Answers

2014-09-07T20:38:14+05:30
f(x)= x^{2} -3x-2
general equation of a quadratic function = x^2 - (α+β)x+αβ
then, in given Eqn. α+β = 3    ---------(1) & αβ = -2    --------------(2)
we know that, (α+β)^2-(α-β)^2 = 2αβ
                       (3)^2 - (α-β)^2 = 2·(-2)
⇒ (α-β)^2 = 13
⇒α-β = √13  -------------(3)
EqN(1) + EqN(3)
α = (3+√13)/2
β = -2/α = -4/(3+√13)

Now.
New Eqn will be ,whose zero's are α/2+β & α+β/2
          x^2- (α/2+β)+αβ/2
⇒       x^2-(α+2β)/2+αβ/2
⇒       x^2-(α+β+β)/2+αβ/2
⇒       x^2 - {3-4/(3+√13)}/2 -1  -----------(α+β)= 3  & β = -4/(3+√13).,, ,αβ= -2
if u solve this eqn then ,u find a equation
2 x^{2} -( \frac{5+- \sqrt{13} }{3+- \sqrt{13} } )x-2

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