Solve sqrt{5x+1}+x=1

$$\sqrt { 5 x + 1 } + x = 1$$

Solve for x

$$x=0$$

Solution Steps

Subtract x from both sides of the equation.

$$\sqrt{5x+1}=1-x$$

Square both sides of the equation.

$$\left(\sqrt{5x+1}\right)^{2}=\left(1-x\right)^{2}$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

$$5x+1=\left(1-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$5x+1=1-2x+x^{2}$$

Subtract 1 from both sides.

$$5x+1-1=-2x+x^{2}$$

Subtract 1 from 1 to get 0.

$$5x=-2x+x^{2}$$

Add 2x to both sides.

$$5x+2x=x^{2}$$

Combine 5x and 2x to get 7x.

$$7x=x^{2}$$

Subtract x^{2} from both sides.

$$7x-x^{2}=0$$

Factor out x.

$$x\left(7-x\right)=0$$

To find equation solutions, solve x=0 and 7-x=0.

$$x=0 x=7$$

Substitute 0 for x in the equation \sqrt{5x+1}+x=1.

$$\sqrt{5\times 0+1}+0=1$$

Simplify. The value x=0 satisfies the equation.

$$1=1$$

Substitute 7 for x in the equation \sqrt{5x+1}+x=1.

$$\sqrt{5\times 7+1}+7=1$$

Simplify. The value x=7 does not satisfy the equation.

$$13=1$$

Equation \sqrt{5x+1}=1-x has a unique solution.

$$x=0$$