Jaqueline Kirby

2022-06-22

Given equation $\sqrt{x+3-2\sqrt{x+2}}+\sqrt{x+27-10\sqrt{x+2}}=4$ , find its solution

upornompe

Step 1
Solving the equation via substitution $t=\sqrt{x+2}$ we get:
$\sqrt{x+3-2t}+\sqrt{x+27-10t}=4$
$x+3-2t=16-8\sqrt{x+27-10t}+x+27-10t$
Here, you have to have
$\sqrt{x+3-2t}=4-\sqrt{x+27-10t}\ge 0$
${t}^{2}-10t+25=x+27-10t$
Here, you have to have
$\sqrt{x+27-10t}=5-t\ge 0$
By the way, the equation can be written as
$\sqrt{\left(\sqrt{x+2}-1{\right)}^{2}}+\sqrt{\left(\sqrt{x+2}-5{\right)}^{2}}=4,$ ,
i.e.
$|\sqrt{x+2}-1|+|\sqrt{x+2}-5|=4$
which should be easy to deal with.

Emmy Dillon

Step 1
Put $t=\sqrt{x+2}$ , so we require $t\ge 0$. Then $\sqrt{x+3-2t}=\sqrt{{t}^{2}+1-2t}=\sqrt{\left(t-1{\right)}^{2}}=|t-1|$ . Similarly, $\sqrt{x+27-10t}=\sqrt{\left(t-5{\right)}^{2}}=|t-5|$ . So we have $|t-1|+|t-5|=4$ . That holds for $t\in \left[1,5\right]$ .
So we must have $1\le \sqrt{x+2}\le 5$ and hence $1\le x+2\le 25$ , so $-1\le x\le 23$ .

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