First slide

Theory of equations

Question

If one root of the quadratic equation ${ax}^{2} + bx + c = 0$ is equal to the n power of the other root,
then the value of ${({ac}^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} =$

Moderate

A

b
B

$- b$
C

$b^{\frac{1}{n + 1}}$
D

$- b^{\frac{1}{n + 1}}$

Solution

Let $α, α^{n}$ be the two roots.
Then
$α + α^{n} = - b / a, {αα}^{n} = c / a Eliminating α, We get {(\frac{c}{a})}^{\frac{1}{n + 1}} + {(\frac{c}{a})}^{\frac{n}{n + 1}} = - \frac{b}{a} \Rightarrow a . a^{- \frac{1}{n + 1}} . c^{\frac{1}{n + 1}} + a . a^{- \frac{n}{n + 1}} . c^{\frac{n}{n + 1}} = - b or {(a^{n} c)}^{\frac{1}{n + 1}} + {({ac}^{n})}^{\frac{1}{n + 1}} = - b$

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