If α & ß are the zeroes of the polynomial 2x² - 4x + 5, then find the value of a.α² + ß²  b. 1/ α + 1/ ß  c. (α - ß)² d. 1/α² + 1/ß²   e.  α³ + ß³ (Ans:-1, 4/5 ,-6, - 4/25 ,-7)

Ans: Let p(x) = 2x² - 4x +5

α+β = - b/a = 4/2 = 2

αβ = c/a = 5/2

a)  α²+β² = (α+β)² - 2αβ

Substitute to get = α²+β² = -1

b) 1/a + 1/β = α + β/αβ

substitute , then we get = 1/a + 1/β  = 4/5

b)  (α-β)²  = (α+β)²  - 4 αβ

Therefore we get, (α-β)² = - 6

d)   1/α ²  +  1/β ² = α ²  + β ²/ αβ ² =  - 1/(5/2)²

∴   1/α ²  +  1/ β ²  = - 4/25

e)  α³+β³  = (α+β)(α²+β² - αβ)

Substitute this,

to get, α³+β³ = -7

Posted Date: 4/8/2013 2:07:57 AM | Location : United States

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