If α and are the zeroes of the polynomial 2x − 5x + k and α − = −1. Find the value of k.
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By using the relationship between the zeroes of the quadratic ploynomial.
We have,
Sum of zeroes = −(coefficient of x)coefficent of x2-coefficient of xcoefficent of x2 and Product of zeroes = constant termcoefficent of x2constant termcoefficent of x2
⇒α+β=−11 and αβ=−21⇒α+β=−1 and αβ=−2Now,(1α−1β)2=(β−ααβ)2⇒α+β=-11 and αβ=-21⇒α+β=-1 and αβ=-2Now,1α-1β2=β-ααβ2
=(α+β)2−4αβ(αβ)2 [∵(β−α)2=(α+β)2−4αβ]=(−1)2−4(−2)(−2)2 [∵α+β=−1 and αβ=−2]=(−1)2−4(−2)4=94=α+β2-4αβαβ2 ∵β-α2=α+β2-4αβ=-12-4-2-22 ∵α+β=-1 and αβ=-2=-12-4-24=94
∵(1α−1β)2=94⇒1α−1β=±32
We have,
Sum of zeroes = −(coefficient of x)coefficent of x2-coefficient of xcoefficent of x2 and Product of zeroes = constant termcoefficent of x2constant termcoefficent of x2
⇒α+β=−11 and αβ=−21⇒α+β=−1 and αβ=−2Now,(1α−1β)2=(β−ααβ)2⇒α+β=-11 and αβ=-21⇒α+β=-1 and αβ=-2Now,1α-1β2=β-ααβ2
=(α+β)2−4αβ(αβ)2 [∵(β−α)2=(α+β)2−4αβ]=(−1)2−4(−2)(−2)2 [∵α+β=−1 and αβ=−2]=(−1)2−4(−2)4=94=α+β2-4αβαβ2 ∵β-α2=α+β2-4αβ=-12-4-2-22 ∵α+β=-1 and αβ=-2=-12-4-24=94
∵(1α−1β)2=94⇒1α−1β=±32
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