Multiple choice

If one root is the $n^{th}$ power of the other root of the equation $x^2-ax+b=0$ then $b^{\dfrac{n}{n+1}}+b^{\dfrac{1}{n+1}}$.

  1. $ab$
  2. $a^n$
  3. $a$
  4. $b^n$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let roots be r and r^n. Sum = r + r^n = a. Product = r * r^n = r^(n+1) = b. Thus r = b^(1/(n+1)). Then r^n = b^(n/(n+1)). The expression is b^(n/(n+1)) + b^(1/(n+1)) = r^n + r = a.

AI explanation

Let the roots of the quadratic equation be r and r^n, so their sum is r + r^n = a and their product is r^(n+1) = b. We need to evaluate b^(n/(n+1)) + b^(1/(n+1)), which equals (r^(n+1))^(n/(n+1)) + (r^(n+1))^(1/(n+1)). Simplifying the exponents gives r^n + r, which is exactly the sum of the roots a. Therefore, the value of the expression is a.

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