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

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

Solve for x

$$x=8$$

Solution Steps

Subtract -x from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

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

Subtract 9 from both sides.

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

Subtract 9 from 1 to get -8.

$$3x-8=-6x+x^{2}$$

Add 6x to both sides.

$$3x-8+6x=x^{2}$$

Combine 3x and 6x to get 9x.

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

Subtract x^{2} from both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-8. To find a and b, set up a system to be solved.

$$a+b=9 ab=-\left(-8\right)=8$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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 -x^{2}+9x-8 as \left(-x^{2}+8x\right)+\left(x-8\right).

$$\left(-x^{2}+8x\right)+\left(x-8\right)$$

Factor out -x in -x^{2}+8x.

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

Factor out common term x-8 by using distributive property.

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

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

$$x=8 x=1$$

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

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

Simplify. The value x=8 satisfies the equation.

$$-3=-3$$

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

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

Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-3$$

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

$$x=8$$