Solve sqrt{x^2+4x+1}-sqrt{2x+4}=0

$$\sqrt { x ^ { 2 } + 4 x + 1 } - \sqrt { 2 x + 4 } = 0$$

Solve for x (complex solution)

$$x=-3$$

$$x=1$$

Solution Steps

Subtract -\sqrt{2x+4} from both sides of the equation.

$$\sqrt{x^{2}+4x+1}=\sqrt{2x+4}$$

Square both sides of the equation.

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

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

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

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

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

Subtract 2x from both sides.

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

Combine 4x and -2x to get 2x.

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

Subtract 4 from both sides.

$$x^{2}+2x+1-4=0$$

Subtract 4 from 1 to get -3.

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

To solve the equation, factor x^{2}+2x-3 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.

$$a+b=2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

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

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

Substitute 1 for x in the equation \sqrt{x^{2}+4x+1}-\sqrt{2x+4}=0.

$$\sqrt{1^{2}+4\times 1+1}-\sqrt{2\times 1+4}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -3 for x in the equation \sqrt{x^{2}+4x+1}-\sqrt{2x+4}=0.

$$\sqrt{\left(-3\right)^{2}+4\left(-3\right)+1}-\sqrt{2\left(-3\right)+4}=0$$

Simplify. The value x=-3 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{x^{2}+4x+1}=\sqrt{2x+4}.

$$x=1 x=-3$$

Solve for x

$$x=1$$

Solution Steps

Subtract -\sqrt{2x+4} from both sides of the equation.

$$\sqrt{x^{2}+4x+1}=\sqrt{2x+4}$$

Square both sides of the equation.

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

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

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

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

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

Subtract 2x from both sides.

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

Combine 4x and -2x to get 2x.

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

Subtract 4 from both sides.

$$x^{2}+2x+1-4=0$$

Subtract 4 from 1 to get -3.

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

To solve the equation, factor x^{2}+2x-3 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.

$$a+b=2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

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

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

Substitute 1 for x in the equation \sqrt{x^{2}+4x+1}-\sqrt{2x+4}=0.

$$\sqrt{1^{2}+4\times 1+1}-\sqrt{2\times 1+4}=0$$

Simplify. The value x=1 satisfies the equation.

$$0=0$$

Substitute -3 for x in the equation \sqrt{x^{2}+4x+1}-\sqrt{2x+4}=0. The expression \sqrt{\left(-3\right)^{2}+4\left(-3\right)+1} is undefined because the radicand cannot be negative.

$$\sqrt{\left(-3\right)^{2}+4\left(-3\right)+1}-\sqrt{2\left(-3\right)+4}=0$$

Equation \sqrt{x^{2}+4x+1}=\sqrt{2x+4} has a unique solution.

$$x=1$$