“Extraneous solution” solves original equation

For the given equation:

$$x - 10 = \sqrt{9x}$$

when one simplifies, through the following steps:

\begin{align*} x^2 - 20x + 100 &= 9x\\ x^2 - 29x + 100 &= 0\\ x &= 25, 4 \end{align*}

we check for extraneous solutions to make sure we have not altered from the set of solutions of the original equation while simplifying.

Hence $25$: \begin{align*} 25-10 &= \sqrt{3 \times 3 \times 5 \times 5}\\ 15 &= 15 \end{align*} Then $4$: \begin{align*} 4-10 &= \sqrt{36}\\ -6 &= \sqrt{36} \end{align*} Since $-6$ is a square root of $36$,

$-6 = -6$. Solutions: $25$, $4$.

However, when I checked this through graphing, it appears that only $25$ is a solution and $4$ is considered extraneous. Why is this so? Square roots accounts for both a positive and a negative value and so the statement $-6 = \sqrt{36}$ should be true.

Looking at the graph of the equation, you could take the positive or the negative of the square root, ending up with two different graphs.

Why did we accept $25$ as a solution but reject $4$?

Any help/explanation would be much appreciated.

• The symbol $\sqrt{a}$ means the positive square root of $a$. Where $a>0$. – Ángel Mario Gallegos Apr 9 '17 at 4:16
• Does notation exist for the definition of the negative square root of a? – AeroFighter76 Apr 9 '17 at 4:17
• If you want the negative square root of $a$ you can write $-\sqrt a$. If you want either the positive or negative one you can write $\pm \sqrt a$ – Ross Millikan Apr 9 '17 at 4:19
• Ok, so then √a is explicitly the positive square root? – AeroFighter76 Apr 9 '17 at 4:20
• @ÁngelMarioGallegos Thanks much for the formatting! – AeroFighter76 Apr 9 '17 at 4:21

This is because you thought that $\sqrt{36}=\pm 6$ though it is not. Look when we do the checking at $x=4$, we get $x-10=4-10=-6$ whereas $\sqrt{9x}=\sqrt{36}=6$ which shows that $x-10=\sqrt{9x}$ does not hold when $x=4$.