Let α and β be the roots of x2-3 x+p=0 and γ and δ be the roots of x2-6 x+q=0. If α β, γ, δ form a geometric progression. Then ratio (2 q+p):(2 q-p) is

Question

Let α and β be the roots of x2-3 x+p=0 and &gamma; and δ be the roots of x2-6 x+q=0. If α β, &gamma;, δ form a geometric progression. Then ratio (2 q+p):(2 q-p) is (A) 3: 1 (B) 33: 31 (C) 9: 7 (D) 5: 3

Tardigrade Admin · Accepted Answer

x2-3 x+p=0<βα α, β, &gamma;, δ in G.P. α+α=3 ldots (1) x2-6 x+q=0<δ&gamma; α r2+α r3=6 ldots(2) (2) ÷(1) r2=2 So , (2 q+p/2 q-p)=(2 r5+r/2 r5-r) =(2 r4+1/2 r4-1)=(9/7)

Q. Let $α$ and $β$ be the roots of $x^{2} - 3 x + p = 0$ and $γ$ and $δ$ be the roots of $x^{2} - 6 x + q = 0$ . If $α$ $β, γ, δ$ form a geometric progression. Then ratio $(2 q + p) : (2 q - p)$ is

3: 1

33: 31

9: 7

5: 3

$x^{2} - 3 x + p = 0 <_{β}^{α}$
$α, β, γ, δ$ in G.P.
$α + α = 3 \dots$ (1)
$x^{2} - 6 x + q = 0 <_{δ}^{γ}$
$α r^{2} + α r^{3} = 6 \dots (2)$
$(2) \div (1)$
$r^{2} = 2$
So $, \frac{2 q + p}{2 q - p} = \frac{2 r ^{5} + r}{2 r ^{5} - r}$
$= \frac{2 r ^{4} + 1}{2 r ^{4} - 1} = \frac{9}{7}$

Q. Let α and β be the roots of x2−3x+p=0 and γ and δ be the roots of x2−6x+q=0. If α β,γ,δ form a geometric progression. Then ratio (2q+p):(2q−p) is