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I've got

$$3\sqrt{x+4}-2\sqrt{x-11}+5\sqrt{x-8}=0$$ and I have to evaluate $x$. I tried \begin{align} &&3\sqrt{x+4}-2\sqrt{x-11}+5\sqrt{x-8}&=&0 \\ &\Leftrightarrow& 3\sqrt{x+4}+5\sqrt{x-8}&=&2\sqrt{x-11} \\ &\Leftrightarrow& (3\sqrt{x+4}+5\sqrt{x-8})^2&=&4(x-11) \\&\Leftrightarrow& 9(x+4)+30(\sqrt{x+4}\sqrt{x-8})+25(x-8)&=&4(x-11)\\ &\Leftrightarrow& 9x+36+30(\sqrt{x^2-4x-32})+25x-200&=&4x-44\\ &\Leftrightarrow& \frac{5x+80}{30}&=&\sqrt{x^2-4x-32} \\ &\vdots&\end{align}

If I continue I don't get a proper solution for $x$, so I would like to know if the first steps are correct or if I should try something else?

Thanks a lot in advance

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@ulead86.. You want to solve for $x$ belonging to what set? Integers? Rationals? Reals? – uforoboa Sep 26 '12 at 9:31
up vote 2 down vote accepted

Let $y=x-11$. Then $x+4=y+15$, $x-8=y+3$, and the equation can be written


or $$3\sqrt{y+15}+5\sqrt{y+3}=2\sqrt{y}\;.$$

But clearly $3\sqrt{y+15}>\sqrt y$ and $5\sqrt{y+3}>\sqrt y$, so in fact


and the equation has no real solution.

If you continue with your approach, restarting at

$$ 9x+36+30(\sqrt{x^2-4x-32})+25x-200=4x-44$$ and avoiding algebra errors, you’ll get

$$\sqrt{x^2-4x-32}=4-x\tag{1}$$ and then $x^2-4x-32=16-8x+x^2$, $4x=48$, $x=12$. But with all that squaring you know that you may have introduced extraneous solutions, and sure enough, $x=12$ doesn’t even satisfy $(1)$, let alone the original equation. (In terms of $y$ this is $y=1$, but $3\sqrt{16}-2\sqrt{1}+5\sqrt4$ is $20$, not $0$: one of the signs has to be made negative for this to work.)

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$\sqrt{x+4}\gt\sqrt{x-11}$ hence $$ 3\sqrt{x+4}-2\sqrt{x-11}+5\sqrt{x-8}\gt\sqrt{x+4}+5\sqrt{x-8}\gt0 $$ and the equation in the post has no solution.

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what if $x\in \Bbb C$ ? – Aang Sep 26 '12 at 9:48
@Avatar: This is unlikely since, if $x$ was complex valued, the first task of the OP would be to define properly a square root function. – Did Sep 26 '12 at 9:53

Hint: it is enough for you to note:

1) The equation is defined (in the reals) only for $\,x\geq 11\,$ ;

2) Taking into account (1) and that we always take the positive square roots (of positive real numbers, of course), show that


From the above it follows at once your equation has no (real) solutions.

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Last step is wrong , it should be $$\frac{-30x+120}{30}=\sqrt{x^2-4x-32}$$ $$\implies -x+4=\sqrt{x^2-4x-32}$$

which can be solved easily.

One more thing,

$$3\sqrt{x+4}+5\sqrt{x-8}=2\sqrt{x-11}\cdots (1)$$ $$ \Leftrightarrow (3\sqrt{x+4}+5\sqrt{x-8})^2=4(x-11)\cdots(2)$$

This is not right.

All solutions of (1) are sloutions of (2) but not vice-versa; After getting solutions of $(2)$, you need to check which satisfies (1).

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Yes, you're right. I changed it in my handwritten solution. Thanks a lot. – ulead86 Sep 26 '12 at 10:01

This equation has no real solution. You can also discover it by letting $u=\sqrt{x+4},v=\sqrt{x-11},w=\sqrt{x-8}$, from which we will get $$ \begin{cases} 3u-2v+5w=0\\ u^2-v^2=15\\ u^2-w^2=12 \end{cases} $$ then $$ \begin{cases} u^2+6uw+5w^2+12=0\\ u^2-w^2=12 \end{cases} $$ then $$ u^2+6uw+5w^2+u^2-w^2=(u+w)(u+2w)=0 $$ But note that $u\ge0,v\ge0,w\ge0$, so no real solution can be found.

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