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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?

ACB
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Robario
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  • Just because some one asks you to solve something doesn't mean there actually is a solution. And squaring both sides doesn't always add an extraneous solution. Sometimes it doesn't add any (Ex. $|x-3| = 4; (x-3)^2=16; x^2-6x-5=0;x=1,4$) and sometimes it adds two Ex. $|x-2| =-7$ then .... well, you can see. – fleablood Nov 20 '22 at 05:51
  • $|x−3|=4; (x−3)^2 = 4^2; x^2-6x+9 = 16; x^2-6x-7 = 0; x^2+x-7x-7=0; x\cdot(x+1) -7\cdot(x+1)=0; (x+1)\cdot(x-7)=0;$ $x=-1$ or $x=7$, One root satisfies $x-3 = 4$ which is $7$ and the extraneous root $-1$ satisfies $x-3 = -4$. An extraneous root is added – Robario Nov 20 '22 at 06:03
  • $x=-1$ is not an extraneous root. One solving $x-3 = 4$ and the other solving $x-3 =-4$ does not make either extraneous. – fleablood Nov 20 '22 at 06:49
  • $x=-1$ does not satisfy the equation $x-3=4$, the definition of an extreneous root: An "extraneous root" may be defined as "a value obtained. for an unknown in the solution of an equation which is not a. root of the equation." – Robario Nov 20 '22 at 06:58
  • But our original equation was not $x-3 =4$. It was $|x-3| = 4$ by your very own definition $x=-1$ is not an extraneous root because $x=-1$ is a solution to $|x-3| = 4$. – fleablood Nov 20 '22 at 07:04
  • If you want another example of an equation not adding any extraneous root consider $x+22 = 6\sqrt{2x+9}$ squaring both sides we get $x=20$ or $x=8$ and they both satisfy $x+22 = 6\sqrt{2x+9}$. No extraneous roots! – fleablood Nov 20 '22 at 07:13

2 Answers2

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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.

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.

lone student
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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.

....

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.

fleablood
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