# Solve the radical equation $x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$

Solve the following equation: $$x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$$

I wanted to solve this equation. First I tried to change the equations under the roots to the complete square to simplify them out, but it just became more complicated.

Can someone help me with this equation, please?

• Solving it by hand? You could multiply both sides by $x\sqrt{x^2+5}\color{red}{-}(2x+1)\sqrt{4x^2+4x+6}$. This will arrive at $-15x^4-32x^3-39x^2-28x-6=0$ which you can attempt to factor (trial and error or using general method for quartics, but that would be incredibly messy). You'll arrive at $-(x+1)(3x+1)(5x^2+4x+6)$ which yields four roots, one of which was introduced with our first changes when we multiplied by the conjugate, the second is our actual solution to the original problem, and the remaining two are complex. – JMoravitz Sep 25 '17 at 0:39
• But how could we factor it? Even the trial and error is very hard to approach and to guess the roots. – Nariman Zendehrooh Sep 25 '17 at 3:38
• @Nariman Zendehrooh:I edited my answer to show how to use the substitution trick to get a factored form. – quasi Sep 25 '17 at 3:42

Let $f\colon \mathbb{R} \to \mathbb{R}$ be given by $f(t) = t\sqrt{t^2+5}$.
Then $f$ is an odd function.
Also, $f$ is strictly increasing, hence $f$ is one-to-one.
Then, letting $u=2x+1$, \begin{align*} &x\sqrt{x^2+5}+(2x+1)\sqrt{4x^2 + 4x + 6}=0 \qquad\qquad\qquad\qquad\;\; \\[4pt] \iff\;&x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt] \iff\;&f(x) + f(u)=0\\[4pt] \iff\;&f(x) = -f(u)\\[4pt] \iff\;&f(x) = f(-u)\qquad\text{[since $f$ is odd]}\\[4pt] \iff\;&x = -u\qquad\qquad\;\;\text{[since $f$ is one-to-one]}\\[4pt] \iff\;&x = -(2x+1)\\[4pt] \iff\;&3x+1 = 0\\[4pt] \iff\;&x = -{\small{\frac{1}{3}}}\\[4pt] \end{align*} As an alternative, using the same trick, you can get a factored form: \begin{align*} &x\sqrt{x^2+5}+u\sqrt{u^2 + 5}=0\\[4pt] \implies\;&x\sqrt{x^2+5}=-u\sqrt{u^2 + 5}\\[4pt] \implies\;&x^2(x^2+5)=u^2(u^2 + 5)\\[4pt] \implies\;&x^4+5x^2=u^4+5u^2\\[4pt] \implies\;&(u^4-x^4)+5(u^2-x^2)=0\\[4pt] \implies\;&(u^2-x^2)(u^2+x^2)+5(u^2-x^2)=0\\[4pt] \implies\;&(u^2-x^2)(u^2+x^2+5)=0\\[4pt] \implies\;&(u-x)(u+x)(u^2+x^2+5)=0\\[4pt] \implies\;&(x+1)(3x+1)(5x^2+4x+6)=0\qquad\text{[replacing $u$ by $2x+1$]} \end{align*} The two candidate real roots, $x=-1,\;x=-{\large{\frac{1}{3}}}\;$need to be verified against the original equation since, when squaring both sides, extraneous real roots were potentially introduced. In this case, as it turns out, the candidate root $x=-{\large{\frac{1}{3}}}$ is ok, but the candidate root $x=-1$ fails, so is not an actual root.
• I recognized that $$4x^2+4x+6 = (4x^2+4x+1) + 5 = (2x+1)^2 + 5$$ and from there, defining $f(t) = t\sqrt{t^2 + 5}$ was a natural move. – quasi Sep 25 '17 at 3:47