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The two roots r1 and r2 of the quadratic equation 5x^2 - px

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The two roots r1 and r2 of the quadratic equation 5x^2 - px [#permalink] New post 29 Jan 2004, 13:05
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The two roots r1 and r2 of the quadratic equation 5x^2 - px + 1 = 0
are such that |r1 - r2| = 1, find 'p'.

a. 2sqrt3
b. -2sqrt3
c. both (a) and (b)
d. 3sqrt5
e. Too busy checking out the hot chick :lol:
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 [#permalink] New post 29 Jan 2004, 13:35
D ?

r1 = ( p+sqrt( p^2-20 ) ) / 10
r2 = ( p-sqrt( p^2-20 ) ) / 10
|r1-r2| = sqrt( p^2-20 )/5 = 1
p^2 = 45
p = 3sqrt(5)
Senior Manager
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 [#permalink] New post 29 Jan 2004, 13:43
Good.

Also, note that the difference between roots is equal to: sqrt of the discriminat over 'a'.
Also note that if a root exceeds the other root by a certain amount 'k', then (a^2)(k^2) = b^2 - 4ac
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Pls include reasoning along with all answer posts.
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Este examen me conduce jodiendo loco

  [#permalink] 29 Jan 2004, 13:43
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The two roots r1 and r2 of the quadratic equation 5x^2 - px

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