What is the solution for $x+22=-6\sqrt{2x+9}$

Question

So, I want to solve for x in the radical equation:

$x + 22 = -6\sqrt{2x+9}$

By Squaring each expression we get:

$(x + 22)^2 = (-6\sqrt{2x+9})^2$

$ x^2 + 44x + 484 = 36\cdot(2x+ 9) $

$ x^2 + 44x + 484 = 72x + 324 $

Now by solving the quadratic equation:

$ x^2 - 28x + 160 = 0 $

$x^2 - 20x - 8x + 160 = 0$

$x\cdot(x-20) -8\cdot(x-20) = 0 $

$ (x-20)\cdot(x-8) = 0 $

$ x = 20 $ or $ x = 8 $

But, none of the values of x satisfies the equation $x + 22 = -6\sqrt{2x+9}$, they satisfy the equation $x + 22 = 6\sqrt{2x+9}$. There should be an extraneous root that satisfies the equation $x + 22 = 6\sqrt{2x+9}$, but boot the roots satisfies this equation and none of them satisfies $x + 22 = -6\sqrt{2x+9}$.

Why is that so? And is there any complex/imaginary solution to the equation?

Answer 1

Sometimes it can be useful to find the domain of the equation before solving equations involving radicals.

By definition of the principal square root, if $\sqrt {f(x)}=g(x)$, then $f(x)\geqslant0\wedge g(x)\geqslant 0$ holds, where $f(x)$ and $g(x)$ are some algebraic expressions.

Thus, you have:

$$ \begin{align}&\begin{cases}\sqrt {2x+9}=\frac {x+22}{-6}\geqslant 0\\ 2x+9\geqslant0\end{cases}\\\\ \implies &\begin{cases}x\leqslant-22\\ x\geqslant-\frac 92\end{cases}\\\\ \implies &x\in\color{#c00}{\emptyset.}\end{align} $$

Therefore, we don't need to solve the equation. Because, the domain of the equation is the empty set. This means, the real solution, indeed, doesn't exist.

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In this part of the answer, we want to prove that the equation $x+22=-6\sqrt{2x+9}$ has no complex roots in general.

Let $\sqrt {2x+9}=z, z\in\mathbb C\setminus \mathbb R$. Then we have:

$$ \begin{align}x+22=\frac {z^2}{2}+\frac {35}{2}=\frac {z^2+35}{2}\end{align} $$

This leads to:

$$ \begin{align}&\frac {z^2+35}{2}=-6z\\ \implies &z^2+12z+35=0\\ \implies &\Delta_z=36-35=1\geqslant 0.\end{align} $$

This implies that $z\in\mathbb R$, which gives a contradiction. This completes the proof.

Answer 2

Just because we can write a question doesn't mean the question has any solutions. For example I can with a perfectly straight face ask you to solve $|x-3|=-7$ (which is impossible as $|x-3| \ge 0 > -7$) and if you solve by squaring both sides to get $(x-3)^2 = (-7)^2$ and you do the work ($x^2-6x + 9 = 49; x^2-6x-40=(x-10)(x+4)=0$ so $x=10$ or $x=-4$) you get only extraneous solutions.

To show there is no solution we can look at the conditions. For $\sqrt{2x+9}$ to exist we must have $2x +9 \ge $ or $x \ge -4\frac 12$. But we must also have $\sqrt{2x+9}\ge 0$ and so $-6\sqrt{2x+9} \le 0$ and must have $x+22 < 0$ so $x < -22$. A contradiction.

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What about complex numbers....

The issue isn't about taking the square root of a negative number. It's about defining which of the two square roots $2x+9$ we are going to consider to be the square root. The only way for $x+22 = -6\sqrt{2x+9}$ is if we allow our selves have either square root be valid.

If we do that we get your extraneous roots.