Solve $x^2 - 3{\sqrt {3x+1} } = 1$

Question

$$x^2 - 3{\sqrt {3x+1} } = 1$$

I found out that there is only one root, and it is between 3 and 4
But I do not know how to simplify the equation and find the root. Can you help me?

Answer 1

The polynomial form of this equation, obtained by bringing the root term to the RHS and squaring, is $$x^4-2x^2-27x-8=0$$ and this happens to factor into two quadratics: $(x^2-3x-1)(x^2+3x+8)$. The real roots of this polynomial are $x=\frac{3\pm\sqrt{13}}2$, but one of these choices does not satisfy the original equation, leaving $x=\frac{3+\sqrt{13}}2$ as the desired solution.

Answer 2

I like the following way.

Let $\sqrt{3x+1}=y$.

Thus, $$x^2=3y+1$$ and $$y^2=3x+1,$$ which gives $$(x-y)(x+y)=3(y-x)$$ and we have: $$x=y$$ or $$x+y=-3.$$ The second is impossible because $x\geq-\frac{1}{3}$ and $y\geq0$.

Id est, we need to solve $$\sqrt{3x+1}=x.$$

Can you end it now?