First slide

Theory of equations

Question

If $α, β$ are roots of $a x^{2} + b x + c = 0$ then the equation $a x^{2} - b x (x - 1) + c (x - 1)^{2} = 0$ has roots

Moderate

A

$\frac{α}{1 - α}, \frac{β}{1 - β}$
B

$\frac{1 - α}{α}, \frac{1 - β}{β}$
C

$\frac{α}{α + 1}, \frac{β}{β + 1}$
D

$\frac{α + 1}{α}, \frac{β + 1}{β}$

Solution

Since $α, β$ are roots of $a x^{2} + b x + c = 0$
$∴ α + β = - b / a, α β = c / a .$
The equation $a x^{2} - b x (x - 1) + c (x - 1)^{2} = 0$ can be written as
$x^{2} (a - b + c) + x (b - 2 c) + c = 0$
Let, $γ, δ$ be its roots. Then
$\begin{array}{l} γ + δ = - \frac{(b - 2 c)}{a - b + c} = \frac{- b + 2 c}{a - b + c} = \frac{- \frac{b}{a} + \frac{2 c}{a}}{1 - \frac{b}{a} + \frac{c}{a}} \\ \Rightarrow γ + δ = \frac{α + β + 2 α β}{1 + α + β + α β} = \frac{α}{α + 1} + \frac{β}{β + 1} \end{array}$
and, $γ δ = \frac{c}{a - b + c} = \frac{\frac{c}{a}}{1 - \frac{b}{a} + \frac{c}{a}} = \frac{α β}{1 + α + β + α β} = \frac{α}{α + 1} \cdot \frac{β}{β + 1}$
Thus, the equation $a x^{2} - b x (x - 1) + c (x - 1)^{2} = 0$ has
$γ = \frac{α}{α + 1}$ and $δ = \frac{β}{β + 1}$ as its two roots.

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