If `alpha` and `beta` are the roots of 5x^2-px+1=0 and `alpha - beta = 1`, then find p?
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The roots of the equation are `alpha` and`beta` . `alpha - beta = 1`
`alpha - beta= 1` or `alpha = 1 + beta`
From the roots of the equation we get `(x - alpha)(x - beta) = 0`
`5x^2 - px + 1 = 0 => x^2 - (p/5)*x + (1/5) = 0`
=> `x^2 - (alpha + beta)*x + alpha*beta = x^2 - (p/5)*x + (1/5)`
This gives `alpha*beta = 1/5 and alpha + beta = p/5`
=> `(1 + beta)*beta = 1/5 and 1 + beta + beta = p/5`
=> `5*beta^2 + 5*beta - 1 = 0`
=> `beta = (-3*sqrt5 - 5)/10 and beta = (-3*sqrt 5 + 5)/10`
`1 + 2*beta = p/5`
=> p = `1 + (5*2*(-3*sqrt5 - 5))/10`
=> p = `-3*sqrt 5 - 4`
or p = `1 + (5*2(-3*sqrt 5 + 5))/10`
=> p = `-3*sqrt 5 + 6`
The value of p can be `-3*sqrt 5 - 4 and -3*sqrt 5 + 6`
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w.k.t.
( α+β)2 = α2 +β2+2αβ--------------1
( α-β)2 = α2 +β2-2αβ
( α-β)2+2αβ= α2 +β2-------------2
Sum of the roots =α+β =p/5
Product of the roots= α β= 1/5
Sub 2 in 1,we get
Therefore, ( α+β)2=( α-β)2+2αβ+2αβ
=( α-β)2+4αβ
GIVEN, α-β =1
(P/5)2 =(1)2+4(1/5)
(P2/25)= 1+(4/5)
(P2/25)=(5+4)/5
P2/5 =9
P2 =45
P = ±3√5
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