Multiple choice

The condition that the equation $x^2 + px + q = 0$, whose one root is the cube of the other root is :

  1. $p = { q }^{ { 1 }/{ 4 } } [1 - {q}^{{1}/{2}}]$
  2. $-p = q^{{1}/{2}} [1 - {q}^{{1}/{4}}]$
  3. $-p = q^{{1}/{4}} [1 + {q}^{{1}/{2}}]$
  4. $p = q^{{1}/{2}} [1 + {q}^{{1}/{4}}]$
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C Correct answer
Explanation

If roots are a and a^3, then a + a^3 = -p and a * a^3 = a^4 = q. Thus a = q^(1/4). Substituting into the first equation: -p = q^(1/4) + (q^(1/4))^3 = q^(1/4) + q^(3/4) = q^(1/4)(1 + q^(1/2)).

AI explanation

Let the roots of x^2 + px + q = 0 be r and r^3. The product of the roots gives r^4 = q, so r = q^(1/4). The sum of the roots gives r + r^3 = -p, which factors to r(1 + r^2) = -p. Substituting r = q^(1/4) and r^2 = q^(1/2) results in the condition -p = q^(1/4) * (1 + q^(1/2)).

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