If α,β are the zeroes of the quadratic polynomial x²-5x+4. Find a quadratic polynomial α+1/β+β+1/α.
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SOLUTION
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SOLUTION
\begin{lgathered}Since \: \alpha \beta \: are \: the \: zeroes \: o f \: the \: quadratic \: polynomial \: {x}^{2} - 5x + 4. \\ Therefore, \\ = > \alpha + \beta = 5 \: and \: \: \alpha \beta = 4 \\ \\ We \: have \: to \: find \: the \: equation \: whose \: roots \: are \: \alpha + \frac{1}{ \beta } \: and \: \beta + \frac{1}{ \alpha } . \\ \\ Let \: us \: denote \: a = \alpha + \frac{1}{ \beta } \: and \: b = \beta + \frac{1}{ \alpha } \\ Then,\: a + b = \alpha + \frac{1}{ \beta } + \beta + \frac{1}{ \alpha } \\ \\ = > ( \alpha + \beta ) + \frac{ \alpha + \beta }{ \alpha \beta } = 5 + \frac{5}{4} = \frac{25}{4} \\ \\ = > a.b = ( \alpha + \frac{1}{ \beta } )( \beta + \frac{1}{ \alpha } ) = ( \frac{ \alpha \beta + 1}{ \beta } )( \frac{ \alpha \beta - 1}{ \alpha } ) \\ \\ = > ( \frac{ \alpha \beta + 1) {}^{2} }{ \alpha \beta } = \frac{25}{4} \\ Hence,\: the \: required \: equation \: whose \: roots \: are \: a = \alpha + \frac{1}{ \beta } and \: b = \beta + \frac{1}{ \alpha } \: is \: given \: by \: {x}^{2} - (a + b)x + ab \\ \\ = > {x}^{2} - \frac{25}{4} x + \frac{25}{4} \\ \\ = > 4 {x}^{2} - 25x + 25\end{lgathered}
Sinceαβarethezeroesofthequadraticpolynomialx
2
−5x+4.
Therefore,
=>α+β=5andαβ=4
Wehavetofindtheequationwhoserootsareα+
β
1
andβ+
α
1
.
Letusdenotea=α+
β
1
andb=β+
α
1
Then,a+b=α+
β
1
+β+
α
1
=>(α+β)+
αβ
α+β
=5+
4
5
=
4
25
=>a.b=(α+
β
1
)(β+
α
1
)=(
β
αβ+1
)(
α
αβ−1
)
=>(
αβ
αβ+1)
2
=
4
25
Hence,therequiredequationwhoserootsarea=α+
β
1
andb=β+
α
1
isgivenbyx
2
−(a+b)x+ab
=>x
2
−
4
25
x+
4
25
=>4x
2
−25x+25