#### Question

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

\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}

(1)

b = \sqrt{3x-5}

(2)

c = \sqrt{4x-7}

(3)

Then

a+b+c = 4x-5

(4)

a^2+b^2+c^2 = x-1+3x-5+4x-7 = 8x-13

(5)

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