# 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`

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