Solve sqrt{4x+1}=x-1

$$\sqrt { 4 x + 1 } = x - 1$$

Solve for x

$$x=6$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x+1}\right)^{2}=\left(x-1\right)^{2}$$

Calculate \sqrt{4x+1} to the power of 2 and get 4x+1.

$$4x+1=\left(x-1\right)^{2}$$

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

$$4x+1=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$4x+1-x^{2}=-2x+1$$

Add 2x to both sides.

$$4x+1-x^{2}+2x=1$$

Combine 4x and 2x to get 6x.

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

Subtract 1 from both sides.

$$6x+1-x^{2}-1=0$$

Subtract 1 from 1 to get 0.

$$6x-x^{2}=0$$

Factor out x.

$$x\left(6-x\right)=0$$

To find equation solutions, solve x=0 and 6-x=0.

$$x=0 x=6$$

Substitute 0 for x in the equation \sqrt{4x+1}=x-1.

$$\sqrt{4\times 0+1}=0-1$$

Simplify. The value x=0 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Substitute 6 for x in the equation \sqrt{4x+1}=x-1.

$$\sqrt{4\times 6+1}=6-1$$

Simplify. The value x=6 satisfies the equation.

$$5=5$$

Equation \sqrt{4x+1}=x-1 has a unique solution.

$$x=6$$