If one root of the quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ is equal to $n^{t h}$ $n^(th)$ power of the other, then show that : ${(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0$ $(ac^n)^(1/(n+1))+(a^n c)^(1/(n+1)) + b=0$ ?

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If one root of the quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ is equal to $n^{t h}$ $n^(th)$ power of the other, then show that : ${(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0$ $(ac^n)^(1/(n+1))+(a^n c)^(1/(n+1)) + b=0$ ?

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Answer 1 · 2017-07-20T11:06:41

If one root of the quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ is equal to $n^{t h}$ $n^(th)$ power of the other, then show that :
${(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0$ $(ac^n)^(1/(n+1))+(a^n c)^(1/(n+1)) + b=0$ ?

Let one root be $α$ $alpha$ then other will be $α^{n}$ $alpha^n$

So we can write

$α^{n} + α = - \frac{b}{a}$ $alpha^n +alpha=-b/a$

and

$α^{n + 1} = \frac{c}{a}$ $alpha^(n+1)=c/a$

$\Rightarrow α = {(\frac{c}{a})}^{\frac{1}{n + 1}}$ $=>alpha=(c/a)^{1/(n+1))$

Now one root of the quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ being $α$ $alpha$ . we can write

$a α^{2} + b α + c = 0$ $aalpha^2+balpha+c=0$

$\Rightarrow a α + b + \frac{c}{α} = 0$ $=>aalpha+b+c/alpha=0$

$\Rightarrow \frac{c}{α} + a α + b = 0$ $=>c/alpha+aalpha+b=0$

$\Rightarrow \frac{c}{{(\frac{c}{a})}^{\frac{1}{n + 1}}} + a {(\frac{c}{a})}^{\frac{1}{n + 1}} + b = 0$ $=>c/(c/a)^{1/(n+1))+a(c/a)^{1/(n+1))+b=0$

$\Rightarrow c {(\frac{a}{c})}^{\frac{1}{n + 1}} + a {(\frac{c}{a})}^{\frac{1}{n + 1}} + b = 0$ $=>c(a/c)^{1/(n+1))+a(c/a)^{1/(n+1))+b=0$

$\Rightarrow {(\frac{c^{n + 1} a}{c})}^{\frac{1}{n + 1}} + {(\frac{a^{n + 1} c}{a})}^{\frac{1}{n + 1}} + b = 0$ $=>((c^(n+1)a)/c)^{1/(n+1))+((a^(n+1)c)/a)^{1/(n+1))+b=0$

$\Rightarrow {(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0$ $=>(ac^n)^{1/(n+1))+(a^nc)^{1/(n+1))+b=0$

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If one root of the quadratic equation $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ is equal to $n^{t h}$ $n^(th)$ power of the other, then show that : ${(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0$ $(ac^n)^(1/(n+1))+(a^n c)^(1/(n+1)) + b=0$ ?

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