Question

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

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

Answer 1

Answer: (D)

Infinitely many solutions

Let $\sqrt{x-1}=t$

This implies a condition that $x\ge 1$

The equation becomes $\sqrt{t^2+4-4t}+\sqrt{t^2+9-6t}=1$

$\Rightarrow |t-2|+|t-3|=1$

When $t\ge 3,t-2+t-3=1$

$\therefore 2t-5=1$ or $t=3$

$\Rightarrow x-1=9$ or $x=10$

When $2\le t\le 3,t-2+3-t=1$

$\Rightarrow t$ can take any value between $2$ and $3$

There are infinitely many solutions to this equation thus.