Solving the equation with three radical terms
$$\sqrt{x-1}+\sqrt{2x-3} +\sqrt{3x-5} = 3x-3 $$

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Solving the equation with three radical terms
$$\sqrt{x-1}+\sqrt{2x-3}   +\sqrt{3x-5} = 3x-3  $$

Uniteasy Admin · Accepted Answer

$$\sqrt{x-1}+\sqrt{2x-3}  +\sqrt{3x-5} = 3x-3 $$
Using the substitution method for the equation with more than three radicals
Let 
$$a = \sqrt{x-1}$$n
$$b = \sqrt{2x-3} $$n
$$c = \sqrt{3x-5} $$n
Then
$$a+b+c = 3x-3$$n
$$a^2+b^2+c^2 = x-1+2x-3+3x-5 = 6x-9$$n
Subtracting 2 times of (4) from (5) gives
$$a^2+b^2+c^2 - 2a-2b-2c = -9+6 = -3$$
Then
$$(a-1)^2+(b-1)^2+(c-1)^2 = 0$$
To make the equation hold true, if and only if
$$a = 1$$
$$b=1 $$
$$c=1 $$
Substitute the values of $$a,b,c$$ to (1), (2), (3) and solve for $$x$$
$$1 = \sqrt{x-1} \implies x = 2$$
$$1 = \sqrt{2x-3} \implies x = 2$$
$$1 = \sqrt{3x-5} \implies x = 2$$
Therefore
The equation has only 1 solution $$x = 2$$

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