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

Beginner2022-06-23Added 20 answers

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

$\sqrt{(\sqrt{x+2}-1{)}^{2}}+\sqrt{(\sqrt{x+2}-5{)}^{2}}=4,$ ,

i.e.

$|\sqrt{x+2}-1|+|\sqrt{x+2}-5|=4$

which should be easy to deal with.

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

$\sqrt{(\sqrt{x+2}-1{)}^{2}}+\sqrt{(\sqrt{x+2}-5{)}^{2}}=4,$ ,

i.e.

$|\sqrt{x+2}-1|+|\sqrt{x+2}-5|=4$

which should be easy to deal with.

Emmy Dillon

Beginner2022-06-24Added 5 answers

Step 1

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

So we must have $1\le \sqrt{x+2}\le 5$ and hence $1\le x+2\le 25$ , so $-1\le x\le 23$ .

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

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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