First slide

Theory of equations

Question

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

Moderate

A

$α β^{- 1}, α^{- 1} β$
B

$α^{- 2}, β^{- 2}$
C

$α^{- 1}, β^{- 1}$
D

$α^{2}, β^{2}$

Solution

$\begin{array}{l} a x^{2} - b x + c = 0 \\ α + β = \frac{b}{a}, α β = \frac{c}{a} \\ Also, (a + c y)^{2} = b^{2} y \\ \Rightarrow c^{2} y^{2} - (b^{2} - 2 a c) y + a^{2} = 0 \\ D i v i d e b y a^{2} \\ \Rightarrow {(\frac{c}{a})}^{2} y^{2} - ({(\frac{b}{a})}^{2} - 2 (\frac{c}{a})) y + 1 = 0 [{(α + β)}^{2} - 2 α β = α^{2} + β^{2}] \\ \Rightarrow (α β)^{2} y^{2} - (α^{2} + β^{2}) y + 1 = 0 \\ D i v i d e b y α^{2} β^{2} \\ \Rightarrow y^{2} - (α^{- 2} + β^{- 2}) y + α^{- 2} β^{- 2} = 0 \\ \Rightarrow (y - α^{- 2}) (y - β^{- 2}) = 0 \\ Hence roots are α^{- 2}, β^{- 2} . \end{array}$

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