf \frac{25 + 15 +25 + 15 }{ 10 }$

$→ \sf \frac{80}{ 10 }$

$→ \fbox{ \bf 8}$

I hope it helps you ...

Answer 2

Answer:

Step-by-step explanation:

Let p( x ) = 2x² - 10x + 5 = 0

α and β are the roots of p ( x ),

Cmparing p ( x ) with ax²+ bx + c = 0,

We have a = 2 , b = - 10 and c = 5.

∴ x = - b ±√ b² - 4ac /2a

= - ( - 10 ) ± √( - 10 )² - 4 ( 2 ) ( 5 ) / 2 ( 2 )

= 10 ± √ 100 - 40 / 4

= 10 ± √60 / 4

= 10 ± √ 4 × 15 / 4

= 10 ± 2√15 / 4

= 2 ( 5 ± √15 ) / 4

= 5 ± √15 / 2

∴ α = 5 + √15 / 2 and β = 5 - √15 / 2.

∴ α / β + β / α = ( 5 + √15 / 2 ) / ( 5 - √15 / 2 ) + ( 5 - √15 / 2 ) / ( 5 + √15 / 2 )

= ( 5 + √15 ) / ( 5 - √15 ) + ( 5 - √15 ) / ( 5 + √15 )

= ( 5 + √15 )² + ( 5 - √15 )² / ( 5 - √15 ) ( 5 + √15 )

= ( 25 + 15 + 10√15 + 25 + 15 - 10√15 ) / ( 25 - 15 )

= ( 80 ) / 10

= 8 is the answer.