First slide

Theory of equations

Question

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

Moderate

A

${αβ}^{- 1}, α^{- 1} β$
B

$α^{- 2}, β^{- 2}$
C

$α^{- 1}, β^{- 1}$
D

$α^{2}, β^{2}$

Solution

$\begin{array}{l} {ax}^{2} - bx + c = 0 \\ α + β = \frac{b}{a}, aβ = \frac{c}{a} \end{array}$
Also, $(a + cy)^{2} = b^{2} y$
$\begin{array}{l} \Rightarrow c^{2} y^{2} - (b^{2} - 2 ac) y + a^{2} = 0 \\ \Rightarrow {(\frac{c}{a})}^{2} y^{2} - ({(\frac{b}{a})}^{2} - 2 (\frac{c}{a})) y + 1 = 0 \end{array}$
$\begin{array}{ll} \Rightarrow (αβ)^{2} y^{2} - (α^{2} + β^{2}) y + 1 = 0 \\ \Rightarrow y^{2} - (α^{- 2} + β^{- 2}) y + α^{- 2} β^{- 2} = 0 \\ \Rightarrow (y - α^{- 2}) (y - β^{- 2}) = 0 \end{array}$
Hence, the roots are $α^{- 2}, β^{- 2}$

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