Solve sqrt{x^2+5x+1}+1=2x

$$\sqrt { x ^ { 2 } + 5 x + 1 } + 1 = 2 x$$

Solve for x

$$x=3$$

Solution Steps

Subtract 1 from both sides of the equation.

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

Square both sides of the equation.

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

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

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

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

$$x^{2}+5x+1=4x^{2}-4x+1$$

Subtract 4x^{2} from both sides.

$$x^{2}+5x+1-4x^{2}=-4x+1$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+5x+1=-4x+1$$

Add 4x to both sides.

$$-3x^{2}+5x+1+4x=1$$

Combine 5x and 4x to get 9x.

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

Subtract 1 from both sides.

$$-3x^{2}+9x+1-1=0$$

Subtract 1 from 1 to get 0.

$$-3x^{2}+9x=0$$

Factor out x.

$$x\left(-3x+9\right)=0$$

To find equation solutions, solve x=0 and -3x+9=0.

$$x=0 x=3$$

Substitute 0 for x in the equation \sqrt{x^{2}+5x+1}+1=2x.

$$\sqrt{0^{2}+5\times 0+1}+1=2\times 0$$

Simplify. The value x=0 does not satisfy the equation.

$$2=0$$

Substitute 3 for x in the equation \sqrt{x^{2}+5x+1}+1=2x.

$$\sqrt{3^{2}+5\times 3+1}+1=2\times 3$$

Simplify. The value x=3 satisfies the equation.

$$6=6$$

Equation \sqrt{x^{2}+5x+1}=2x-1 has a unique solution.

$$x=3$$