First slide

Theory of equations

Question

If $α$ and $β$ are the roots of $6 x^{2} - 6 x + 1 = 0,$ then the value of $\frac{1}{2} [a + bα + {cα}^{2} + {dα}^{3}] + \frac{1}{2} [a + bβ + {cβ}^{2} + {dβ}^{3}]$ is

Moderate

A

$\frac{1}{4} (a + b + c + d)$
B

$\frac{a}{1} + \frac{b}{2} + \frac{c}{3} + \frac{d}{4}$
C

$\frac{a}{2} - \frac{b}{2} + \frac{c}{3} - \frac{d}{4}$
D

None of these

Solution

$α, β$ are the roots of the equation $6 x^{2} - 6 x + 1 = 0$
$\Rightarrow α + β = 1, αβ = 1 / 6$
$∴ \frac{1}{2} [a + bα + {cα}^{2} + {dα}^{3}] + \frac{1}{2} [a + bβ + {cβ}^{2} + {dβ}^{3}]$
$= a + \frac{1}{2} b (α + β) + \frac{1}{2} c (α^{2} + β^{2}) + \frac{1}{2} d (α^{3} + β^{3}) - 3 α β (α + β)$
$= a + \frac{b}{2} + \frac{1}{2} c [{(1)}^{2} - 2 . \frac{1}{6}] + \frac{1}{2} d [{(1)}^{3} - 3 . \frac{1}{6}]$
$= \frac{a}{1} + \frac{b}{2} + \frac{c}{3} + \frac{d}{4} .$

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