First slide

Theory of equations

Question

If $α \neq β but α^{2} = 5 α - 3$ and $β^{2} = 5 β - 3$ , then the equation whose roots are $α / β$ and $β / α$ is

Moderate

A

$3 x^{2} - 25 x + 3 = 0$
B

$x^{2} + 5 x - 3 = 0$
C

$x^{2} - 5 x + 3 = 0$
D

$3 x^{2} - 19 x + 3 = 0$

Solution

$α^{2} - 5 α + 3 = 0 . . . (i)$
$β^{2} - 5 β + 3 = 0 . . . . (ii)$
From $(i) - (ii),$
$\Rightarrow (α^{2} - β^{2}) - 5 α + 5 β = 0$
$\Rightarrow α^{2} - β^{2} = 5 (α - β) \Rightarrow α + β = 5$
From $(i) + (ii),$
$\Rightarrow (α^{2} + β^{2}) - 5 (α + β) + 6 = 0$
$\Rightarrow (α^{2} + β^{2}) - 5.5 + 6 = 0 \Rightarrow α^{2} + β^{2} = 19$
Then ${(α + β)}^{2} = α^{2} + β^{2} + 2 αβ$
$\Rightarrow 25 = 19 + 2 αβ \Rightarrow αβ = 3$
then the equation, whose roots are $\frac{α}{β} and \frac{β}{α}$ , is
$x^{2} - x (\frac{α}{β} + \frac{β}{α}) + \frac{α}{β} . \frac{β}{α} = 0$
$\Rightarrow x^{2} - x (\frac{α^{2} + β^{2}}{αβ}) + 1 = 0$
$\Rightarrow x^{2} - x . \frac{19}{3} + 1 = 0 \Rightarrow 3 x^{2} - 19 x + 3 = 0$

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