Given α and β are the roots of the quadratic equation x2-4 x+k=0(k ≠ 0). If α β, α β2+α2 β, α3+β3 are in geometric progression then the value of ' k prime equals

Question

Given α and β are the roots of the quadratic equation x2-4 x+k=0(k ≠ 0). If α β, α β2+α2 β, α3+β3 are in geometric progression then the value of ' k prime equals (A) 4 (B) (16/7) (C) (3/7) (D) 12

Tardigrade Admin · Accepted Answer

Given α β ; α β(α+β) ; α3+β3 are in G.P. α+β=4 ; α β= k ; α β2+α2 β=α β(α+β)=4 k α3+β3=(α+β)3-3 α β(α+β) =64-3 k (4)=4(16-3 k ) ∴ k ; 4 k ; 4(16-3 k ) text are in G.P. 16 k 2=4 k (16-3 k ) 4 k (4 k -16+3 k )=0 k =0 ; k =(16/7)

Q. Given $α$ and $β$ are the roots of the quadratic equation $x^{2} - 4 x + k = 0 (k \neq = 0)$ . If $α β, α β^{2} + α^{2} β$ , $α^{3} + β^{3}$ are in geometric progression then the value of ' $k^{'}$ equals

4

$\frac{16}{7}$

$\frac{3}{7}$

12

Q. Given α and β are the roots of the quadratic equation x2−4x+k=0(k=0). If αβ,αβ2+α2β, α3+β3 are in geometric progression then the value of ' k′ equals

4

716​

73​

12

Given αβ;αβ(α+β);α3+β3 are in G.P. α+β=4;αβ=k;αβ2+α2β=αβ(α+β)=4k α3+β3=(α+β)3−3αβ(α+β) =64−3k(4)=4(16−3k) ∴k;4k;4(16−3k) are in G.P. 16k2=4k(16−3k) 4k(4k−16+3k)=0 k=0;k=716​

Q. Given $α$ and $β$ are the roots of the quadratic equation $x^{2} - 4 x + k = 0 (k \neq = 0)$ . If $α β, α β^{2} + α^{2} β$ , $α^{3} + β^{3}$ are in geometric progression then the value of ' $k^{'}$ equals

$\frac{16}{7}$

$\frac{3}{7}$