Question

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

• A. $b$
• B. $-b$
• C. $\frac{1}{\left(b{x}^{n+1}\right)}$
• D. $-\frac{1}{\left(b{x}^{n+1}\right)}$

Answer 1

Answer: (A), (B)

$-b$

Explanation for the correct option.

Step 1: Assumptions and information for the solution

Let one root of the equation be $\alpha$ then its other root will be ${\alpha }^n$ as per the given condition.

Now, we know that

$$\begin{gathered} \mathbf{\text{Sum of the roots}}\mathbf{=}\mathbf{-}\frac{\mathbf{\text{coefficient of}}\mathbf{}\mathbf{x}}{\mathbf{\text{coefficient of}}\mathbf{}{\mathbf{x}}^{\mathbf{2}}} \\ \mathbf{\text{Product of the roots}}\mathbf{=}\frac{\mathbf{\text{constant}}}{\mathbf{\text{coefficient of}}\mathbf{}{\mathbf{x}}^{\mathbf{2}}} \end{gathered}$$

Step 2: Solution of the given expression.

Then, by the second formula,

$\begin{array}{rcl}\therefore \alpha {\alpha }^n& =& \frac{c}{a}\\ & \Rightarrow & {\alpha }^{n+1}=\frac{c}{a}\\ & \Rightarrow & \alpha ={\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}\end{array}$

Now, by the first formula,

$\begin{array}{rcl}\therefore \alpha +{\alpha }^n& =& -\frac{b}{a}\\ & \Rightarrow & {\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}+{\left(\frac{c}{a}\right)}^{\frac{n}{n+1}}=-\frac{b}{a}\left[\because \alpha ={\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}\right]\\ & \Rightarrow & a{\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}+a{\left(\frac{c}{a}\right)}^{\frac{n}{n+1}}=-b\end{array}$

Solving the LHS separately,

$\begin{array}{rcl}\therefore a{\left(\frac{c}{a}\right)}^{\frac{1}{n+1}}+a{\left(\frac{c}{a}\right)}^{\frac{n}{n+1}}& =& a{a}^{-\frac{1}{n+1}}{\left(c\right)}^{\frac{1}{n+1}}+a{a}^{-\frac{n}{n+1}}{\left(c\right)}^{\frac{n}{n+1}}\\ & =& {\left(a\right)}^{\frac{n+1-1}{n+1}}{\left(c\right)}^{\frac{1}{n+1}}+{\left(a\right)}^{\frac{n+1-n}{n+1}}{\left(c^n\right)}^{\frac{1}{n+1}}\\ & =& {\left(a^n\right)}^{\frac{1}{n+1}}{\left(c\right)}^{\frac{1}{n+1}}+{\left(a\right)}^{\frac{1}{n+1}}{\left(c^n\right)}^{\frac{1}{n+1}}\\ & =& {\left(a^{n}c\right)}^{\frac{1}{n+1}}+{\left(a{c}^n\right)}^{\frac{1}{n+1}}\end{array}$

Then we have,

$\begin{array}{rcl}{\left(a^{n}c\right)}^{\frac{1}{n+1}}+{\left(a{c}^n\right)}^{\frac{1}{n+1}}& =& -b\end{array}$

Hence, the correct option is (B).