Dear Uncle Colin

I’m told that $x$ satisfies $x^3 - x-1=0$ , and I need to find the value of $\sqrt{3x^2 - 4x} +x \sqrt{2x^2 +3x + 2}$. Obviously, I could solve the cubic and plug the answer in, but I don’t think that’s in the spirit of it.

- Can Anyone Reduce Down Algebraic Nonsense?

Hi, CARDAN, and thanks for your message!

The first part drops out very nicely if you spot that the radicand ((the bit under the root)) is the difference between a binomial expansion and your equation: $(x^3 - x - 1) - (3x^2 - 4x) = x^3 - 3x^2 + 3x -1$, or $(x-1)^3$.

So $0 - (3x^2-4x) = (x-1)^3$, and $\sqrt{3x^2-4x} = 1-x$. We’re off to a flyer!

Can we do something similar with the other root? Almost. I’m going to see if I can finagle it into the form of a fourth power.

First, I’m going to multiply in the $x$, which becomes $x^4$ under the square root: the radicand is now $2x^6 + 3x^5 + 2x^4$.

Now, $x^6 = x^2 + 2x + 1$, just squaring the constraint, and I can sub that in: the radicand becomes $3x^5 + 2x^4 + 2x^2 + 4x + 2$.

The $x^5$ is less straightforward; I’m going to make it $x^3 + x^2$, so the radicand is $2x^4 + 3x^3 + 5x^2 + 4x + 2$.

Ideally, I’m only going to have one $x^4$ under the root, so I’ll replace one of them with $x^2 + x$. Now I have $x^4 + 3x^3 + 6x^2 + 5x + 2$.

That’s excitingly close! I know that $(x+1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1$. The difference between the two? $x^3 - x - 1$, which is zero!

So our radicand is $(x+1)^4$ and the second term is simply $x+1$.

The whole expression is $(1-x) + (x+1)$, or… drumroll… two.

Hope that helps!

- Uncle Colin