If the roots of the equation, x 3+ Px 2+ Qx -19=0 are each one more than the roots of the equation, x 3- Ax 2+ Bx - C =0, where A , B , C , P Q are constants then the value of A + B + C =

Question

If the roots of the equation, x 3+ Px 2+ Qx -19=0 are each one more than the roots of the equation, x 3- Ax 2+ Bx - C =0, where A , B , C , P Q are constants then the value of A + B + C = (A) 18 (B) 19 (C) 20 (D) none

Tardigrade Admin · Accepted Answer

Let roots of x 3- Ax 2+ Bx - C =0 are α, β, &gamma; ⇒ α+β+&gamma;= A , Sigma α β= B , α β &gamma;= C &(α+1)(β+1)(&gamma;+1)=19 ⇒(α+β+&gamma;)+(α β+β &gamma;+&gamma; α)+α β &gamma;+1=19 A + B + C =18

Q. If the roots of the equation, $x^{3} + P x^{2} + Q x - 19 = 0$ are each one more than the roots of the equation, $x^{3} - A x^{2} + B x - C = 0$ , where $A, B, C, P & Q$ are constants then the value of $A + B + C =$

18

19

20

none

Let roots of $x^{3} - A x^{2} + B x - C = 0$ are $α, β, γ$
$\Rightarrow α + β + γ = A, Σ α β = B, α β γ = C$
$& (α + 1) (β + 1) (γ + 1) = 19$
$\Rightarrow (α + β + γ) + (α β + β γ + γ α) + α β γ + 1 = 19$
$A + B + C = 18$

Q. If the roots of the equation, x3+Px2+Qx−19=0 are each one more than the roots of the equation, x3−Ax2+Bx−C=0, where A,B,C,P&Q are constants then the value of A+B+C=

18

19

20

none

Let roots of x3−Ax2+Bx−C=0 are α,β,γ ⇒α+β+γ=A,Σαβ=B,αβγ=C &(α+1)(β+1)(γ+1)=19 ⇒(α+β+γ)+(αβ+βγ+γα)+αβγ+1=19 A+B+C=18

Q. If the roots of the equation, $x^{3} + P x^{2} + Q x - 19 = 0$ are each one more than the roots of the equation, $x^{3} - A x^{2} + B x - C = 0$ , where $A, B, C, P & Q$ are constants then the value of $A + B + C =$

Let roots of $x^{3} - A x^{2} + B x - C = 0$ are $α, β, γ$
$\Rightarrow α + β + γ = A, Σ α β = B, α β γ = C$
$& (α + 1) (β + 1) (γ + 1) = 19$
$\Rightarrow (α + β + γ) + (α β + β γ + γ α) + α β γ + 1 = 19$
$A + B + C = 18$