If α, β and γ are three consecutive terms of a non-constant G.P. Such that the equations α x ^ 2 + 2 + γ = 0 and x ^ 2 + x - 1 = 0 have a common root, then α (β + γ) is equal to:

Correct answer: A. βγ Given α, β, γ are in G.P. ⇒ β ^ 2 = αγ For the equation α x ^ 2 + 2 + γ = 0 D = 4 β ^ 2 - 4αγ = 0 Hence roots of the equation are equal and equals to – (β)/(α) = - (γ)/(β) Since given equation have common roots, hence (- γ)/(β) must be a root of x ^ 2 + x - 1 = 0 ⇒ γ ^ 2 β ^ 2 - (γ)/(β) - 1 = 0 ⇒ γ ^ 2 - γβ - β ^ 2 = 0 ⇒ γ ^ 2 = β(β + γ) ⇒ γ . β ^ 2 α = β(β + γ) ⇒ α(β + γ) = βγ

JEE Main 201912 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2019 · 12 April · Evening Shift · Mathematics · Quadratic Equation

If α, β and γ are three consecutive terms of a non-constant G.P....

\alpha, \beta and

\gamma

are three consecutive terms of a non-constant G.P. Such that the equations

\alpha{x}^{2} + 2\betax + \gamma = 0

and

{x}^{2} + x - 1 = 0

have a common root, then

\alpha

\left(\beta + \gamma\right)

is equal to:

Options

A $\beta\gamma$
B $\alpha\beta$
C $\alpha\gamma$
D $0$

Correct answer

A. $\beta\gamma$

Step-by-step solution

Given

\alpha, \beta, \gamma

are in G.P.

⇒ {\beta}^{2} = \alpha\gamma

For the equation

\alpha{x}^{2} + 2\betax + \gamma = 0

D = 4{\beta}^{2} - 4\alpha\gamma = 0

Hence roots of the equation are equal and equals to

– \dfrac{\beta}{\alpha} = - \dfrac{\gamma}{\beta}

Since given equation have common roots, hence

\dfrac{- \gamma}{\beta}

must be a root of

{x}^{2} + x - 1 = 0

⇒ \dfrac{{\gamma}^{2}}{{\beta}^{2}} - \dfrac{\gamma}{\beta} - 1 = 0

⇒ {\gamma}^{2} - \gamma\beta - {\beta}^{2} = 0

⇒ {\gamma}^{2} = \beta\left(\beta + \gamma\right)

⇒ \gamma . \dfrac{{\beta}^{2}}{\alpha} = \beta\left(\beta + \gamma\right)

⇒ \alpha\left(\beta + \gamma\right) = \beta\gamma

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