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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
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5 Answers 5

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

$$3\sqrt{y+15}-2\sqrt{y}+5\sqrt{y+3}=0\;,$$

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

$$3\sqrt{y+15}+5\sqrt{y+3}>2\sqrt{y}\;,$$

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
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@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
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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

$$3\sqrt{x+4}\,,\,5\sqrt{x-8}>2\sqrt{x-11}\,$$

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
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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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