Question:

If one root of the quadratic equation $a x^{2}+b x+c=0$ is equal to $n^{\text {th }}$ power of the other root, then the value of: $a^{\frac{n}{n-1}} C^{\frac{1}{n-1}}+c^{\frac{n}{n-1}} a^{\frac{1}{n-1}}$ is equal to

Updated On: Aug 22, 2023
  • $b$
  • $-b$
  • $\frac{1}{b^{n+1}}$
  • $-b^{n+1}$
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The Correct Option is B

Solution and Explanation

Let one root be $\alpha$
Then the other root is $\alpha^{ n }$
So,product of roots $=\frac{ c }{ a }$
$\therefore(\alpha)\left(\alpha^{ n }\right)=\frac{ c }{ a }$
$\therefore \alpha^{ n +1}=\frac{ c }{ a }$
$\therefore \alpha=\left(\frac{ c }{ a }\right)^{\frac{1}{n+1}}$...(1)
sum of roots $=-\frac{b}{a}$
$\therefore \alpha+\alpha^{ n }=-\frac{ b }{ a }$
Substituting the value of $\alpha$ from equation (1), we get
$\therefore\left(\frac{ c }{ a }\right)^{\frac{1}{n+1}}+\left(\frac{ c }{ a }\right)^{\frac{ n }{n+1}}=-\frac{ b }{ a }$
$\therefore a ^{\frac{n}{n+1}} C ^{\frac{1}{n+1}}+ a ^{\frac{1}{n+1}} C ^{\frac{n}{n+1}}=- b$
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Concepts Used:

Quadratic Equations

A polynomial that has two roots or is of degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b, and c are the real numbers

Consider the following equation ax²+bx+c=0, where a≠0 and a, b, and c are real coefficients.

The solution of a quadratic equation can be found using the formula, x=((-b±√(b²-4ac))/2a)

Two important points to keep in mind are:

  • A polynomial equation has at least one root.
  • A polynomial equation of degree ‘n’ has ‘n’ roots.

Read More: Nature of Roots of Quadratic Equation

There are basically four methods of solving quadratic equations. They are:

  1. Factoring
  2. Completing the square
  3. Using Quadratic Formula
  4. Taking the square root