Question

plz answer this question

Answer 1

hope it helps u......

Answer 2

Answer:

2p+q+5 = 0

Step-by-step explanation:

f(x) = $x^{3}+px^{2}+qx+2$

Since, α,β and Y are the roots of f(x)

Then, by properties of roots of cubic equation, ($ax^{3}+bx^{2}+cx+d$)

α+ β+ Y= -b/a = -p

αβY = -d/a = -2

αβ + βY+ αY = c/a =q

Putting these values,

2p+q+5 = 2[-(α+ β+ Y)] + αβ + βY+ αY+5

            =   -2α-2β-2Y +αβ +βY+Yα +1+4

            =   -2α-2β-2Y +βY+Yα +4 (Since, 1+αβ=0)

           =   -2α +Yα -2β+βY-2Y+4

           = α(Y-2) +β(Y-2)-2(Y-2)

            = (α+β-2)(Y-2)

            = (α+β-2)(Y+αβY)

           = (α+β-2)Y(1+αβY)

            = 0  (Since, 1+αβ=0)