Let $\alpha$ and $\beta$ be roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0$. If $\alpha, \beta$, $\gamma, \delta$, form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is:

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Quadratic Equations and Inequalities

Let

α

and

β

be 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:

Answer: (D)

Let the four numbers in geometric progression be

α, α r, α r^{2}, α r^{3}

, with

α, α r

the roots of

x^{2} - 3 x + p = 0

and

α r^{2}, α r^{3}

the roots of

x^{2} - 6 x + q = 0

. Then

α + α r = 3

and

α r^{2} + α r^{3} = 6

; dividing these gives

r^{2} = 2

. The products of the roots give

p = α^{2} r

and

q = α^{2} r^{5}

. Hence

\frac{2 q + p}{2 q - p} = \frac{2 α ^{2} r ^{5} + α ^{2} r}{2 α ^{2} r ^{5} - α ^{2} r} = \frac{2 r ^{4} + 1}{2 r ^{4} - 1} = \frac{2 ( 4 ) + 1}{2 ( 4 ) - 1} = \frac{9}{7}

, a ratio of

9 : 7

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option