Question

In the real number system, the equation
$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$

• A. No solution
• B. Exactly two distinct solutions
• C. Exactly four distinct solutions
• D. Infinitely many solutions

Answer 1

The correct option is D Infinitely many solutions

$\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$

$\sqrt{(x-1)-2\left(2\right)\sqrt{x-1}+4}+\sqrt{(x-1)-2\left(3\right)\sqrt{x-1}+9}=1$

$\sqrt{(\sqrt{x-1}{)}^2-2(2)\sqrt{x-1}+4}+\sqrt{(\sqrt{x-1}{)}^2-2(3)\sqrt{x-1}+9}=\mathrm{1....}\left(1\right)$

(1) can be written as $\sqrt{(\sqrt{x-1}-2{)}^2}+\sqrt{(\sqrt{x-1}-3{)}^2}=1$

$(\sqrt{x-1}-2)+(\sqrt{x-1}-3)=1\to x=10$

Also (1) can be written as $\sqrt{(2-\sqrt{x-1}{)}^2}+\sqrt{(3-\sqrt{x-1}{)}^2}=1$

$(2-\sqrt{x-1})+(3-\sqrt{x-1})=1\to x=5$

Also (1) can be written as $\sqrt{(2-\sqrt{x-1}{)}^2}+\sqrt{(\sqrt{x-1}{)}^2-3}=1$

$(2-\sqrt{x-1})+(\sqrt{x-1}-3)=1\to -1\ne 1$

Also (1) can be written as $\sqrt{(\sqrt{x-1}{)}^2-2}+\sqrt{(3-\sqrt{x-1}{)}^2}=1$

$(\sqrt{x-1}-2)+(3-\sqrt{x-1})=1\to 1=1$ But we also need to check domain, that is as in square root so always positive so should be positive after removing it $\sqrt{x-1}-2\ge 0,3-\sqrt{x-1}\ge 0$

$\to x\ge 5,x\le 10\to xϵ[5,10]$