Question

if α and β are zeras of P(x)= x² +x-1, then find (i)α²β+αβ² (ii)α³β+αβ³​

Answer 1

Answer:If  α and β are the roots of Quadratic Equation ax² + bx + c = 0 then :

Sum of the roots : α + β = -b/a

Product of the roots : α × β = c/a

Given that α and β are the zeros of Polynomial P(x) = x² - x - 4

⇒ α and β are the Roots of Quadratic Equation :  x² - x - 4 = 0

Here a = 1 and b = -1 and c = -4

⇒ Sum of the Roots : α + β = -b/a = -(-1)/1 = 1

⇒ Product of the Roots : αβ = c/a = -4/1 = -4

The Question is Find the Value of α²β + αβ²

α²β + αβ² can be written as αβ(α + β)

⇒ α²β + αβ² = -4(1) = -4

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Answer 2

Answer:

Step-by-step explanation:

let a and b are the roots

sum of roots=a+b=-1

product of roots=ab=-1

a^2+b^2=(a+b)^2-2ab=(-1)^2-2(-1)=1+2=3

i) a^2b+ab^2=ab(a+b)=(-1)(-1)=1

ii)a^3b+b^3a=ab(a^2+b^2)=(-1)(3)=-3