Question

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

• A. \(-2b\)
• B. \(-b\)
• C. \(b-1\)
• D. \(b+1\)

Hint:

For a quadratic equation \[ ax^2+bx+c=0, \] always use \[ \text{sum of roots}=-\frac{b}{a} \] and \[ \text{product of roots}=\frac{c}{a} \] to connect the given root relation with coefficients.

Answer 1

The Correct Option is B

Step 1: Assume the roots.
Let the two roots of the equation be \[ \alpha \quad \text{and} \quad \alpha^n \] For the quadratic equation \[ ax^2+bx+c=0, \] sum of roots is \[ \alpha+\alpha^n=-\frac{b}{a} \] and product of roots is \[ \alpha\cdot \alpha^n=\frac{c}{a} \] So, \[ \alpha^{n+1}=\frac{c}{a} \]

Step 2: Find \(\alpha\).
From \[ \alpha^{n+1}=\frac{c}{a}, \] we get \[ \alpha=\left(\frac{c}{a}\right)^{\frac{1}{n+1}} \] and \[ \alpha^n=\left(\frac{c}{a}\right)^{\frac{n}{n+1}} \]

Step 3: Use sum of roots.
Since \[ \alpha+\alpha^n=-\frac{b}{a}, \] multiplying both sides by \(a\), we get \[ a\alpha+a\alpha^n=-b \] Now, \[ a\alpha = a\left(\frac{c}{a}\right)^{\frac{1}{n+1}} \] \[ = a^{\frac{n}{n+1}}c^{\frac{1}{n+1}} \] \[ = (a^nc)^{\frac{1}{n+1}} \] Also, \[ a\alpha^n = a\left(\frac{c}{a}\right)^{\frac{n}{n+1}} \] \[ = a^{\frac{1}{n+1}}c^{\frac{n}{n+1}} \] \[ = (ac^n)^{\frac{1}{n+1}} \]

Step 4: Substitute in the required expression.
Therefore, \[ (ac^n)^{\frac{1}{n+1}}+(a^nc)^{\frac{1}{n+1}} = a\alpha^n+a\alpha \] \[ = a(\alpha+\alpha^n) \] \[ = a\left(-\frac{b}{a}\right) \] \[ =-b \]

Step 5: Final conclusion.
Hence, \[ \boxed{-b} \]