If one root of the equation ${ x }^{ 2 }+px+q=0$ is the square of the other root, then
- ${ p }^{ 3 }+{ q }^{ 2 }-q(3p+1)=0$
- ${ p }^{ 3 }+{ q }^{ 2 }+q(1+3p)=0$
- ${ p }^{ 3 }+{ q }^{ 2 }+q(3p-1)=0$
- ${ p }^{ 3 }+{ q }^{ 2 }+q(1-3p)=0\quad $
Let roots be a and a^2. Sum: a + a^2 = -p. Product: a^3 = q. Substituting a = q^(1/3) into the sum equation: q^(1/3) + q^(2/3) = -p. Cubing both sides: q + q^2 + 3*q^(1/3)q^(2/3)(q^(1/3)+q^(2/3)) = -p^3. q + q^2 + 3q(-p) = -p^3. p^3 + q^2 + q(1-3p) = 0.
Let the roots be r and r^2. The sum of the roots gives r + r^2 = -p, and the product gives r^3 = q. Factoring the sum gives r(r+1) = -p, and cubing both sides results in r^3(r+1)^3 = -p^3. Expanding (r+1)^3 yields r^3(r^3 + 3r^2 + 3r + 1) = -p^3. Substituting r^3 = q, we get q(q + 3r(r^2) + 3r + 1) = -p^3. Replacing 3r^2 + 3r with 3(r^2+r) = -3p gives q(q - 3p + 1) = -p^3, which simplifies to p^3 + q^2 + q(1 - 3p) = 0.