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

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

Solve for x (complex solution)

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=1-\sqrt{3}\approx -0.732050808$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 2x from both sides.

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)}}{2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8}}{2}$$

Add 4 to 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is plus. Add 2 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+2}{2}$$

Divide 2+2\sqrt{3} by 2.

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

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 2.

$$x=\frac{2-2\sqrt{3}}{2}$$

Divide 2-2\sqrt{3} by 2.

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

The equation is now solved.

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

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

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

Simplify. The value x=\sqrt{3}+1 satisfies the equation.

$$\left(3+2\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(2\times 3^{\frac{1}{2}}+3\right)^{\frac{1}{2}}$$

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

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

Simplify. The value x=1-\sqrt{3} satisfies the equation.

$$i\left(-\left(3-2\times 3^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}=i\left(-\left(3-2\times 3^{\frac{1}{2}}\right)\right)^{\frac{1}{2}}$$

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

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

Solve for x

$$x=\sqrt{3}+1\approx 2.732050808$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

Subtract 2x from both sides.

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

Subtract 1 from both sides.

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

Subtract 1 from -1 to get -2.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-2\right)}}{2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-2\right)}}{2}$$

Multiply -4 times -2.

$$x=\frac{-\left(-2\right)±\sqrt{4+8}}{2}$$

Add 4 to 8.

$$x=\frac{-\left(-2\right)±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{-\left(-2\right)±2\sqrt{3}}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is plus. Add 2 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+2}{2}$$

Divide 2+2\sqrt{3} by 2.

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

Now solve the equation x=\frac{2±2\sqrt{3}}{2} when ± is minus. Subtract 2\sqrt{3} from 2.

$$x=\frac{2-2\sqrt{3}}{2}$$

Divide 2-2\sqrt{3} by 2.

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

The equation is now solved.

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

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

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

Simplify. The value x=\sqrt{3}+1 satisfies the equation.

$$\left(3+2\times 3^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(2\times 3^{\frac{1}{2}}+3\right)^{\frac{1}{2}}$$

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

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

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

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