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Theory of Equations: Roots Equation Roots Equation

JEE Maths question with a full step-by-step solution.

Question

If roots of the equation

a(x-\ell)(x-m) + n = 0

are

\alpha, \beta

, then roots of the equation

a\!\left(\dfrac{x}{x-k} - \alpha\right)\!\left(\dfrac{x}{x-k} - \beta\right) - n = 0

are

\dfrac{\ell}{\ell+1},\, \dfrac{m}{m+1}

\dfrac{k(\ell+1)}{\ell},\, \dfrac{k(m+1)}{m}

\dfrac{k\ell}{\ell-1},\, \dfrac{km}{m-1}

correct

\dfrac{\ell-1}{\ell},\, \dfrac{m-1}{m}

Solution

Step 1: Since

\alpha, \beta

are roots of

a(t-\ell)(t-m)+n = 0

, we have

a(t-\ell)(t-m)+n = a(t-\alpha)(t-\beta)

. Therefore

a(t-\alpha)(t-\beta) - n = a(t-\ell)(t-m)

. Step 2: Set

t = \dfrac{x}{x-k}

. The equation becomes:

a\!\left(\frac{x}{x-k}-\alpha\right)\!\left(\frac{x}{x-k}-\beta\right) - n = 0 \implies a\!\left(\frac{x}{x-k}-\ell\right)\!\left(\frac{x}{x-k}-m\right) = 0

Step 3:

\dfrac{x}{x-k} = \ell \implies x = \ell(x-k) \implies x = \dfrac{k\ell}{\ell-1}

. Similarly:

\dfrac{x}{x-k} = m \implies x = \dfrac{km}{m-1}

. Correct answer: (3)

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