If one root of the quadratic equation ax^2 + | vedclass

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(B) Let the two roots be $\alpha$ and $\alpha^n$.From the properties of quadratic equations,the sum of roots is $\alpha + \alpha^n = -\frac{b}{a}$ and

Vedclass · Accepted Answer

$-b$

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(B) Let the two roots be $\alpha$ and $\alpha^n$.From the properties of quadratic equations,the sum of roots is $\alpha + \alpha^n = -\frac{b}{a}$ and the product of roots is $\alpha \cdot \alpha^n = \alpha^{n+1} = \frac{c}{a}$.From the product of roots,we have $\alpha = (\frac{c}{a})^{\frac{1}{n+1}}$.Substituting this into the sum of roots equation:$(\frac{c}{a})^{\frac{1}{n+1}} + ((\frac{c}{a})^{\frac{1}{n+1}})^n = -\frac{b}{a}$$(\frac{c}{a})^{\frac{1}{n+1}} + (\frac{c}{a})^{\frac{n}{n+1}} = -\frac{b}{a}$Multiply both sides by $a$:$a(\frac{c}{a})^{\frac{1}{n+1}} + a(\frac{c}{a})^{\frac{n}{n+1}} = -b$$a^{1 - \frac{1}{n+1}} c^{\frac{1}{n+1}} + a^{1 - \frac{n}{n+1}} c^{\frac{n}{n+1}} = -b$$a^{\frac{n}{n+1}} c^{\frac{1}{n+1}} + a^{\frac{1}{n+1}} c^{\frac{n}{n+1}} = -b$$(a^n c)^{\frac{1}{n+1}} + (a c^n)^{\frac{1}{n+1}} = -b$.

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

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