Solve the radical equation
$$\sqrt{x-1}+\sqrt{3x-5}+\sqrt{4x-7} = 4x-5 $$

Question

Solve the radical equation
$$\sqrt{x-1}+\sqrt{3x-5}+\sqrt{4x-7} = 4x-5      $$

Uniteasy Admin · Accepted Answer

$$\sqrt{x-1}+\sqrt{3x-5}+\sqrt{4x-7} = 4x-5   $$
Using the substitution method for the equation with more than three radicals
Let 
$$a = \sqrt{x-1}$$n
$$b = \sqrt{3x-5} $$n
$$c = \sqrt{4x-7} $$n
Then
$$a+b+c = 4x-5 $$n
$$a^2+b^2+c^2 = x-1+3x-5+4x-7 = 8x-13$$n
Subtracting 2 times of (4) from (5) gives
$$a^2+b^2+c^2 - 2a-2b-2c = -13+10 = -3$$
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 $$x$$
$$1 = \sqrt{x-1} \implies x = 2$$
$$1 = \sqrt{3x-5} \implies x = 2$$
$$1 = \sqrt{4x-7} \implies x = 2$$
Therefore
The equation has only 1 solution $$x = 2$$

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