Solve x=3-sqrt{3x+1}

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

Solve for x

$$x=1$$

Solution Steps

Subtract 3 from both sides of the equation.

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

Square both sides of the equation.

$$\left(x-3\right)^{2}=\left(-\sqrt{3x+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-6x+9=\left(-\sqrt{3x+1}\right)^{2}$$

Expand \left(-\sqrt{3x+1}\right)^{2}.

$$x^{2}-6x+9=\left(-1\right)^{2}\left(\sqrt{3x+1}\right)^{2}$$

Calculate -1 to the power of 2 and get 1.

$$x^{2}-6x+9=1\left(\sqrt{3x+1}\right)^{2}$$

Calculate \sqrt{3x+1} to the power of 2 and get 3x+1.

$$x^{2}-6x+9=1\left(3x+1\right)$$

Use the distributive property to multiply 1 by 3x+1.

$$x^{2}-6x+9=3x+1$$

Subtract 3x from both sides.

$$x^{2}-6x+9-3x=1$$

Combine -6x and -3x to get -9x.

$$x^{2}-9x+9=1$$

Subtract 1 from both sides.

$$x^{2}-9x+9-1=0$$

Subtract 1 from 9 to get 8.

$$x^{2}-9x+8=0$$

To solve the equation, factor x^{2}-9x+8 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-9 ab=8$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -9.

$$a=-8 b=-1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-8\right)\left(x-1\right)$$

To find equation solutions, solve x-8=0 and x-1=0.

$$x=8 x=1$$

Substitute 8 for x in the equation x=3-\sqrt{3x+1}.

$$8=3-\sqrt{3\times 8+1}$$

Simplify. The value x=8 does not satisfy the equation because the left and the right hand side have opposite signs.

$$8=-2$$

Substitute 1 for x in the equation x=3-\sqrt{3x+1}.

$$1=3-\sqrt{3\times 1+1}$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

Equation x-3=-\sqrt{3x+1} has a unique solution.

$$x=1$$