Solve the equation sqrt(3x - 8) - sqrt(2x - 3) + sqrt(x - 1) = 0.

Step-by-step solution

1. 1

   Find the domain before anything else. All three radicands must be non-negative: $3x \ge 8$, $2x \ge 3$, $x \ge 1$. The binding condition is $x \ge \tfrac{8}{3} \approx 2.6667$, so any answer below that is automatically invalid.

2. 2

   Regroup so both sides are non-negative. As written, the left side mixes signs and squaring it would be a mess. Move the negative term to the right:

   $\sqrt{3x-8} + \sqrt{x-1} = \sqrt{2x-3}$

   Now both sides are sums of principal roots, hence $\ge 0$, and squaring cannot introduce a sign error.

3. 3

   Square once. Using $(u+v)^2 = u^2 + 2uv + v^2$:

   $(3x-8) + 2\sqrt{(3x-8)(x-1)} + (x-1) = 2x-3$

   $4x - 9 + 2\sqrt{(3x-8)(x-1)} = 2x - 3 \;\Longrightarrow\; \sqrt{(3x-8)(x-1)} = 3 - x$

4. 4

   Record the new sign condition, then square again. The left side is $\ge 0$, so we need $3 - x \ge 0$, i.e. $x \le 3$. Combined with the domain, any solution lies in $\left[\tfrac{8}{3},\,3\right]$ — a very narrow window. Squaring:

   $3x^2 - 11x + 8 = x^2 - 6x + 9 \;\Longrightarrow\; 2x^2 - 5x - 1 = 0$

5. 5

   Solve and filter. By the quadratic formula:

   $x = \frac{5 \pm \sqrt{25+8}}{4} = \frac{5 \pm \sqrt{33}}{4}$

   That is $x \approx 2.6861$ or $x \approx -0.1861$. The negative value is far outside the domain (its radicands are negative), so only $x = \tfrac{5+\sqrt{33}}{4}$ can work — and note how tightly it sits inside $[2.6667,\,3]$.

6. 6

   Verify in the original equation. At $x = 2.6861407$: $\sqrt{3x-8} = \sqrt{0.058422} = 0.241706$, $\sqrt{2x-3} = \sqrt{2.372281} = 1.540221$, $\sqrt{x-1} = \sqrt{1.686141} = 1.298515$. Then $0.241706 - 1.540221 + 1.298515 = 0.000000$ ✓. Symbolically the residual simplifies to exactly $0$, so $x = \tfrac{5+\sqrt{33}}{4}$ is the unique solution.