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I have a question about how to solve the following equation: $$ \sqrt{2x-8} + 8 = x $$

It doesn't seem so difficult as we can just isolate the radical and be left with $\sqrt{2x-8} = x - 8$. further we can square both sides to get rid of the radical all together leaving us with $2x-8=x^2+64$. This however is where I get lost. I could try to factor this but it doesn't break down to pairs of binomials. I have the two answers this will generate which is $6$ and $12$, could anyone show me how to proceed from here?

Thanks!

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  • $\begingroup$ Plugged 6 and 12 back into original and it seems 6 does not satisfy, but 12 does. $\endgroup$ – klorzan Nov 14 '15 at 21:05
  • $\begingroup$ Are you familiar with these facts? 1) $\sqrt{a}$ exists in $\mathbb R$ if and only if $a\ge 0$; 2) if $a\ge 0$, then $\sqrt{a}\ge 0$; 3) if $a,b\ge 0$, then $\sqrt{a}=b\iff a=b^2$ $\endgroup$ – user228113 Nov 14 '15 at 21:10
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Be careful that $$ (x-8)^2=x^2-16x+64 $$ where you forgot the $-16x$ term.

So after squaring you get $$ 2x-8=x^2-16x+64 $$ that simplifies to $$ x^2-18x+72=0 $$ which has $6$ and $12$ as roots.

However $6$ is not a solution, because it would lead to the false equality $\sqrt{4}+8=6$.

You can get a priori conditions by setting $2x-8\ge0$ (that is $x\ge4$) for the existence of the square root, and also $x-8\ge0$ (that is $x\ge8$) before squaring, because $\sqrt{2x-8}\ge0$ by definition, so also $x-8$ must be non negative.

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  • $\begingroup$ $x^{2} - 18 x + 72 = (x - 6) (x - 12)$. $\endgroup$ – Andreas Caranti Nov 14 '15 at 21:07
  • $\begingroup$ 9 is not a solution... $\endgroup$ – YoTengoUnLCD Nov 14 '15 at 21:07
  • $\begingroup$ @AndreasCaranti Oops! ;-) Fixed $\endgroup$ – egreg Nov 14 '15 at 21:09
  • $\begingroup$ I'm confused how you are getting a -16x in there, how is it that after the isolation you can get a -16x? (x-8)^2 could only come out to be x^2+64 with my understanding. Is there perhaps a rule I'm missing? Also thank you for the extreme markup and detail you put in, its really appreciated! $\endgroup$ – Argent Drake Nov 14 '15 at 21:14
  • $\begingroup$ @ArgentDrake $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$. $\endgroup$ – egreg Nov 14 '15 at 21:14
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Hint: A radical equation is equivalent to the system of an equation without radical and an inequation.

$$\sqrt A=B\iff B\ge0\enspace\text{and}\enspace A=B^2.$$ Some details:

This equation is equivalent to $$\begin{cases}2x-8=(x-8)^2\\ x-8\ge 0 \end{cases}\iff\begin{cases}x^2-18x+72\\ x\ge8 \end{cases} $$ The quadratic equation has an obvious interger root, $6$, hence the other root is $72/6=12$. On ly the second root satisfies the inequality. Thus the radical equation has one solution, $12$.

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  • $\begingroup$ I don't understand how this even helps OP. $\endgroup$ – YoTengoUnLCD Nov 14 '15 at 21:06
  • $\begingroup$ It explains how to remove radicals in a safe way. Note I use $\iff$, not $\implies$. $\endgroup$ – Bernard Nov 14 '15 at 21:07
  • $\begingroup$ I'm terrible at math but I really appreciate the straightforward formula. though i'm not sure what <==> or ==> means. $\endgroup$ – Argent Drake Nov 14 '15 at 21:12
  • $\begingroup$ "A=> B" means "A implies B" or "If A is true then B must be true" but NOT necessarily the other way around. "A<==>B" means "If A is true then B must be true and if B is true then A must be true". $\endgroup$ – user247327 Nov 14 '15 at 21:21
  • $\begingroup$ Let me add that in practice, it means both sides have the same solution, instead of one set set of solutions being included in the other set. It will well known that solving radical equations just squaring often introduces parasitic solutions, and when you have ‘solved’ the equation you have to check the solutions found are indeed solutions directly in the original equation. $\endgroup$ – Bernard Nov 14 '15 at 21:43
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First of all it must be $2x-8\ge 0$ for the root to exist; this gives $x\ge 4$. Then, since $$ \sqrt{2x-8}=x-8, $$ and since the square root is non-negative itself for all $x\ge 4$, it must also be $x-8\ge 0$, this is $x\ge 8$. Now, squaring left and right: \begin{align} 2x-8&=x^2-16x+64\\ x^2-18x+72&=0\\ (x-6)(x-12)&=0 \end{align} The solution $x=12$ can be accepted, while $x=6$ does not satisfy $x\ge 8$.

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