One root of the equation $ \displaystyle ax^{2}+bx+c=0 $ is square of the other if
- $ \displaystyle a^{2}c+b^{3}+ac^{2}=3abc $
- $ \displaystyle a^{3}+b^{3}+c^{3}=3abc $
- $ \displaystyle b^{2}c+c^{2}a+a^{2}b=abc $
- $ \displaystyle a^{2}+b^{2}+c^{2}-ab-bc-ca=0 $
If roots are alpha and alpha^2, then alpha + alpha^2 = -b/a and alpha * alpha^2 = c/a. From the second, alpha^3 = c/a. Substituting into the first: alpha^2 + alpha = -b/a. Cubing both sides: (alpha^2 + alpha)^3 = (-b/a)^3. This leads to the condition a^2c + b^3 + ac^2 = 3abc.
Let the roots be t and t^2. Using Vieta's formulas for ax^2 + bx + c = 0, the sum of the roots gives a(t + t^2) = -b, and the product of the roots gives a(t^3) = c. Since t is not zero (otherwise c would be 0, making the equation non-quadratic), we can divide the two equations to get t^2 / (t + t^2) = ac / (-b). Simplifying this yields t / (1 + t) = -ac / b. From the sum equation, t = -b/a - t^2. Substituting the product relations and cubing the sum relation a^3(t + t^2)^3 = -b^3, we expand to find a^3(t^3 + t^6 + 3t^3(t + t^2)) = -b^3. Substituting a*t^3 = c and a(t + t^2) = -b into the expansion yields a^2 c + c^2 + 3abc = -b^3. Rearranging this gives the required condition a^2 c + b^3 + ac^2 = 3abc.