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Cube Roots

Evaluate without a calculator: (5 sqrt2 + 7)^{1/3} - (5 sqrt2 - 7)^1/3}.

Conjugate Tracker

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

This solution is from Tiffany of Island School. Andrei of Tudor Vianu National College, Romania also sent in a good solution.

In the equation $x^2 + px + q = 0$ if we fix $q$ and vary $p$ and observe the complex solutions which occur (for $p^2 - 4q < 0$) then, as $p$ changes, the complex roots of the equation also change. If these roots are plotted on an Argand diagram then you will see that the roots lie on a circle.

To prove that the complex roots do lie on a circle we use the fact that the product of the roots of the quadratic equation $x^2 + px + q = 0$ is $q$.

This equation has roots $z_1$ and $z_2$ given by $${-p \pm \sqrt{p^2 - 4q} \over 2}$$ so $$z_1 = {-p + \sqrt{p^2 - 4q} \over 2} = u + iv $$ and $$z_2 = {-p - \sqrt{p^2 - 4q} \over 2} = u - iv $$ where $u+iv$ and $u-iv$ are complex conjugates.

The product of the complex conjugate roots is given by $$z_1z_2 = (u+iv)(u-iv) = u^2 + v^2 = q.$$ Hence as the quadratic changes keeping $q$ fixed and varying $p$ the locus of the complex roots in the $(u,v)$ plane is the circle with radius $\sqrt q$ centre at the origin.