In the real number system, the equation x + 3 - 4 x - 1 + x + 8 - 6...

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

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

$⇒ \sqrt{( \sqrt{x - 1} - 2{)}^{2}} + \sqrt{( \sqrt{x - 1} - 3{)}^{2}} = 1$

$⇒ | \sqrt{x - 1} - 2 | + | \sqrt{x - 1} - 3 | = 1$

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

We know that,

If $| x - a | + | x - b | = \left|( x - a ) - ( x - b )\right|$

then $( x - a ) ( x - b ) < 0$

$∴ ( \sqrt{x - 1} - 2 ) ( \sqrt{x - 1} - 3 ) < 0$

$⇒ 2 < \sqrt{x} - 1 < 3 ⇒ 5 < x < 10$

$∴$ Equation has& infinite many solutions.